Answer:
[tex]L=6.21m[/tex]
Explanation:
For the simple pendulum problem we need to remember that:
[tex]\frac{d^{2}\theta}{dt^{2}}+\frac{g}{L}sin(\theta)=0[/tex],
where [tex]\theta[/tex] is the angular position, t is time, g is the gravity, and L is the length of the pendulum. We also need to remember that there is a relationship between the angular frequency and the length of the pendulum:
[tex]\omega^{2}=\frac{g}{L}[/tex],
where [tex]\omega[/tex] is the angular frequency.
There is also an equation that relates the oscillation period and the angular frequeny:
[tex]\omega=\frac{2\pi}{T}[/tex],
where T is the oscillation period. Now, we can easily solve for L:
[tex](\frac{2\pi}{T})^{2}=\frac{g}{L}\\\\L=g(\frac{T}{2\pi})^{2}\\\\L=9.8(\frac{5}{2\pi})^{2}\\\\L=6.21m[/tex]
a projectile is shot at an inclination of 45 frin tge horizontial with a speed of 250 m/s. how far will it travek ub the horizontal direction
Answer:
6250 m
Explanation:
When an object is projected into the air, the distance along the horizontal direction is called the Range.
The Range of a projectile is expressed as;
[tex]R = \frac{u^{2}sin2\alpha }{g}[/tex]
Where,
R is the range of the projectile
α is the angle of inclination with the horizontal
g is the acceleration due to gravity = 9.8 m/s ≈ 10 m/s
Given; α =45° , u = 250 m/s
[tex]R = \frac{250^{2}sin2(45)}{10}[/tex]
[tex]R = \frac{62500sin90}{10}[/tex]
[tex]R = \frac{62500}{10}[/tex]
R = 6250 m
The range is 6250 m
In a projectile motion, the given object travel 6377.55 m in the horizontal direction.
In a projectile motion, the distance of the object along the horizontal direction is called the Range.
The Range of a projectile
[tex]\bold {R = \dfrac {u^2 sin2\alpha }{g}}[/tex]
Where,
R - range of the projectile
u - initial speed = 250 m/s
α - angle of inclination with the horizontal = 45°
g - gravitational acceleration = 9.8 m/s
Put the values in the formula,
[tex]\bold {R = \dfrac {(250)^2 sin2(45) }{9.8}}\\\\\bold {R = \dfrac {62500\ sin 90 }{9.8}}\\\\\bold {R =6377.55}[/tex] Since, sin 90 = 1
Therefore, in a projectile motion, the given object travel 6377.55 m in the horizontal direction.
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The sound source of a ship’s sonar system operates at a frequency of 22.0 kHz . The speed of sound in water (assumed to be at a uniform 20∘C) is 1482 m/s . What is the difference in frequency between the directly radiated waves and the waves reflected from a whale traveling straight toward the ship at 4.95 m/s ? Assume that the ship is at rest in the water.
Your target variable is Δf, the magnitude of the difference in frequency between the waves emitted from the sonar device and the waves received by the device after reflecting off the whale. Write an expression for Δf in terms of the relevant frequencies using the subscript notation introduced above.
Express your answer in terms of some or all of the variables fLe, fLr, fSe, and fSr.
Δf =
Answer:
[tex]\Delta f=f_{Lr}-f_{Se}[/tex]
147.45 Hz
Explanation:
v = Speed of sound in water = 1482 m/s
[tex]v_w[/tex] = Speed of whale = 4.95 m/s
Frequency of the wave in stationary condition
[tex]f_{Lr}=f\dfrac{v+v_w}{v-v_w}[/tex]
Ship's frequency which is reflected back
[tex]f_{Se}=f\dfrac{v}{v-v_w}[/tex]
The difference in frequency is given by
[tex]\Delta f=f\dfrac{v+v_w}{v-v_w}-f\dfrac{v}{v-v_w}\\\Rightarrow \Delta f=f_{Lr}-f_{Se}[/tex]
[tex]\mathbf{\Delta f=f_{Lr}-f_{Se}}[/tex]
[tex]f_{Lr}=22\times \dfrac{1482+4.95}{1482-4.95}\\\Rightarrow f_{Lr}=22.14745\ kHz[/tex]
[tex]f_{Se}=22\ kHz[/tex]
[tex]\Delta f=f_{Lr}-f_{Se}\\\Rightarrow \Delta f=22.14745-22\\\Rightarrow \Delta f=0.14745\ kHz\\\Rightarrow \Delta f=147.45\ Hz[/tex]
The difference in wavelength is 147.45 Hz
To get up on the roof, a person (mass 92.0 kg) places a 5.60 m aluminum ladder (mass 14.0 kg) against the house on a concrete pad with the base of the ladder 2.00 m from the house. The ladder rests against a plastic rain gutter, which we can assume to be frictionless. The center of mass of the ladder is 2 m from the bottom. The person is standing 3 m from the bottom. What are the magnitudes (in N) of the forces on the ladder at the top and bottom?
