The energy change in an electron's transition in a hydrogen atom can be calculated using the Bohr formula. This energy corresponds to the energy of the emitted photon in the transition, and can be used to calculate the wavelength of the light emitted during this transition.
Explanation:The energy change associated with the transition of an electron between two energy levels in a hydrogen atom can be calculated using the equation E = 13.6 eV/n². In this equation, 'E' represents the energy of the electron, and 'n' is the quantum number of the orbit that the electron occupies. If we consider a transition of an electron from an orbital with n = 10 to an orbital with n = 8, we can calculate the energy change (∆E) using the difference in energies of these two states. This energy change corresponds to the energy of the emitted photon when the electron transition occurs. The energy of the photon can be connected to its wavelength through the equation E = hf, where 'h' is Planck's constant, and 'f' is the frequency, which is related to the wavelength (λ) by the speed of light (c) as f = c/λ. Therefore, you can first calculate ∆E using the given energy equation, then use the result to find 'f' using E = hf, and finally substitute 'f' in f = c/λ to find λ, the wavelength of the emitted light.
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When a sound wave travels directly toward a hard wall, the incoming and reflected waves can combine to produce a standing wave. There is an antinode right at the wall, just as at the end of a closed tube, so the sound near the wall is loud. You are standing beside a brick wall listening to a 85 Hz tone from a distant loudspeaker.How far from the wall must you move to find the first quiet spot?
The student should position themselves approximately 1.009 meters from the brick wall to find the first quiet spot, which is the location of the first node in the standing wave pattern created by the 85 Hz sound wave reflecting off the wall.
When a sound wave travels directly toward a hard wall, it reflects and can combine with the incoming wave to create a standing wave pattern. At the position of a node in this pattern, there is destructive interference, and the sound is minimized or goes silent. Since the wall acts like the closed end of a tube, we know there's an antinode at the wall, and the distance from the wall to the first node would be one quarter of the wavelength of the sound.
The wavelength (λ) can be found using the speed of sound in air (approximately 343 meters per second) and the frequency of the tone. The equation v = f λ gives us λ = v / f. Plugging in the values, we get λ = 343 m/s / 85 Hz = 4.035 meters. Thus, the distance from the wall to the first quiet spot, which is one quarter of the wavelength, is λ / 4 = 4.035 m / 4 ≈ 1.009 meters.
The student should move approximately 1.009 meters away from the wall to find the first quiet spot where there is a node of a standing wave created by the 85 Hz tone.
A waterfall is 145 m high.
What is the increase in water temperature at the bottom of the falls if all the initial potential energy goes into heating the water? (g = 9.8 m/s2, cw = 4 186 J/kg⋅°C)
a. 0.16°C
b. 0.34°C
c. 0.69°C
d. 1.04°C
Answer:
option B
Explanation:
height, h = 145 m
cw = 4186 J/kg °C
g = 9.8 m/s^2
According to the conservation of energy
Potential energy = thermal energy
m x g x h = m x c x ΔT
where, ΔT is the rise in temperature
9.8 x 145 = 4186 x ΔT
ΔT = 0.34°C
Answer:[tex]\Delta T=0.339^{\circ}C[/tex]
Explanation:
Given
height from which water is falling [tex]h=145 m[/tex]
heat capacity of water [tex]c_w=4186 J/kg-^{\circ}C[/tex]
here Potential Energy is converted to heat the water
i.e. [tex]P.E.=mc_w\Delta T[/tex]
[tex]mgh=mc_w\Delta T[/tex]
[tex]9.8\times 145=4186\times \Delta T[/tex]
[tex]\Delta T=0.339^{\circ}C[/tex]
Which is true for spectroscopy of a mixture?
A. The absorbance for a mixture at a particular wavelength is the sum of the absorbances for the components that absorb at the particular wavelength.
B. Spectrophotometers can differentiate between mixture components that absorb at the same wavelength.
C. Each component in a mixture has the same molar absorptivity at the same wavelength.
D. The concentration for each component in the mixture is easily calculated with a set of simultaneous equations if there is significant overlap of individual spectra.
E. The concentration for each component in a mixture is easily calculated by least squares for guesses of each component when the individual spectra are well resolved.
Absorbance of a mixture solution is additive. That mean the total absorbance at a perpendicular wavelength is the sum of the absorbance of its components at that wavelength.
This can be demonstrated through the sum of the absorbance in the solutions for which it is necessary to add to each specific absorbance its respective wavelength. Mathematically this is:
[tex]A_{total,\lambda} = \sum\limit_i A_{i,\lambda}[/tex]
[tex]A_{total,\lambda} = A_{1,\lambda}+A_{2,\lambda}+A_{3,\lambda}+...[/tex]
Therefore the correct option is A.
Final answer:
The true statement for spectroscopy of a mixture is that the absorbance at a particular wavelength is the sum of the absorbances of the components that absorb at that wavelength, in line with Beer's Law.
Explanation:
When analyzing a mixtute using spectroscopy, it is true that the absorbance for a mixture at a particular wavelength is the sum of the absorbances for the components that absorb at that wavelength as stated by Beer's Law. Spectrophotometers cannot differentiate between mixture components that absorb at the same wavelength - distinction typically relies on different wavelengths where individual compounds absorb maximally. Each component in a mixture does not necessarily have the same molar absorptivity at the same wavelength, as molar absorptivity is a characteristic of the molecular structure and its interaction with specific wavelengths of light.
When an overlap in the spectra of the analytes is significant, it can be challenging to calculate their concentrations directly. In such cases, methods like multiwavelength linear regression analysis might be used to improve accuracy and precision. This involves comparing the absorbance of the mixture to that of standard solutions containing known concentrations of the analytes at multiple wavelengths.
A puddle holds 150 g of water. If 0.50 g of water evaporates from the surface, what is the approximate temperature change of the remaining water? (Lv = 540 cal/g)
a. +1.8 C°
b. -1.8 C°
c. +0.18 C°
d. −0.18 C°
Answer:
The correct answer is b
Explanation:
In this exercise we must assume that no heat from the environment enters, the initial heat of the puddle is distributed in the heat of the evaporated water and the heat of liquid water remaining
Let's look for the heat to evaporate the water
Q₁ = m L
Q₁ = 0.50 540
Q₁ = 270 cal
The remaining water is
m = 150 -0.50 = 149.5 gr
The heat for this water is
Q2 = m ce DT
The amount of heat should be conserved, the heat assigned is equal to minus the heat absorbed
Q₁ = - Q₂
Q₁ = -m ce ΔT
ΔT = -Q₁ / m ce
Δt = -270 / (149.5 1)
ΔT = -1.8 ° C
The water has cooled 1.8 ° c
The correct answer is b
While helping an astronomy professor, you discover a binary star system in which the two stars are in circular orbits about the system's center of mass. From their color and brightness, you determine that each star has the same mass as our Sun. The orbital period of the pair is 21.3days , based on the oscillation of brightness observed as one star occludes (hides) the other. From this information you are able to ascertain the distance between the stars.
Part A
Calculate the distance between the stars.