Answer:
Down
F1ₓ = 219.6N
[tex]F1_{y}[/tex] = 1038.8 N
Top
F2ₓ = 219.6 N
[tex]F2_{y}[/tex] = 0
Explanation:
For this exercise we must make a free body diagram of the ladder, see attached, then use the balance equations on each axis
Transnational Balance
X axis
F1ₓ -F2ₓ = 0
F1ₓ = F2ₓ
Y Axis
[tex]F1_{y}[/tex] - [tex]F2_{y}[/tex] - W - W_man = 0 (1)
Rotational balance
The reference system is placed at the bottom of the stairs and we can turn the anti-clockwise direction of rotation as positive
F2ₓ y - [tex]F2_{y}[/tex] x - W x - W_man x_man = 0
Let us write the data they give, the masses of the ladder (m = 14.0 kg), the mass of man (m_man = 92 kg), the center of mass of the ladder that is 2m from the bottom (the height) and the position of the man which is 3 m high
Let's look with trigonometry for distances
The angle of the stairs is
cos θ = x / L
θ = cos⁻¹ x / L
θ = cos⁻¹ 2 / 5.6
θ = 69⁰
Height y
tan 69 = y / x
y = x tan 69
y = 2 tan 69
y = 5.21 m
Distance x
tan 69 = 2 / x
x = 2 / tan 69
x = 0.7677 m
The distance x_man
x_man = 3 / tan 69
x_man = 1,152 m
They indicate that between the scalars and the support there is no friction so the vertical force at the top is zero
[tex]F2_{y}[/tex] = 0
Let's replace in the translational equilibrium equation
F2ₓ y - [tex]F2_{y}[/tex] x - W x - W_man x_man = 0
F2ₓ 5.21 -0 - 14.0 9.8 0.7677 - 92.0 9.8 1,152 = 0
F2ₓ = 1143.97 / 5.21
F2ₓ = 219.6 N
We use equation 1
[tex]F1_{y}[/tex] + 0 - W - W_man = 0
[tex]F1_{y}[/tex] = W + W_man
[tex]F1_{y}[/tex] = (m + m_man) g
[tex]F1_{y}[/tex] = (14 +92) 9.8
[tex]F1_{y}[/tex] = 1038.8 N
We can write the force on each part of the ladder
Down
F1ₓ = 219.6N
[tex]F1_{y}[/tex] = 1038.8 N
Top
F2ₓ = 219.6 N
[tex]F2_{y}[/tex] = 0
Julie drives 100 mi to Grandmother's house. On the way to Grandmother's, Julie drives half the distance at 44.0mph and half the distance at 65.0 . On her return trip, she drives half the time at 44.0mph and half the time at 65.0mph .
1)What is Julie's average speed on the way to Grandmother's house?
2)What is her average speed on the return trip?
*Express your answer with the appropriate units.
Answer:
52.47706 mph
54.5 mph
Explanation:
The average speed is given by
[tex]V_{av}=\dfrac{Distance}{Time}[/tex]
[tex]V_{av}=\dfrac{100}{\dfrac{50}{44}+\dfrac{50}{65}}\\\Rightarrow V_{av}=52.47706\ mph[/tex]
Julie's average speed on the way to Grandmother's house is 52.47706 mph
[tex]V_{av}=\dfrac{44+65}{2}\\\Rightarrow V_{av}=54.5\ mph[/tex]
Average speed on the return trip is 54.5 mph
Two point sources produce waves of the same wavelength that are in phase. At a point midway between the sources, you would expect to observea) constructive or destructive interference depending on the wavelength.b) destructive interferencec) no interferenced) constructive interferencee) alternating constructive and destructive interference interference
Answer:
Constructive Interference.
Explanation:
Constructive Interference.
Definition:
Two waves meet in such a way their highs(Crests) combine to form a new waves whose magnitude is the sum of magnitude of combining waves.
Since two waves have same wavelength and are in phase so when they combine they well form a way which has the magnitude equal to the sum of the magnitude of both waves.
Reasons why it is Constructive Interference:Waves
Have Same wavelengthAre in phase (Can have phase difference of 2πHave crests aligned with each other (Appear at same point)A(n) _________ is a line joining the points of equal magnetic declination. Contour line isohyet isotherm isomag isogonic line
Answer:
Isogonic line is a line joining the points of equal magnetic declination.
Explanation:
Isogonal line is a line that joins the places of equal declination. Also isogonal line is known as the line which connects the point having the same magnetic declination.
An isogonic line joins points of equal magnetic declination on a map. This is important for navigation, as it reflects the angle difference between true north and magnetic north. Isotherms, isohyets, and isobars are other types of isolines used in geography. Thus option 5. isogonic line is correct.
A isogonic line is a line joining points of equal magnetic declination. Magnetic declination is the angle between magnetic north (the direction the north end of a compass needle points) and true north. These lines are important for navigational purposes and are often shown on special maps known as isogonic charts.
Other types of isolines include:
Isotherm: connects points of equal temperature.Isohyet: connects points of equal precipitation.Isobar: connects points of equal atmospheric pressure.Understanding these different lines helps in various geographical and meteorological analyses, making it easier to interpret maps and forecasts.
Complete question.
A(n) _________ is a line joining the points of equal magnetic declination.
Contour lineisohyet line isotherm lineisomag lineisogonic lineou plan to excite electrons in a material by exposing it laser radiation. If you want to jump electrons from the 2 shell to the 6 shell, what wavelength of laser should you use
Answer:
[tex]410.2 nm[/tex]
Explanation:
We are given that
[tex]n_1=2,n_2=6[/tex]
We have to find the wavelength of laser should you used.
We know that
[tex]\frac{1}{\lambda}=R(\frac{1}{n^2_1}-\frac{1}{n^2_2})[/tex]
Where [tex]R=1.097\times 10^7/m[/tex]=Rydberg constant
[tex]\lambda[/tex]=Wavelength
Using the formula
[tex]\frac{1}{\lambda}=1.097\times 10^7(\frac{1}{2^2}-\frac{1}{6^2})[/tex]
[tex]\frac{1}{\lambda}=1.097\times 10^7(\frac{1}{4}-\frac{1}{36})[/tex]
[tex]\frac{1}{\lambda}=1.097\times 10^7(\frac{9-1}{36}=1.097\times 10^7\times \frac{8}{36}[/tex]
[tex]\frac{1}{\lambda}=\frac{1.097\times 10^7}{4}[/tex]
Using identity:[tex]\frac{1}{a^x}=a^{-x}[/tex]
[tex]\lambda=\frac{4}{1.097}\times 10^{-7}[/tex]=[tex]4.102\times 10^{-7} m[/tex]
1 nm=[tex]10^{-9} m[/tex]
[tex]\lambda=4.102\times 100 \times 10^{-9}=410.2\times 10^{-9} [/tex] m=410.2 nm
Hence, the wavelength of laser=[tex]410.2 nm[/tex]
A positively charged particle initially at rest on the ground accelerates upward to 160 m/s in 2.10 s. The particle has a charge-to-mass ratio of 0.100 C/kg and the electric field in this region is constant and uniform.What are the magnitude and direction of the electric field? Express your answer to two significant digits and include the appropriate units.
Answer:
The magnitude of the electric field is [tex]8.6\times10^{2}\ N/C[/tex]
Explanation:
Given that,
Time t = 2.10 s
Speed = 160 m/s
Specific charge =Ratio of charge to mass = 0.100 C/kg
We need to calculate the acceleration
Using equation of motion
[tex]a=\dfrac{v-u}{t}[/tex]
Put the value into the formula
[tex]a=\dfrac{160-0}{2.10}[/tex]
[tex]a=76.19\ m/s^2[/tex]
We need to calculate the magnitude of the electric field
Using formula of electric field
[tex]E=\dfrac{F}{q}[/tex]
[tex]E=\dfrac{ma}{q}[/tex]
[tex]E=\dfrac{a+g}{\dfrac{q}{m}}[/tex]
Put the value into the formula
[tex]E=\dfrac{76.19+9.8}{0.100}[/tex]
[tex]E=8.6\times10^{2}\ N/C[/tex]
The direction is upward.
Hence, The magnitude of the electric field is [tex]8.6\times10^{2}\ N/C[/tex]
Answer:
E = 8.6 x 10² N/C
Explanation:
given,
initial speed of charge,u = 0 m/s
final speed of charge,v = 160 m/s
time,t = 2.1 s
charge-to-mass ratio = 0.100 C/kg
Electric field of the region = ?
Acceleration of the charge
[tex]a = \dfrac{v-u}{t}[/tex]
[tex]a = \dfrac{160 - 0}{2.1}[/tex]
a = 76.19 m/s²
specific charge = [tex]\dfrac{q}{m} = 0.1[/tex]
now,
Electric field,
[tex]E = \dfrac{F}{q}[/tex]
charge is moving upwards so,
[tex]E = \dfrac{(a + g)}{\dfrac{q}{m}}[/tex]
[tex]E = \dfrac{(76.19+9.8)}{0.1}[/tex]
E = 860 N/C
electric field , E = 8.6 x 10² N/C
hence, the magnitude of electric field is equal to E = 8.6 x 10² N/C
Consider a steam power plant that operates on the ideal reheat Rankine cycle. The plant maintains the boiler at 17.5 MPa, the reheater at 2 MPa, and the condenser at 50 kPa. The temperature is 550°C at the entrance of the high-pressure turbine, and 300°C at the entrance of the low-pressure turbine. Determine the thermal efficiency of this system.
The ideal thermal efficiency for the steam power plant operating on the reheat Rankine cycle is approximately 57.0%.
To determine the thermal efficiency of the steam power plant operating on the ideal reheat Rankine cycle, we will follow the steps in the Problem-Solving Strategies for Thermodynamics:
Identify the Temperatures and Pressures
Boiler Pressure (P1): 17.5 MPa (High-pressure turbine inlet)
Reheater Pressure (P2): 2 MPa (Low-pressure turbine inlet)
Condenser Pressure (P3): 50 kPa
High-Pressure Turbine Inlet Temperature (T1): 550°C = 823 K
Low-Pressure Turbine Inlet Temperature (T2): 300°C = 573 K
Calculate Maximum Efficiency (Carnot Efficiency)
The maximum theoretical efficiency (Carnot efficiency) for a heat engine operating between two temperatures is given by the formula:
[tex]\eta_{max} = 1 - \frac{T_c}{T_h}[/tex]
where:
[tex]T_h[/tex] = Highest temperature (at the boiler, T1) = 823 K
[tex]T_c[/tex] = Lowest temperature (at the condenser) = Convert 50 kPa to temperature
Using steam tables or Mollier diagrams, the saturation temperature corresponding to 50 kPa is approximately 81°C = 354 K.