Express your answer to two significant digits and include the appropriate units.
d =
Answer:
Explanation:
Convert orbital period into seconds.
21.3 days = 21.3x 24 x 60x60 = 1840320 s
Write the expression for the gravitational force balanced by the linear speed to calculate the distance between the stars.
[tex]\frac{Gm^2}{d^2}=\frac{mv^2}{r}[/tex] [tex]d=2r;v=\frac{2\pi r}{T}[/tex]
[tex]\frac{Gm^2}{(2r)^2}=\frac{m(\frac{2\pi r}{T})^2}{r}\\\\\frac{Gm^2}{4r^2}=\frac{m(\frac{4\pi^2 r^2}{T^2})}{r}\\\\r^3=(\frac{GmT^2}{16\pi^2})[/tex]
[tex]r=(\frac{(6.67\times 10^{-11})(1.99\times 10^{30})(1840320)^2}{16(3.14)^2})^{\frac{1}{3}}\\\\=1.42\times 10^{10}m[/tex]
Calculate the distance between the stars.
[tex]d=2r\\\\=2(1.42\times 10^{10}m)\\\\=2.83\times 10^{10}m[/tex]
The distance between the two stars in the binary system can be calculated using Kepler's third law as approximately 0.150 Astronomical Units (AU).
Explanation:In physics, we can determine the distance between the stars using Kepler's third law which relates the orbital period of an object in a binary system to the distance between the two objects. The law is expressed as T² = R³ where T is the orbital period in years and R is the separation distance in Astronomical Units (AU). So first, we'll need to convert the given period of 21.3 days into years, which gives us approximately 0.058 years. Plugging the orbital period into Kepler's third law gives us a radius cubed of (0.058)² = 0.003364. Taking the cube root of both sides gives us a separation of about 0.150 AU between the stars.
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A particle moves through an xyz coordinate system while a force acts on it. When the particle has the position vector r with arrow = (2.00 m)i hat − (3.00 m)j + (2.00 m)k, the force is F with arrow = Fxi hat + (7.00 N)j − (5.80 N)k and the corresponding torque about the origin is vector tau = (3.40 N · m)i hat + (2.80 N · m)j + (0.800 N · m)k. Determine Fx.
Answer:
-4.40
Explanation:
explanation is in attachment
To determine Fx, the x-component of the force, we consider the torque formula τ = r × F, solve for Fx using the given torque and position vectors, and by calculating the determinant involving the unit vectors and the components of r and F.
Explanation:The student is asked to calculate the value of Fx from a given torque value and the position vector of the particle. In physics, particularly in the study of dynamics, the torque about a point is the cross product of the position vector (r) and the force (F) applied. The torque (τ) vector formula is:
τ = r × F
Given the torque vector τ = (3.40 N · m)i + (2.80 N · m)j + (0.800 N · m)k, and the position vector r = (2.00 m)i - (3.00 m)j + (2.00 m)k, we can set up the equation to solve for the torque:
τ = r × F ==> (3.40 N · m)i + (2.80 N · m)j + (0.800 N · m)k = (2.00 m)i - (3.00 m)j + (2.00 m)k × (Fxi + 7.00 Nj - 5.80 Nk)
We can calculate it using the determinant of a 3x3 matrix consisting of the unit vectors i, j, and k, the components of r, and the components of F. Solving for Fx involves calculating the determinant and equating each corresponding component of the torque vector to form a system of equations. The x-component of the torque provides the equation to solve for Fx, as the other components give us irrelevant information for this specific calculation.
The magnitude of the magnetic field in a magnetic resonance imaging (MRI) machine can be as great as B = 2.0 T . Under normal circumstances, this field cannot be shut off by just flipping a switch. Instead the magnitude needs to be carefully decreased to zero. In an emergency, however, the magnet can be "quenched" so that B reduces to zero in 20 s. Such a quench can cost thousands of dollars and likely damages the magnets. Assume that the magnetic field exists inside a cylinder of radius R = 400 mm and length ℓ = 300 mm .
1. How much magnetic potential energy is dissipated when the magnetic field is quenched in this way?2. What is the average rate at which energy is dissipated?
Answer
given,
B = 2 T
time = 20 s
Radius = R = 400 mm
L= 300 mm
apply the formula for magnetic energy density
[tex]\eta_0 = \dfrac{energy}{volume}[/tex]
[tex]\eta_0 = \dfrac{B^2}{2\mu}[/tex]
[tex]\eta_0 = \dfrac{2^2}{2\times 4 \pi 10^{-7}}[/tex]
energy density = 1.59 x 10⁶ J/m³
Now , in the cylinder ,
Energy = energy density x volume
E = 1.59 x 10⁶ x π x r² x h
E = 1.59 x 10⁶ x π x 0.4² x 0.5
E = 4 x 10⁵ J
now, average rate of energy dissipated
rate of dissipation of energy = Energy/time
= 4 x 10⁵ /20
= 2 x 10⁴ W
the average rate at which energy is dissipated is 2 x 10⁴ W
1. The energy be "4 × 10⁵ J".
2. The average rate be "2 × 10⁴ W"
Magnetic fieldAccording to the question,
Time, t = 20 s
Radius, R = 400 mm
Length, L = 300 mm
We know the formula,
Magnetic energy density,
→ [tex]\eta_0[/tex] = [tex]\frac{Energy}{Volume}[/tex]
= [tex]\frac{B^2}{2 \mu}[/tex]
By substituting the values, we get
= [tex]\frac{(2)^2}{2\times 4 \pi 10^{-7}}[/tex]
= 1.59 × 10⁶ J/m³
Now,
1.
Energy will be:
= Energy density × Volume
= 1.59 × 10⁶ × π × (0.4)² × 0.5
= 4 × 10⁵ J
2.
We know,
The rate of dissipation = [tex]\frac{Energy}{Time}[/tex]
= [tex]\frac{4\times 10^5}{20}[/tex]
= 2 × 10⁴ W
Thus the above response is correct.