Now substituting:
[tex]\eta_{max} = 1 - \frac{354}{823} \approx 0.570 \text{ or } 57.0\%[/tex]
Based on the scaling laws, how many times greater is the strength-to-weight ratio of a nanotube (D = 10 nm) than the leg of a flea (D = 100 μm)? Than the leg of an elephant (D = 2m)?
To solve this problem we will apply the laws of proportion between the resistance / weight ratio that may occur depending on the cases mentioned. The strength-to-weight ratio for each object is,
For the nano tube,
[tex]\frac{S}{W} = \frac{1}{10*10^{-9}} =1*10^{8}[/tex]
For the leg of a flea,
[tex]\frac{S}{W}= \frac{1}{100*10^{-6}}=10000[/tex]
For the elephant,
[tex]\frac{S}{W} = \frac{1}{2}= 0.5[/tex]
From this we can conclude that the resistance / weight ratio of the nano tube is 10 thousand times better than the flea's leg and [tex]10 ^ 8[/tex] times better than the elephant's leg.
A neutron star is the remnant left after certain supernovae (explosions of giant stars). Typically, neutron stars are about 17 km in diameter and have around the same mass as our sun. What is a typical neutron star density in g/cm3
To solve this concept we will apply the mathematical equations concerning the calculation of Volume in a sphere and the relation of density as a function of mass and volume, that is
The volume of the neutron star is
[tex]V = \frac{4\pi }{3}R^3[/tex]
[tex]V = \frac{4 \pi}{3} (\frac{17*10^{5}cm}{2})^3[/tex]
[tex]V = 25.72^{17}cm^3[/tex]
Now the density of the neutron star is
[tex]\rho = \frac{M}{V}[/tex]
[tex]\rho = \frac{1.989*10^{30}kg(\frac{10^3g}{1kg})}{25.72*10^{17}cm^3}[/tex]
[tex]\rho = 7.733*10^{14}g/cm^3[/tex]
Therefore the density of the neutron star is [tex]\rho = 7.733*10^{14}g/cm^3[/tex]
The Sun orbits the Milky Way galaxy once each 2.60 x 10^8 y, with a roughly circular orbit averaging 3.00 x 10^4 light years in radius. (A light year is the distance traveled by light in 1 y.) Calculate the centripetal acceleration of the Sun in its galactic orbit..
Final answer:
To calculate the centripetal acceleration and average speed of the Sun in its galactic orbit, we utilize relevant formulas and astronomical data. The concept of centripetal acceleration is fundamental in understanding circular motion in celestial bodies like the Sun as it orbits the Milky Way galaxy.
Explanation:
Centripetal acceleration is the acceleration directed toward the center of a circular path. To calculate the centripetal acceleration of the Sun in its galactic orbit, we use the formula: a = v^2 / r, where v is the speed of the Sun and r is the radius of its orbit. Given that the Sun's orbit radius is 3.00 x 10^4 light years and it takes 2.60 x 10^8 years to orbit the Milky Way galaxy, we can also calculate the average speed of the Sun in its galactic orbit. To determine if a nearly inertial frame of reference can be located at the Sun, we need to consider the motion relative to the galactic center.
Suppose a balloons was laying by the window at night. The next day, when the sun came up, it warmed the gas (air) that was in the balloon. What would be true about the density of the air in the balloon?
As the gas (air) in the balloon warms, it expands due to thermal expansion, causing its density to decrease. This lower density compared to the surrounding air leads to buoyancy, making the balloon rise.
Explanation:When the sun warms the gas (air) inside a balloon, the air expands due to an increase in temperature. This process, known as thermal expansion, causes the molecules in the air to move faster and spread out more, occupying a larger volume. As a result, the density of the air inside the balloon decreases because density is defined as mass per unit volume, and while the mass of the air remains constant, its volume increases. In the context of a hot air balloon, or any closed system where air is heated, this decrease in density compared to the cooler surrounding air leads to buoyancy. Buoyancy is the force that makes things float, which in the case of the balloon, causes it to rise since the hot air inside it is less dense than the cooler external environment.
A small drop of water is suspended motionless in air by a uniform electric field that is directed upward and has a magnitude of 11000 N/C. The mass of the water drop is 3.37 × 10-9 kg. How many excess electrons or protons reside on the drop?
To solve this problem we will apply the concepts related to the electric field such as the smelting of the Force and the load (In this case the force is equivalent to the weight). Later we will apply the ratio of the total charge as a function of the multiplication of the number of electrons and their individual charge.
[tex]E = \frac{mg}{q}[/tex]
Here,
m = mass
g = Acceleration due to gravity
Rearranging to find the charge,
[tex]q = \frac{mg}{E}[/tex]
Replacing,
[tex]q = \frac{(3.37*10^{-9})(9.8)}{11000}[/tex]
[tex]q = 3.002*10^{-12}C[/tex]
Since the field is acting upwards the charge on the drop should be negative to balance it in air. The equation to find the number of electrons then is
[tex]q = ne[/tex]
Here,
n = Number of electrons
e = Charge of each electron
[tex]n = \frac{q}{e}[/tex]
Replacing,
[tex]n = \frac{3.002*10^{-12}}{1.6*10^{-19}}[/tex]
[tex]n = 2.44*10^7[/tex]
Therefore the number of electrons that reside on the drop is [tex]2.44*10^7[/tex]
To determine the number of excess electrons or protons residing on the water drop, we can use the principle of electrostatics. When an electric field is applied to a charged object, the electrostatic force acting on it can be calculated using the equation:
[tex]F = qE[/tex]
Where:
F is the electrostatic force
q is the charge on the object
E is the electric field strength
In this case, the electrostatic force acting on the water drop is balanced by the gravitational force, so we have:
F = mg
Where:
m is the mass of the water drop
g is the acceleration due to gravity
We can equate these two forces and solve for the charge q:
qE = mg
From this equation, we can isolate the charge q:
q = mg / E
Now we can calculate the charge on the water drop:
m = 3.37 × 10^(-9) kg
g = 9.8 m/s^2
E = 11000 N/C
Substituting the values into the equation:
q = (3.37 × 10^(-9) kg * 9.8 m/s^2) / 11000 N/C
Calculating this expression:
q = 3.037 × 10^(-15) C
The elementary charge of an electron or proton is approximately 1.602 × 10^(-19) C. To find the number of excess electrons or protons, we can divide the calculated charge by the elementary charge:
Number of excess electrons or protons = q / elementary charge
Number of excess electrons or protons = (3.037 × 10^(-15) C) / (1.602 × 10^(-19) C)
Calculating this expression:
Number of excess electrons or protons ≈ 1.895 × 10^(4)
Therefore, the water drop has approximately 18,950 excess electrons or protons.
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If the magnitude of the resultant force acting on the eyebolt is 570 N and its direction measured clockwise from the positive x axis is θ = 33 ∘, determine the magnitude of
Answer:
F1 = 1210.65 N
Q = 65.7081 degrees
Explanation:
Sum of Forces in x - direction:
F1 * cos (Q) + F2*sin(30) - F3*(3/5) = Fres*cos (theta)
F1 * cos (Q) + (500)*(0.5) - 450*(3/5) = 570*cos(33)
F1*cos (Q) = 498.0422237 N .... Eq1
Sum of Forces in y - direction:
F1 * sin (Q) - F2*sin (60) - F3 * (4/5) = Fres*sin (theta)
F1 * sin (Q) - 500*sin(60) - 450 * (4/5) = 570*sin (33)
F1 * sin (Q) = 1103.45692 N .... Eq 2
Divide Eq 2 by Eq 1
tan (Q) = 2.21558916
Q = arctan (2.21558916) = 65.7081 degrees
F1 = 1210.65 N
If the neutral atom of an element has only 5 valence electrons it must be in which group? 1. VIIA 2. VA 3. IVB 4. VIA 5. IVA 6. IIIA
Answer:
2. VA
Explanation:
The valency electron or outer electron of a neutral atom of an element determines the group at which an element belong. The electronic configuration of an atom that have valency of 5 can be represented as 2 5 , 2 8 5 etc.
The electronic configuration 2 5 represent Nitrogen atom while 2 8 5 represent phosphorus atom. The valency 5 depicts the element belongs to group 5A(VA).
This means atoms of valency electron of 3 belong to group IIIA, valency electrons of 4 belongs to group IVA, valency electron of 6 belong to group VIA and valency electrons of 7 belong to group VIIA.
A 1.50 kg rock whose density is 4700 kg/m3 is suspended by a string such that half of the rock's volume is under water.
What is the tension in the string?(In N)
Answer:
Tension T = 13.14N
Explanation:
Given:
Mass of rock m = 1.50kg
Density of rock p = 4700kg/m^3
Volume of rock V = mass/density = m/p
V = 1.50kg/4700kg/m3 = 3.19×10^-4m3
Taking the summation of forces acting on the rock;
T-W+Fb = 0
T = W - Fb .....1
T = tension
W = weight of rock
Fb = buoyant force
Fb = pw(0.5V)g = density of water × Volume under water×™ acceleration due to gravity
g = 9.8m/s^2
T = mg - pw(0.5V)g
T = 1.50×9.8 - 1000kg/m^3 ×0.5(3.19 × 10^-4) × 9.8
T = 13.14N
Final answer:
To find the tension in the string, we can use the concept of buoyancy. By equating the buoyant force and weight of the rock, we can solve for the tension. The tension will be equal to the weight of the rock minus the buoyant force.
Explanation:
In order to find the tension in the string, we can use the concept of buoyancy. Since half of the rock's volume is under water, it experiences an upward buoyant force equal to the weight of the water displaced by that volume. The buoyant force can be found using the following equation:
Buoyant force = density of water * volume of water displaced * acceleration due to gravity
The weight of the rock is equal to its mass multiplied by the acceleration due to gravity.
By equating the buoyant force and weight of the rock, we can solve for the tension in the string. The tension will be equal to the weight of the rock minus the buoyant force.
Assume this field is generated by a point charge of Q = 5 × 10-9 C. How far away is this charge located? Give your answer in meters.
Final answer:
The distance from the charge is 30 meters.