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Ifthe amount of ice on the planet increased to the amount that was present during the last glacial maximum, how muchwould it cause global sea level to fall? If you do not remember the approximate size of this rise, the volume of additionalice is 51000000 km3 and the surface area of the world's ocean is 361,000,000 km;A. less than 10 centimetersB. 10 centimeters to 1 meterC. 1 meter to 10 metersD. 10 meters to 50 metersE. more than 50 meters
Answer:
h = 12.95 m , The correct answer is D
Explanation:
For this exercise we use the definition of density
ρ= m V
For the water
[tex]\rho_{w}[/tex] = m / [tex]V_{w}[/tex]
For ice
[tex]\rho_{i}[/tex] = m / [tex]V_{i}[/tex]
The tabulated water density is [tex]\rho_{w}[/tex] = 997 Kg / m³ average between temperatures, the density near the freezing point that is 1000 kg/m³ can also be used, let's use the latter; The density of ice is 916.8 kg / m³
The mass of the water that is frozen is equal to the mass of ice that is formed, therefore, we can clear the doughs in the two equals formulas
Water
m = [tex]\rho_{w}[/tex] [tex]V_{w}[/tex]
Ice
m = [tex]\rho_{i}[/tex] [tex]V_{w}[/tex]
[tex]\rho_{w}[/tex] Vw = [tex]\rho_{i}[/tex] [tex]V_{i}[/tex]
Vw = [tex]\rho_{i}[/tex] / [tex]\rho_{w}[/tex] [tex]V_{i}[/tex]
Reduce to SI system
[tex]V_{i}[/tex] = 5.1 107 km³ (10³m / 1 km)³ = 5.1 10¹⁶ m³
Let's calculate
[tex]V_{w}[/tex] = 916.8 / 1000 5.1 10¹⁶
[tex]V_{w}[/tex] = 4,676 10¹⁶ m³
It indicates the surface of the ocean, so it is volume
V = A h
Where A is the surface and h is the height
A = 3.61 109 km² (10³ m / 1km)² = 3.61 10¹⁵ m²
h = V / A
h = 4.676 10¹⁶ / 3.61 10¹⁵
h = 12.95 m
The correct answer is D
A penny rides on top of a piston as it undergoes vertical simple harmonic motion with an amplitude of 4.0cm. If the frequency is low, the penny rides up and down without difficulty. If the frequency is steadily increased, there comes a point at which the penny leaves the surface.
1) At what point in the cycle does the penny first lose contact with the piston? (answer choices:)
A. midpoint moving up
B. midpoint losing down
C. highest point
D. lowest point
2) What is the maximum frequency for which the penny just barely remains in place for the full cycle?
1) At the moment of being at the top, the piston will not only tend to push the penny up but will also descend at a faster rate at which the penny can reach in 'free fall', in that short distance. Therefore, at the highest point, the penny will lose contact with the piston. Therefore the correct answer is C.
2) To solve this problem we will apply the equations related to the simple harmonic movement, hence we have that the acceleration can be defined as
[tex]a = -\omega^2 A[/tex]
Where,
a = Acceleration
A = Amplitude
[tex]\omega[/tex]= Angular velocity
From a reference system in which the downward acceleration is negative due to the force of gravity we will have to
[tex]a = -g[/tex]
[tex]-\omega^2 A = -g[/tex]
[tex]\omega = \sqrt{\frac{g}{A}}[/tex]
From the definition of frequency and angular velocity we have to
[tex]\omega = 2\pi f[/tex]
[tex]f = \frac{1}{2\pi} \sqrt{\frac{g}{A}}[/tex]
[tex]f = \frac{1}{2\pi} \sqrt{\frac{9.8}{4*10^{-2}}}[/tex]
[tex]f = 2.5Hz[/tex]
Therefore the maximum frequency for which the penny just barely remains in place for the full cycle is 2.5Hz
Final answer:
The penny loses contact with the piston at the highest point in the cycle, and the maximum frequency for which the penny remains in place is at its lowest point.
Explanation:
In simple harmonic motion, the penny loses contact with the piston at the highest point in the cycle. This is because at the highest point, the acceleration due to gravity is greater than the restoring force provided by the piston. Therefore, the correct answer is C. highest point.
The maximum frequency for which the penny just barely remains in place for the full cycle is when the acceleration due to gravity is equal to the restoring force provided by the piston. At this frequency, the penny is in equilibrium and does not leave the surface. Therefore, the maximum frequency is when the penny is at its lowest point in the cycle. So, the correct answer is D. lowest point.
Use technology and the given confidence level and sample data to find the confidence interval for the population mean muμ. Assume that the population does not exhibit a normal distribution. Weight lost on a diet:Weight lost on a diet: 95 % confidence95% confidence n equals 41n=41 x overbar equals 3.0 kgx=3.0 kg s equals 5.8 kgs=5.8 kg What is the confidence interval for the population mean muμ? 1.21.2 kgless than
Answer:
a. μ[tex]_{95%} = [/tex] 3 ± 1.8 = [1.2,4.8]
b. The correct answer is option D. No, because the sample size is large enough.
Explanation:
a. The population mean can be determined using a confidence interval which is made up of a point estimate from a given sample and the calculation error margin. Thus:
μ[tex]_{95%} = x_[/tex]±(t*s)/sqrt(n)
where:
μ[tex]_{95%} = [/tex] = is the 95% confidence interval estimate
x_ = mean of the sample = 3
s = standard deviation of the sample = 5.8
n = size of the sample = 41
t = the t statistic for 95% confidence and 40 (n-1) degrees of freedom = 2.021
substituting all the variable, we have:
μ[tex]_{95%} = [/tex] 3 ± (2.021*5.8)/sqrt(41) = 3 ± 1.8 = [1.2,4.8]
b. The correct answer is option D. No, because the sample size is large enough.
Using the the Central Limit Theorem which states that regardless of the distribution shape of the underlying population, a sampling distribution of size which is ≥ 30 is normally distributed.
A hydrogen atom in the ground state absorbs energy and makes a transition to the n = 3 state. The atom returns to the ground state by emitting two photons. What are the wavelengths of the two photons?
(A) 6.56 x 10-7m, 1.22 x 10-7m
(B) 6.56 x 10-7m, 5.45 x 10-7m
(C) 3.43 x 10-7m, 1.22 x 10-7m
(D) 5.45 x 10-7m, 3.43 x 10-7m
Answer:
at n= 3 λ = 656 nm
at n= 2 λ = 121.58 nm
Explanation:
Given details
transition of hydrogen atom from n = 2 to n = 3 state
Difference in energy between n = 3 state and n = 2 state :
[tex]= 2.18*10^{-18} \times [1/4 - 1/9] J = 3.03 \times 10^{-19} J[/tex]
so, energy of photon is given as [tex]=\frac{h\timesc}{\lambda} [/tex]
[tex]E = 3.03*10^{-19}
[/tex] So solve for wavelength
so, λ [tex]= \frac{6.626*10^{-34}*3*10^8}{3.03*10^{-19}}m[/tex]
=[tex] 6.56*10^{-7} m[/tex]
= 656 nm
for second transition,
energy transmitted is given asΔE [tex]=\frac{h\times c}{\lambda}[/tex]
and it is calculated as [tex]= 2.18*10^{-18}*[1/1 -1/4] J[/tex]
E = 1.635*10-18 J solving for wavelength in ENERGY equation we get
so, [tex]\lambda' = 1.2158*10^{-7}[/tex] m
= 121.58 nm
You know that you sound better when you sing in the shower. This has to do with the amplification of frequencies that correspond to the standing-wave resonances of the shower enclosure. A shower enclosure is created by adding glass doors and tile walls to a standard bathtub, so the enclosure has the dimensions of a standard tub, 0.75 m wide and 1.5 m long. Standing sound waves can be set up along either axis of the enclosure. What are the lowest two frequencies that correspond to resonances on each axis of the shower
Answer:
a) L = 0.75m f₁ = 113.33 Hz , f₃ = 340 Hz, b) L=1.50m f₁ = 56.67 Hz , f₃ = 170 Hz
Explanation:
This resonant system can be simulated by a system with a closed end, the tile wall and an open end where it is being sung
In this configuration we have a node at the closed end and a belly at the open end whereby the wavelength
With 1 node λ₁ = 4 L
With 2 nodes λ₂ = 4L / 3
With 3 nodes λ₃ = 4L / 5
The general term would be λ_n= 4L / n n = 1, 3, 5, ((2n + 1)
The speed of sound is
v = λ f
f = v / λ
f = v n / 4L
Let's consider each length independently
L = 0.75 m
f₁ = 340 1/4 0.75 = 113.33 n
f₁ = 113.33 Hz
f₃ = 113.33 3
f₃ = 340 Hz
L = 1.5 m
f₁ = 340 n / 4 1.5 = 56.67 n
f₁ = 56.67 Hz
f₃ = 56.67 3
f₃ = 170 Hz
The lowest two frequencies of resonance for the shower enclosure are 229 Hz and 114.33 Hz. The second harmonics for these lengths would be 458 Hz and 228.67 Hz respectively.