Explanation:
The distance from the point charge can be calculated using the equation E = kQ/r^2, where E is the electric field strength, k is the electrostatic constant (9 x 10^9 Nm^2/C^2), Q is the charge, and r is the distance. In this case, the electric field strength is the same at any point 5.00 mm away from the charge. Therefore, we can set up the equation as follows:
E = kQ/r^2
5.00 x 10^-3 m = (9 x 10^9 Nm^2/C^2)(5 x 10^-9 C)/r^2
By rearranging the equation and solving for r, we can find the distance:
r^2 = (9 x 10^9 Nm^2/C^2)(5 x 10^-9 C)/(5.00 x 10^-3 m)
r^2 = 900 m^2
r = sqrt(900 m^2) = 30 m
Therefore, the charge is located 30 meters away.
A circular test track for cars has a circumference of 3.5 kmkm . A car travels around the track from the southernmost point to the northernmost point.
-What distance does the car travel? (In km)
-What is the car's displacement from its original position? (In km)
Answer:
(A) Distance will be equal to 1.75 km
(B) Displacement will be equal to 1.114 km
Explanation:
We have given circumference of the circular track = 3.5 km
Circumference is given by [tex]2\pi r=3.5[/tex]
r = 0.557 km
(a) It is given that car travels from southernmost point to the northernmost point.
For this car have to travel the distance equal to semi perimeter of the circular track
So distance will be equal to [tex]=\frac{3.5}{2}=1.75km[/tex]
(b) If car go along the diameter of the circular track then it will also go from southernmost point to the northernmost point. and it will be equal to diameter of the track
So displacement will be equal to d = 2×0.557 = 1.114 m
a cat is on a merry-go-round that completes 1 full rotation in 6 seconds. The cat sits at a radius of 8.4 metres from the centre.
Find the minimum coefficient of friction to prevent the cat from sliding off.
To solve this problem we will use the concepts related to the uniform circular movement from where we will obtain the speed of the object. From there we will go to the equilibrium equations so that the friction force must be equal to the centripetal force. We will clear the value of the coefficient of friction sought.
The velocity from the uniform circular motion can be described as
[tex]v = \frac{2 \pi r}{T}[/tex]
Here,
r = Radius
T = Period
Replacing,
[tex]v = \frac{2\pi (8.4)}{6}[/tex]
[tex]v =8.7964 m/s[/tex]
From equilibrium to stay in the circle the friction force must be equivalent to the centripetal force, therefore
[tex]F_f = F_c[/tex]
[tex]\mu N = \frac{mv^2}{r}[/tex]
Here,
[tex]\mu =[/tex] Coefficient of friction
N = Normal Force
m = mass
v = Velocity
r = Radius
The value of the Normal force is equal to the Weight, then
[tex]\mu(mg) = \frac{mv^2}{r}[/tex]
Rearranging to find the coefficient of friction
[tex]\mu = \frac{v^2}{gr}[/tex]
Replacing,
[tex]\mu = \frac{(8.7964)^2}{(9.8)(8.4)}[/tex]
[tex]\mu =0.9399[/tex]
Therefore the minimum coefficient of friction to prevent the cat from sliding off is 0.9399
Assume that you have a rectangular tank with its top at ground level. The length and width of the top are 14 feet and 7 feet, respectively, and the bottom of the tank is 15 feet beneath ground level. How much work W does it take to empty the tank by pumping the liquid back to ground level once the tank is full?
To solve this problem we will use the work theorem, for which we have that the Force applied on the object multiplied by the distance traveled by it, is equivalent to the total work. From the measurements obtained we have that the width and the top are 14ft and 7ft respectively. In turn, the bottom of the tank is 15ft. Although the weight of the liquid is not given we will assume this value of [tex] 62 lb / ft ^ 3 [/tex] (Whose variable will remain modifiable until the end of the equations subsequently presented to facilitate the change of this, in case be different). Now the general expression for the integral of work would be given as
[tex]W = \gamma A * \int_0^15 dy[/tex]
Basically under this expression we are making it difficult for the weight of the liquid multiplied by the area (Top and widht) under the integral of the liquid path to be equivalent to the total work done, then replacing
[tex]W = (62)(14*7)\int^{15}_0 (15-y)dy[/tex]
[tex]W = (14*7*62)\big [15y-\frac{y^2}{2}\big ]^{15}_0[/tex]
[tex]W = (14*7*62)[15(15)-\frac{(15)^2}{2}][/tex]
[tex]W = 683550ft-lbs[/tex]
Therefore the total work in the system is [tex]683550ft-lbs[/tex]
What is the electric potential energy of a system that consists of two protons 2.1×10−15 m apart-as might occur inside a typical nucleus? Express your answer using two significant figures.
The electric potential energy of a system comprising two protons 2.1 x 10^-15 m apart can be calculated using Coulomb's law, which gives us a result of approximately 0.68 MeV.
Explanation:The electric potential energy of a system depends upon the charge of the components and the distance between them. In this case, we can calculate the electric potential energy using Coulomb's law which states that the electric potential energy 'V' between two charges is given by the equation V = k*q1*q2/r where 'k' is Coulomb's constant (8.99 × 10^9 N m^2/C^2), 'q1' and 'q2' are the two charges, and 'r' is the distance between the charges. Given that the charges are two protons, they both have the same charge (1.6 × 10^-19 Coulombs). The distance 'r' is given as 2.1 × 10^-15 m. Substituting these values in, we get: V = (8.99 × 10^9 N m^2/C^2)* (1.6 × 10^-19 C) * (1.6 × 10^-19 C) / (2.1 × 10^-15 m) which results in an electric potential energy of approximately 0.68 MeV (mega electron volts).
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A 372-g mass is attached to a spring and undergoes simple harmonic motion. Its maximum acceleration is 17.6 m/s2 , and its maximum speed is 1.75 m/s. a)Determine the angular frequency. b)Determine the amplitude. c)Determine the spring constant.
Answer with Explanation:
We are given that
Mass , m=372 g=[tex]\frac{372}{1000}=0.372 Kg[/tex]
1 kg=1000g
Maximum acceleration, a=[tex]17.6 m/s^2[/tex]
Maximum speed ,v=1.75 m/s
a.We know that
Maximum acceleration, a=[tex]A\omega^2[/tex]
Maximum speed, v=[tex]\omega A[/tex]
[tex]17.6=A\omega^2[/tex]
[tex]1.75=A\omega[/tex]
[tex]\frac{17.6}{1.75}=\frac{A\omega^2}{A\omega}=\omega[/tex]
Angular frequency,[tex]\omega=10.06 rad/s[/tex]
b.Substitute the value of angular frequency
[tex]1.75=A(10.06)[/tex]
[tex]A=\frac{1.75}{10.06}=0.17 m[/tex]
Hence, the amplitude=0.17 m
c.Spring constant,k=[tex]m\omega^2[/tex]
Using the formula
[tex]k=0.372\times (10.06)^2[/tex]
Hence, the spring constant,k=37.6 N/m
Final answer:
The angular frequency is approximately 7.686 rad/s, the amplitude is approximately 0.227 m, and the spring constant is approximately 6.5472 N/m.
Explanation:
To determine the angular frequency, we can use the formula:
ω = √(k/m)
where ω is the angular frequency in radians per second, k is the spring constant in Newtons per meter, and m is the mass in kilograms.
Given that the maximum acceleration is 17.6 m/s^2 and the mass is 372 g (or 0.372 kg), we can calculate the spring constant:
k = m * a
k = 0.372 kg * 17.6 m/s^2 = 6.5472 N/m
Now we can find the angular frequency:
ω = √(6.5472 N/m / 0.372 kg) ≈ 7.686 rad/s
To determine the amplitude, we can use the formula:
A = vmax / ω
where A is the amplitude, vmax is the maximum speed, and ω is the angular frequency.
Given that the maximum speed is 1.75 m/s and the angular frequency is 7.686 rad/s, we can calculate the amplitude:
A = 1.75 m/s / 7.686 rad/s ≈ 0.227 m
Therefore, the angular frequency is approximately 7.686 rad/s, the amplitude is approximately 0.227 m, and the spring constant is approximately 6.5472 N/m.
If we know the moon's position in the sky and its phase, we can estimate the ____________. In general, knowing any two of the following three things allows us to estimate the third:
Answer:
we can estimate the _time__.
The three things are;
1. Moon's position in the sky
2. The moon phase
3. The time
Explanation:
In general, knowing any two of the following three things; (the moon's position in the sky, the moon phase, and the time) will allows us to estimate the third. Yes, this statement is true because, position and phase of moon is used to determine the hour of the day and night most especially in the morning. Example of this is the determination of prayer time by Muslims community as it was the major time determinant in the past before the advent of clock or watch.
Also, if will know the time and moon position, we can determine the phase of the moon likewise using time and moon phase to know the moon position.
The different phases of the moon cause changes in the size of the moon. If we know the moon's position in the sky and its phase. we can easily estimate the time.
What is the moon phase?The moon changes shape every day. This is due to the fact that the celestial body has no light of its own and can only reflect sunlight.
Only the side of the moon facing the sun can reflect this light and seem bright. The opposite side appears black. this is a full moon.
We can only see the black section when it lies between the sun and the earth when a new moon occurs. We witness intermediate phases like a half-moon and crescent in between these two extremes.
In general, we may estimate the third by knowing any two of the following three things
(1) the moon's position in the sky
(2) the moon phase
(3) the time
Yes, this statement is correct since the moon's location and phase are utilized to define the time of day and night, particularly in the morning.
Also, if we know the time and moon position, we can figure out the moon phase by utilizing the time and moon phase to figure out the moon position.
Hence If we know the moon's position in the sky and its phase. we can easily estimate the time.
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Which of the following is a small electronic component made up of transistors (tiny switches) and other miniaturized parts?
a.
Peripheral
b.
Integrated circuit (IC)
c.
Tablet
d.
Mouse
e.
Vacuum tube
Answer:
Integrated circuit (IC)
Explanation:
An integrated circuit ( IC ) is a semiconductor which contains multiple electronic components
interconnected to form a complete electronic function. Integrated circuits are the most essential part of all electronic products.
Modern integrated circuits contain as much as billions of circuit components such as transistors , diodes , resistors , and capacitors
onto a single monolithic die .