Explanation:The lowest two frequencies that correspond to resonances in your shower enclosure, which is 0.75 m wide and 1.5 m long, can be found using the formula for the fundamental frequency of a standing wave: f = v/2L, where v is the speed of sound in the air (~343 m/s), and L is the length or width of the enclosure, respectively.
For the axis parallel to the width (0.75 m), the lowest frequency is f = 343/(2*0.75) = 229 Hz. For the second harmonic along this axis, the frequency would be twice the fundamental, so 2*229 = 458 Hz.
For the axis parallel to the length (1.5 m), the lowest frequency is f = 343/(2*1.5) = 114.33 Hz. For the second harmonic along this axis, the frequency would be twice the fundamental, so 2*114.33 = 228.67 Hz.
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What can you conclude about the nature of vertical acceleration for a freely falling projectile?
Answer:
For a free falling projectile the acceleration of the body is always equal to the acceleration due to gravity and in the downward direction.
Explanation:
The acceleration of a body falling freely will experience the acceleration always in the downward direction.When the height of free fall is of the order of radius of the earth then we have a variable acceleration with respect to its height. This is given as:[tex]a=g.\frac{R^2}{(R+h)^2}[/tex]
where:
g = acceleration due to gravity on the earth surface
R = radius of the earth
h = height of the body above the earth surface
Final answer:
In a freely falling projectile, vertical acceleration is constant at 9.8 m/s² due to gravity, with velocity decreasing on the ascent and increasing on the descent, independent of horizontal motion.
Explanation:
Understanding Vertical Acceleration in Freely Falling Projectiles
When examining the characteristics of a freely falling projectile, we can make specific conclusions about its vertical acceleration. This acceleration is caused by the force of gravity, and its value is constant at 9.8 m/s² in the downward direction, irrespective of any horizontal motion the object may have.
This means that during the upward flight, the magnitude of the vertical velocity decreases, becoming zero at the projectile's highest point. Once the projectile starts its descent, the velocity increases again at the same rate due to the consistent force of gravity acting on it vertically.
The unique aspect of projectile motion is the independence of vertical motion from horizontal motion. Despite any horizontal velocity, the vertical acceleration remains unaffected and maintains a steady value of 9.8 m/s². This is a fundamental concept that was of great interest historically for both scientific and military applications and still holds significant relevance in physics.
An ice-making machine inside a refrigerator operates in a Carnot cycle. It takes heat from liquid water at 0.0 ∘C and rejects heat to a room at a temperature of 20.3 ∘C. Suppose that liquid water with a mass of 82.0 kg at 0.0 ∘C is converted to ice at the same temperature. Take the heat of fusion for water to be Lf = 3.34×105 J/kg .
Answer:
Energy required = [tex]J=2.73\times10^7 \ J.[/tex]
Explanation:
We know according to zeroth law of Thermodynamics, all bodies in a close system remains in thermal equilibrium.
Means, Energy gain by one part of system = energy loss by one another part of system.
Here, energy is given to surrounding and it is lost by water.
Therefore, its temperature decreases.
Energy required to convert 82 kg liquid, E = [tex]m\times L[/tex].
Here, [tex]m=82 \ kg[/tex]
[tex]L=larent \ heat \ of \ fusion=3.34\times 10^5J/kg.[/tex].
E=[tex]82 \times 3.34\times 10^5\ J=2.73\times10^7 \ J.[/tex]
A web page designer creates an animation in which a dot on a computer screen has position ⃗r=[4 cm+(2.5cm/s 2 )t 2 ] ^ i+(5cm/s)t ^ j .
a) Find the magnitude and direction of the dot’s average velocity between t=0 and t=2 s.
b) Find the magnitude and direction of the instantaneous velocity at t=0, t=1 s, and t=2 s. and show the velocities calculated in pan (b).
Answer:
Explanation:
a )
r = (4 + 2.5 t² )i + 5 t j
When t = 0
r₁ = 4 cm i
When t = 2s
r₂ = 14 i + 10 j
Displacement
= r₂ - r₁
=14 i + 10 j -4i
= 10i + 10j
average velocity
= displacement / time
=( 10i + 10j )/2
= 5i + 5j
b )
r = (4 + 2.5 t² )i + 5 t j
dr / dt = 5 t i + 5 j
Instantaneous velocity at t = o
(dr / dt)₀ = 5 j
Instantaneous velocity at t = 1s
(dr / dt)₁ = 5i + 5j
Instantaneous velocity at t = 2s
(dr / dt)₂ = 10i + 5 j
(a) The change in position or displacement (x) divided by the time intervals (t) in which the displacement happens is the average velocity. the magnitude and direction of the dot’s average velocity between t=0 and t=2 s will be 5 m/sec, 5 m/sec.
(b) the velocity of an object at a given point in time. . the magnitude and direction of the instantaneous velocity at t=0, t=1 s, and t=2 s. will be 10 m/sec and 5 m/sec.
(a) What is average velocity?The change in position or displacement (x) divided by the time intervals (t) in which the displacement happens is the average velocity
given,
[tex]\vec{r}= (4+2.5t^{2})\hat{i}+5t\hat{j}[/tex]
At time t=0
[tex]r_1= 4 \hat{i}[/tex][tex]\vector{r_2} -\vector{r_1}= 14 \hat{i}+10\hat{j}-4\hat{i}[/tex]
at (t=2)
[tex]\vector{r_2}= 14 \hat{i}+10\hat{j}[/tex]
[tex]\vector{r_2} -\vector{r_1}= 10 \hat{i}+10\hat{j}[/tex]
Average velocity = [tex]\frac{\vector{r_2} -\vector{r_1}}{t_2-t_1}[/tex]
Average velocity =[tex]5\hat{i}+5 \hat{j}[/tex]
hence the magnitude and direction of the dot’s average velocity between t=0 and t=2 s will be 5 m/sec, 5 m/sec.
(b)What is instantaneous velocity?instantaneous velocity is the velocity of an item in motion at a single point in time.
Given,
[tex]\vec{r}= (4+2.5t^{2})\hat{i}+5t\hat{j}[/tex]
[tex](\frac{ \del{r}}{\del{t}})_0= 5\hat{j}[/tex][tex](\frac{ \del{r}}{\del{t}})_2= 10\hat{i}+5\hat{j}[/tex]
[tex](\frac{ \del{r}}{\del{t}})_1= 5\hat{j}+5\hat{J}[/tex]
Hence the magnitude and direction of the instantaneous velocity at t=0, t=1 s, and t=2 s. will be 10 m/sec and 5 m/sec.
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A simply supported wood beam of rectangular cross section and span length 1.2 m carries a concentrated load P at midspan in addition to its own weight (see gure). The cross section has width 140 mm and height 240 mm. The weight density of the wood is 5.4 kN/m3 . Calculate the maximum permissible value of the load P if (a) the allowable bending stress is 8.5 MPa and (b) the allowable shear stress is 0.8 MPa.
Answer:
(a) P = 37.97 kN
(b) P = 35.62 kN
Explanation:
In the given problem, using the parameters and figure provided, we have:
A = b*h = 0.14*0.24 = 0.0336 m^2
S = (b*h^2)/6 = (0.14*0.24*0.24)/6 = 0.001344 m^3
q = 5400*0.0336 = 181.44 N/m
(a) The load P when σ (allowable bending stress) = 8.5 MPa = 8.5*10^6 Pa
σ = M/S
Thus: M = σ*S = 0.001344*8.5*10^6 = 11424 Nm
In addition,
M = (P*L/4) + (q*L^2)/8
11424 = (P*1.2/4) + (181.44*1.2^2)/8
11424 = 0.3*P + 32.6592
P = (11424-32.6592)/0.3 = 37.97 kN
(b) The value of P if τ (allowable shear stress) = 0.8 MPa = 0.8*10^6 Pa
(2*A*τ/3) = P/2 + q*L/2
(2*0.0336*0.8*10^6)/3 = P/2 + 181.44*1.2/2
17920 = P/2 + 108.864
P = (17920 - 108.864)*2 = 35.62 kN
To calculate the maximum permissible value of the load P, we need to consider both the bending stress and the shear stress.
Explanation:To calculate the maximum permissible value of the load P, we need to consider both the bending stress and the shear stress.
For (a), the maximum allowable bending stress is 8.5 MPa. Using the formula for bending stress, we can calculate the moment of inertia and then find the maximum permissible load P.
For (b), the maximum allowable shear stress is 0.8 MPa. Using the formula for shear stress, we can calculate the cross-sectional area and then find the maximum permissible load P.
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A machinist turns the power on to a grinding wheel, which is at rest at time t = 0.00 s. The wheel accelerates uniformly for 10 s and reaches the operating angular velocity of The wheel is run at that angular velocity for 37 s and then power is shut off. The wheel decelerates uniformly at until the wheel stops. In this situation, the time interval of angular deceleration (slowing down) is closest to:
Complete Question:
A machinist turns the power on to a grinding wheel, which is at rest at time t = 0.00 s. The wheel accelerates uniformly for 10 s and reaches the operating angular velocity of 25 rad/s. The wheel is run at that angular velocity for 37 s and then power is shut off. The wheel decelerates uniformly at 1.5 rads/s2 until the wheel stops. In this situation, the time interval of angular deceleration (slowing down) is closest to
Answer:
t= 16.7 sec.
Explanation:
As we are told that the wheel is accelerating uniformly, we can apply the definition of angular acceleration to its value:
γ = (ωf -ω₀) / t
If the wheel was at rest at t-= 0.00 s, the angular acceleration is given by the following equation:
γ = ωf / t = 25 rad/sec / 10 sec = 2.5 rad/sec².
When the power is shut off, as the deceleration is uniform, we can apply the same equation as above, with ωf = 0, and ω₀ = 25 rad/sec, and γ = -1.5 rad/sec, as follows:
γ= (ωf-ω₀) /Δt⇒Δt = (0-25 rad/sec) / (-1.5 rad/sec²) = 16.7 sec
A large crate filled with physics laboratory equipment must be moved up an incline onto a truck The crate is at rest on the incline. What can you say about the force of friction acting on the crate?A. the frictional force points up the inclineB. the frictional force points down the inclineC. the frictional force is zero
Answer:
A. The frictional force points up the incline.
Explanation:
If the crate is at rest, this means that no net force is acting on it.
In absence of friction, we have two forces acting on the crate; the normal force (which prevents that the crate fall through the incline surface), which is always perpendicular to the surface and upward) and gravity force (always downward).
If no net force acting, the 2nd Newton's Law can be expressed as follows:
Fnet = m*a = 0
This is a vector equation, which can be expressed as two algebraic equations, just decomposing both forces along two axes ,perpendicular each other.
As normal force (Fn) has no component along the incline, we can choose as our axes, one parallel to the incline, and another perpendicular, as we will only need to decompose Fg in two perpendicular components.
If the net force is 0, both components must be zero also:
Fx (calling x-axis to the parallel to the incline) = 0
Along the incline. we have a component of the gravity force only, as follows:
Fx = m*g*sin θ = 0
In the perpendicular direction, we have:
Fy = N-m*g*cosθ = 0 ⇒ N = m*g*cos θ (choosing the upward direction as positive)
Returning to Fx, it is clear, that for any angle θ other than 0º, there must be another force, that added to this force, gives 0 as result.
This force can't be another than friction force, which always opposes to the relative movement between the surfaces in contact each other.
As the component of gravity force along the incline tries to accelerate the crate downwards the incline, friction force must point up the incline, i.e., A is the right choice.
When moving a crate up an incline, the frictional force points down the incline to oppose the motion.
The frictional force resists motion, so when the crate is being moved up an incline, the frictional force points down the incline to oppose the motion. Therefore, option B: the frictional force points down the incline.
Snorkelers breathe through tubes that extend above the surface of the water. In prin- ciple, a snorkeler could go deeper with a longer tube, but the extra pressure at the greater depth would make breathing too difficult. If a snorkeler can develop a pressure in her lungs that is 10 kPa below the pressure outside her body, what is the longest snorkel she could use in the ocean?
Answer:
h = 1.02 m
Explanation:
This is a fluid mechanics exercise, where the pressure is given by
P = [tex]P_{atm}[/tex] + ρ g h
The gauge pressure is
P - [tex]P_{atm}[/tex] = ρ g h
In this case the upper part of the tube we have the atmospheric pressure. and the diver can exert a pressure 10 KPa below the outside pressure, this must be the gauge pressure
[tex]P_{m}[/tex] = P - [tex]P_{atm}[/tex]
[tex]P_{m}[/tex] = ρ g h
h =[tex]P_{m}[/tex] / ρ g
calculate
h = 10 103 / (1000 9.8)
h = 1.02 m
This is the depth at which man can breathe
A snorkeler can use a snorkel to go to a depth of approximately 0.099 atmospheres or 3.267 feet.
Explanation:The pressure a snorkeler experiences while underwater increases with depth. To calculate the longest snorkel a person can use, we need to consider the pressure difference between the inside and outside of their body. If a snorkeler can develop a pressure in their lungs that is 10 kPa below the pressure outside their body, we can find the depth they can dive to by converting the pressure difference to atmospheres using the conversion 1 atm = 101.325 kPa. Then, we can use the relationship that every 33 feet of seawater represents 1 atmosphere of pressure in addition to the 1 atmosphere of pressure from the atmosphere at sea level. By dividing the pressure difference in atmospheres by this value, we can determine the maximum depth a snorkeler can go.
Let's calculate:
(10 kPa * 1 atm/101.325 kPa) / (33 ft * 1 atm / (33 ft))
≈ 0.099 atm
So, a snorkeler can use a snorkel to go to a depth of approximately 0.099 atmospheres or 3.267 feet.
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11–8 Consider a heavy car submerged in water in a lake with a flat bottom. The driver’s side door of the car is 1.1 m high and 0.9 m wide, and the top edge of the door is 10 m below the water surface. Determine the net force acting on the door (normal to its surface) and the location of the pressure center if (a) the car is well-sealed and it contains air at atmospheric pressure and (b) the car is filled with water.
Answer:
Explanation:
position of centre of mass of door from surface of water
= 10 + 1.1 / 2
= 10.55 m
Pressure on centre of mass
atmospheric pressure + pressure due to water column
10 ⁵ + hdg
= 10⁵ + 10.55 x 1000 x 9.8
= 2.0339 x 10⁵ Pa
the net force acting on the door (normal to its surface)
= pressure at the centre x area of the door
= .9 x 1.1 x 2.0339 x 10⁵
= 2.01356 x 10⁵ N
pressure centre will be at 10.55 m below the surface.
When the car is filled with air or it is filled with water , in both the cases pressure centre will lie at the centre of the car .
The net force acting on the door (normal to its surface) and the location of the pressure should be [tex]2.01356 \times 10^5 N[/tex]
Calculation of net force & pressure location:The position of center of mass of door from the surface of water should be
[tex]= 10 + 1.1 \div 2[/tex]
= 10.55 m
Now
Pressure on center of mass
= atmospheric pressure + pressure due to water column
[tex]= 10^5 + 10.55 \times 1000 \times 9.8\\\\= 2.0339 \times 10^5 Pa[/tex]
Now
the net force acting on the door (normal to its surface)
[tex]= .9 \times 1.1 \times 2.0339 \times 10^5\\\\= 2.01356 \times 10^5 N[/tex]
So,
pressure centre will be at 10.55 m below the surface.
Therefore,
When the car is filled with air or it is filled with water , in both the cases pressure center should be lie at the centre of the car .
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The orbital quantum number for the electron in a hydrogen atom is l=5. What is the smallest possible value (algebraically) for the total energy of this electron? Give your answer in electron volts.
The smallest possible value for the total energy of an electron with l=5 in a hydrogen atom is -0.37778 eV.
Explanation:The smallest possible value for the total energy of an electron in a hydrogen atom can be found by using the energy formula for Bohr's model. The energy is given by:
E = -13.6 eV / n^2
where n is the principal quantum number. In this case, since l=5, the smallest possible value for the total energy is when n=6. Plugging this value into the energy formula, we get:
E = -13.6 eV / 6^2 = -0.37778 eV
A pendulum, comprising a light string of length L and a small sphere, swings in the vertical plane. The string hits a peg located a distance d below the point of suspension (Fig. P8.68). (a) Show that if the sphere is released from a height below that of the peg, it will return to this height after the string strikes the peg. (b) Show that if the pendulum is released from rest at the horizontal position (u 5 908) and is to swing in a complete circle centered on the peg, the minimum value of d must be 3L/5.
Answer:
the demonstrations are in the description.
Explanation:
First of all, reference is made to a picture that may be the one shown attached.
when the string hits the pin, the sphere will continue to swing to the right until a height h'.
a) Initially, the system has only potential energy because it is released with zero velocity from a height h.
if we consider that the system has no energy losses, then all that potential energy must be conserved. Thus, this energy is transformed in principle into kinetic energy and once the kinetic energy is maximum, it will be transformed again into potential energy.
The potential energy is given by:
[tex]E_p = mgh[/tex]
As we can see, potential energy only depends on mass, height and gravity. Since mass and gravity are constant, then for potential energy to be conserved, heights must be the same.
b) the statement probably has a transcription error in question b. Corrected, it would look like:
Show that if the pendulum is released from the horizontal position (∡ = 90) and is to swing in a complete circle centered on the peg, the minimum value of d must be 3L/5.
podemos asumir que la altura cero está dada en el punto de reposo del péndulo, por lo cual la energía inicial (que es potencial) está dada por:
[tex]E_0=MGL[/tex]
For the pendulum to swing, the tension must always be positive, and as a consequence, the centripetal force must be greater or at least equal to the weight. Mathematically
[tex]mV^2 /(L-d) = mg[/tex] (1)
Where:
[tex]mV^2 /(L-d) [/tex] It is the centripetal force that the sphere experiences.
Now, for the sphere to describe the circular motion, the minimum kinetic energy must be:
[tex]\frac{mV^2}{2} = mgL-2mg(L-d)[/tex] (2)
Then, replacing equation 1 in 2:
[tex]\frac{mg(L-d)}{2} = mgL-2mg(L-d)[/tex]
Simplifying:
[tex]L-2L+2d= \frac{L}{2} - \frac{d}{2}[/tex]
[tex]\frac{5}{2}d= \frac{3}{2}L[/tex]
Which means
[tex]d=\frac{3L}{5}[/tex]
Final answer:
When a pendulum hits a peg, it will return to its original height if released from below the peg. The minimum value of d, in order for a pendulum to swing in a complete circle centered on a peg, must be 3L/5.
Explanation:
A pendulum comprising a light string and a small sphere swings in the vertical plane. When the string hits a peg located below the point of suspension, the sphere will return to its original height if it is released from a height below that of the peg. The reason for this is that when the string hits the peg, the tension in the string abruptly changes direction, causing the sphere to swing back upwards.
If the pendulum is released from rest at the horizontal position and is to swing in a complete circle centered on the peg, the minimum value of d must be 3L/5. This is because when the pendulum is at the lowest point, the tension in the string must be equal to or greater than the weight of the sphere in order to maintain circular motion. The minimum value of d ensures that this condition is satisfied.
A coil consists of 200 turns of wire. Each turn is a square of side d = 18 cm, and a uniform magnetic field directed perpendicular to the plane of the coil is turned on. If the field changes linearly from 0 to 0.50 T in 0.80 s, what is the magnitude of the induced emf in the coil while the field is changing?
Answer:
Induced emf in the coil, [tex]\epsilon=4.05\ volts[/tex]
Explanation:
Given that.
Number of turns in the coil, N = 200
Side of square, d = 18 cm = 0.18 m
The field changes linearly from 0 to 0.50 T in 0.80 s.
To find,
The magnitude of the induced emf in the coil while the field is changing.
Solution,
We know that due to change in the magnetic field, an emf gets induced in the coil. The formula of induced emf is given by :
[tex]\epsilon=\dfrac{d\phi}{dt}[/tex]
[tex]\phi[/tex] = magnetic flux
[tex]\epsilon=-\dfrac{d(NBA)}{dt}[/tex]
A is the ares of square
[tex]\epsilon=AN\dfrac{d(B)}{dt}[/tex]
[tex]\epsilon=AN\dfrac{B_f-B_i}{t}[/tex]
[tex]\epsilon=(0.18)^2\times 200 \times \dfrac{0.5-0}{0.8}[/tex]
[tex]\epsilon=4.05\ volts[/tex]
So, the induced emf in the coil is 4.05 volts. Hence, this is the required solution.
To calculate the induced emf in a 200-turn coil with a changing magnetic field, use Faraday's Law of Induction. The magnitude of the induced emf is 4.05 V. This result is obtained through calculations involving the number of turns, area of each turn, and the rate of change of the magnetic field.
The problem involves calculating the induced emf in a coil with 200 turns of wire, where each turn is a square of side d = 18 cm (or 0.18 m). The magnetic field (B) changes linearly from 0 to 0.50 T over a time interval of 0.80 s.
We use Faraday's Law of Induction, which states that the induced emf (ε) in a coil is given by:
ε = -N(dΦ/dt)
where:
N is the number of turns in the coil (200 turns)Φ is the magnetic flux through one turn of the coilThe magnetic flux (Φ) through one turn is calculated as:
Φ = B × A
where A is the area of one turn of the coil:
A = d × d = 0.18 m × 0.18 m = 0.0324 m²
Since the magnetic field changes linearly, the rate of change of the magnetic field (dB/dt) is:
dB/dt = (0.50 T - 0 T) / 0.80 s = 0.625 T/s
Now, we can calculate the rate of change of the magnetic flux (dΦ/dt):
dΦ/dt = A × dB/dt = 0.0324 m² × 0.625 T/s = 0.02025 Wb/s
Finally, we use Faraday's Law to find the induced emf:
ε = -N(dΦ/dt) = -200 (0.02025 Wb/s) = -4.05 V
The magnitude of the induced emf in the coil while the field is changing is 4.05 V.
A solenoid coil with 22 turns of wire is wound tightly around another coil with 330 turns. The inner solenoid is 21.0 cm long and has a diameter of 2.30 cm . At a certain time, the current in the inner solenoid is 0.140 A and is increasing at a rate of 1800 A/s .
Answer:
The average magnetic flux through each turn of the inner solenoid is [tex]11.486\times10^{-8}\ Wb[/tex]
Explanation:
Given that,
Number of turns = 22 turns
Number of turns another coil = 330 turns
Length of solenoid = 21.0 cm
Diameter = 2.30 cm
Current in inner solenoid = 0.140 A
Rate = 1800 A/s
Suppose For this time, calculate the average magnetic flux through each turn of the inner solenoid
We need to calculate the magnetic flux
Using formula of magnetic flux
[tex]\phi=BA[/tex]
[tex]\phi=\dfrac{\mu_{0}N_{2}I}{l}\times\pi r^2[/tex]
Put the value into the formula
[tex]\phi=\dfrac{4\pi\times10^{-7}\times330\times0.140}{21.0\times10^{-2}}\times\pi\times(\dfrac{2.30\times10^{-2}}{2})^2[/tex]
[tex]\phi=11.486\times10^{-8}\ Wb[/tex]
Hence, The average magnetic flux through each turn of the inner solenoid is [tex]11.486\times10^{-8}\ Wb[/tex]
As part of a science project you investigate the effect of punching holes in a plastic container of water and observing the flow of liquid out of the different holes. If the container has a water level 20 cm above the base and you punch a hole in the side 12 cm high, calculate the speed at which water will initially flow out of the hole.
a.1.25 m/s
b.15.3 m/s
c.12.5 m/s
d. 1.50 m/s
Final answer:
Using Bernoulli's principle and Torricelli's law, the initial speed of water flowing out of a hole in a container can be calculated. The speed is determined by the height of water above the hole and the acceleration due to gravity, resulting in an initial speed of approximately 1.25 m/s for the given problem.
Explanation:
The question involves calculating the initial speed at which water will flow out of a hole in a plastic container, using principles of fluid dynamics found in physics. To determine this speed, Bernoulli's principle and Torricelli's law can be applied. These principles state that the speed (v) of the fluid exiting the hole can be calculated using the equation v = √(2gh), where g is the acceleration due to gravity (9.81 m/s²) and h is the height of the water column above the hole.
In this case, h = 20 cm - 12 cm = 8 cm = 0.08 m. Plugging this into the equation gives v = √(2 * 9.81 m/s² * 0.08 m), which results in v ≈ 1.25 m/s. Therefore, the initial speed of the water flowing out of the hole is approximately 1.25 m/s, making option a the correct answer.
Two semiconductors are identical except that one has a band gap of 1.2 eV, while the other has a band gap of 1.1 eV. The room temperature (T = 300 K) intrinsic carrier density of the 1.1 eV material is ni1(300K) = 1.0×1019 m−3 and that of the 1.2 eV material is ni2(300K).1)Calculate the ratio of intrinsic carrier densities for the two materials at room temperature. ni2/ni1=7.56×10−21
To solve this problem it is necessary to apply the relationship given by the intrinsic carrier concentration, in each of the phases.
The intrinsic carrier concentration is the number of electrons in the conduction band or the number of holes in the valence band in intrinsic material. This number of carriers depends on the band gap of the material and on the temperature of the material.
In general, this can be written mathematically as
[tex]\eta_i = \sqrt{N_cN_v}e^{-\frac{E_g}{2KT}}[/tex]
Both are identical semiconductor but the difference is band gap which is:
[tex]E_{g1} = 1.1eV[/tex]
[tex]n_{i1} = 1*10^{19}m^{-3}[/tex]
[tex]E_{g2} = 1.2eV[/tex]
[tex]T=300K[/tex]
The ratio between the two phases are given as:
[tex]\frac{\eta_{i1}}{\eta_{i2}} = \frac{e^{-\frac{E_{g1}}{2KT}}}{e^{-\frac{E_{g2}}{2KT}}}[/tex]
[tex]\frac{\eta_{i1}}{\eta_{i2}} = e^{\frac{E_{g2}-E_{g1}}{2KT}}[/tex]
[tex]\frac{\eta_{i1}}{\eta_{i2}} =e^{\frac{(1.2-1.1)(1.6*10^{-19})}{2(1.38*10^{-23})(300)}}[/tex]
[tex]\frac{\eta_{i1}}{\eta_{i2}} =e^{-1.932367}[/tex]
[tex]\frac{\eta_{i1}}{\eta_{i2}} =0.145[/tex]
Therefore the ratio of intrinsic carrier densities for the two materials at room temperature is 0.145
Two identical bowling balls are rolling on a horizontal floor without slipping. The initial speed of both balls is v = 10 m/s. Ball A encounters a frictionless ramp, reaching a maximum vertical height HA above the floor. Ball B on the other hand rolls up a regular ramp (i.e. without slipping), reaching a maximum vertical height HB above the floor. Which ball goes higher and by how much? Show the steps of your calculations.
Answer:
The difference between frictionless ramp and a regular ramp is that on a frictionless ramp the ball cannot roll it can only slide, but on a regular ramp the ball can roll without slipping.
We will use conversation of energy.
[tex]K_A_1 + U_A_1 = K_A_2 + U_A_2\\\frac{1}{2}I\omega^2 + \frac{1}{2}mv^2 + 0 = 0 + mgH_A[/tex]
Note that initial potential energy is zero because the ball is on the bottom, and the final kinetic energy is zero because the ball reaches its maximum vertical distance and stops.
For the ball B;
[tex]K_B_1 + U_B_1 = K_B_2 + U_B_2[/tex]
[tex]\frac{1}{2}I_B\omega^2 + \frac{1}{2}mv^2 + 0 = 0 + mgH_B[/tex]
The initial velocities of the balls are equal. Their maximum climbing point will be proportional to their final potential energy. Since their initial kinetic energies are equal, their final potential energies must be equal as well.
Hence, both balls climb the same point.
Explanation:
Both balls A and B will climb up to the same height according to the conservation of energy.
Calculating the height:
Since the balls are identical, let both have mass m and velocity v.
Also the moment of inertia I and the angular speed ω will be the same.
Ball A encounters a frictionless ramp:
On a frictionless ramp, the ball slides down the ramp since it cannot roll as there is no friction present. Since there is o frictional force, there is no dissipation of energy. Energy is conserved.
According to the law of conservation of energy, the total energy of the system must be conserved.
KE(initial) + PE(initial) = KE(final) + PE(final)
[tex]\frac{1}{2}I\omega^2+\frac{1}{2}mv^2+0=0+mgH_A\\\\H_A=\frac{1}{mg} [\frac{1}{2}I\omega^2+\frac{1}{2}mv^2][/tex]
initial KE has rotational and translational kinetic energy and the initial PE is zero since the ball is on the ground, also the final KE is zero since the velocity at the highest point will be zero.
Ball B encounters a regular ramp:
On a regular ramp, the ball can roll without sliding due to friction.
[tex]\frac{1}{2}I\omega^2+\frac{1}{2}mv^2+0=0+mgH_B\\\\H_B=\frac{1}{mg} [\frac{1}{2}I\omega^2+\frac{1}{2}mv^2][/tex]
The height will be the same since the velocity is the same as ball A.
Also, the translational or rotational velocity will be zero at the highest point.
Hence, both balls climb the same height.
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The tape in a videotape cassette has a total length 191 m and can play for 1.9 h. As the tape starts to play, the full reel has an outerradius of 35 mm and an inner radius of 11 mm. At some point during the play, both reels will have the same angular speed.What is this common angular speed?
The common angular speed is 9090.91 rad/h.
Explanation:To find the common angular speed, we need to first find the initial and final radii of the full reel. The initial radius is 35 mm and the final radius is 11 mm. The angular speed can be found using the equation v = rω, where v is the linear speed, r is the radius, and ω is the angular speed. We can solve for ω by dividing the linear speed by the radius. Since both reels will have the same angular speed, we can use the same equation to find the angular speed when the tape has unspooled a distance of 191 m.
First, let's convert the radii from millimeters to meters:
Initial radius: 35 mm = 0.035 mFinal radius: 11 mm = 0.011 mNext, we'll find the linear speeds at the initial and final radii. Since the length of the tape is 191 m and it plays for 1.9 hours, the linear speed can be calculated by dividing the length of the tape by the playtime:
Linear speed at initial radius: (191 m)/(1.9 h) = 100 m/hLinear speed at final radius: (191 m)/(1.9 h) = 100 m/hFinally, we can find the common angular speed by dividing the linear speed by the radius:
Initial angular speed: (100 m/h)/(0.035 m) = 2857.14 rad/hFinal angular speed: (100 m/h)/(0.011 m) = 9090.91 rad/hTherefore, the common angular speed is 9090.91 rad/h.
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Scuba divers are warned that if they must make a rapid ascent, they should exhale on the way up. If a diver rapidly ascends to the surface with lungs full of air, his lungs could be damaged. Explain why this is so.
Air for a diver comes out of a high pressure tank at - Same- pressure compared to the water around the diver (metered by the regulator).
This means the lungs are inflated with - Highly pressurized- gas.
This does not adversely affect the diver when deep underwater, because the entire environment around the diver is at -Same - pressure.
If the diver suddenly surface, the air in the alveoli in the lungs will still be at - a higher - pressure compared to the air around the diver, which will be at - a lower - pressure.
The gas in the diver's lungs will - expand - and can damage the alveoli.
Estimate the kinetic energy necessary for a projectile with mass m can "escape" from the surface of a planet if it is launched vertically upward. Express your estimation in terms of g which is the acceleration due to gravity at the planet's surface and Rp which is the planet's radius. Ignore air resistance.
A. The kinetic energy is about than 2mgRp.
B. The kinetic energy greater than mgRp.
C. The kinetic energy is about than 2mR2p/g.
D. The kinetic energy greater than mR2p/g.
Answer:
B. The kinetic energy has to be greater than mgRp.
Explanation:
Hi there!
Since there is no friction and because of the conservation of energy, all the initial kinetic energy will be converted into gravitational potential energy. At a height equal to the radius of the planet, the gravitational potential energy will be:
EP = mgRp
Where:
m = mass of the projectile.
g = acceleration due to gravity.
Rp = height = Planet´s radius.
To reach that height, the initial kinetic energy has to be equal to that potential energy (remember that at the maximum height, the potential energy will be equal to the initial kinetic energy because there is no energy dissipation by heat because there is no air friction).
Then, to escape from the surface of the planet, the initial kinetic energy has to be greater than mgRp (Answer B).
To solve this problem it is necessary to apply the concepts related to the conservation of kinetic energy and potential energy.
As there is an increase in gravitational potential energy, there is a decrease in kinetic energy - which 'generates' the movement - on the particle.
Mathematically this can be expressed as
[tex]KE = PE_g[/tex]
[tex]KE = \frac{GMm}{R_p}[/tex]
Where,
G = Gravitational Universal Constant
M = Mass of Earth
m = Mass of object
R = Radius
Acceleration due to gravity we know that it is defined as
[tex]g = \frac{GM}{R_p^2}[/tex]
From the kinetic energy formula we can then re-adjust it mathematically as
[tex]KE = \frac{GMm}{R_p}[/tex]
[tex]KE = \frac{GMm}{R_p}*\frac{R_p}{R_p}[/tex]
[tex]KE = \frac{GM}{R_p^2}*(R_p)(m)[/tex]
[tex]KE = g R_p m[/tex]
Finally we can observe that the kinetic energy must be at least equivalent to [tex]mgR_p[/tex] (Correct Answer is B), in order to escape.