The correct answer is option b. An Integrated Circuit (IC), or microchip, is a small electronic component consisting of transistors and miniaturized parts etched onto a piece of silicon, pivotal in the development and miniaturization of electronic devices
The small electronic component made up of transistors (tiny switches) and other miniaturized parts is called an Integrated Circuit (IC). An Integrated Circuit, sometimes referred to as a microchip, is an electronic circuit of transistors etched onto a small piece of silicon. This technology allows for complex circuitry to be compacted into a tiny chip, which is crucial for the functionality of modern electronic devices like computers and cell phones. The invention of the IC was pivotal in launching the modern computer revolution because it significantly reduced the size and complexity of electronic devices, replacing bulky vacuum tubes and complicated wiring with a compact, efficient solution.
Integrated Circuits are designed to handle both analog and digital signals, but in the realm of digital electronics, they are essential for managing binary code, the series of ones and zeroes that computers use to process data. This is achieved through the hundreds, thousands, or even millions of transistors that can act as on-off switches within a single IC. The miniaturization and efficiency of ICs have been fundamental in advancing the technological capabilities of electronic devices, making them smaller, faster, and more accessible to the general public.
When the ambulance passes the person (switching from moving towards him to moving away from him), the perceived frequency of the sound he receives decreases by 9.27 percent. How fast is the ambulance driving?
To develop this problem we will apply the considerations made through the concept of Doppler effect. The Doppler effect is the change in the perceived frequency of any wave movement when the emitter, or focus of waves, and the receiver, or observer, move relative to each other. At first the source is moving towards the observer. Than the perceived frequency at first
[tex]F_1 = F \frac{{343}}{(343-V)}[/tex]
Where F is the actual frequency and v is the velocity of the ambulance
Now the source is moving away from the observer.
[tex]F_2 = F\frac{343}{(343+V)}[/tex]
We are also so told the perceived frequency decreases by 11.9%
[tex]F_2 = F_1 - 9.27\% \text{ of } F_1[/tex]
[tex]F_2 = F_1-0.0927F_1[/tex]
[tex]F_2 = 0.9073F_1[/tex]
Equating,
[tex]F\frac{343}{(343+V)}= 0.9073(F\frac{343}{(343-V)})[/tex]
[tex]\frac{1}{(343+V)}= 0.9073\frac{1}{(343-V)}[/tex]
[tex]0.9073(343+V) = 343-V[/tex]
[tex](0.9073)(343)+(0.9073)V = 343-V[/tex]
[tex]V+0.9073V = 343-(0.9073)(343)[/tex]
Solving for V,
[tex]V = 16.67 m/s[/tex]
A 1100 kg car rounds a curve of radius 68 m banked at an angle of 16 degrees. If the car is traveling at 95 km/h, will a friction force be required? If so, how much and in what direction?
Final answer:
Determining if a friction force is required for a car rounding a banked curve depends on the car's speed relative to the curve's ideal speed. At 95 km/h, a frictional force may be needed if this speed is not the ideal speed for the 68 m radius curve banked at 16 degrees.
Explanation:
When a 1100 kg car rounds a curve of radius 68 m banked at an angle of 16 degrees, we need to determine if a friction force is required when the car is traveling at 95 km/h. If the car is traveling at the correct banked curve speed, it could complete the turn without any frictional force. However, if the car travels at a speed higher or lower than this optimal speed, a frictional force will be necessary either to prevent the car from slipping outward or to prevent it from falling inward towards the center of the curve.
To find out whether a friction force is needed, we first need to calculate the ideal speed for this banked turn. This involves calculating the speed at which the components of the normal force provide enough centripetal force for the turn. The ideal speed is reached when no friction force is needed to keep the car on the path, meaning the force of gravity, the normal force, and the centripetal force are in perfect balance.
However, if the car is indeed traveling at 95 km/h, faster or slower than this ideal speed, then either a static frictional force acting upwards along the bank or a static frictional force opposite to the car's direction would be required to maintain its circular path without slipping.
A boat is traveling at 3.2 m/s in the same direction as an ocean wave of wavelength 30 m and speed 6.8 m/s. If the boat is on the crest of a wave, how much time will elapse until the boat is next on a crest?
To solve this problem we apply the kinematic equations of linear motion. For which the speed is described as the distance traveled in a time interval. This would be,
[tex]v = \frac{x}{t} \rightarrow t = \frac{x}{v}[/tex]
Our values are given as,
[tex]\text{The speed of the boat} = v_b = 3.8m/s[/tex]
[tex]\text{The speed of the ocean} = V = 6.8m/s[/tex]
[tex]\text{The wave length of the wave is the same distance traveled by boat} = d = \lambda = 30m[/tex]
[tex]\text{The relative speed of the boat} = v_r = -3.8 +6.8 = 3m/s[/tex]
Replacing we have,
[tex]t = \frac{d}{v_r}[/tex]
[tex]t = \frac{30m}{3m/s}[/tex]
[tex]t = 10s[/tex]
Therefore will take until the boat is next on a crest around to 10s
You are driving down the highway at 65 m p h, which is 29 m/s. Your tires have a radius of 0.30 m. a. How many times per second does each tire rotate? b. What is the speed of a point at the top of a tire, relative to the ground?
To solve this problem we will apply the concepts related to linear velocity and angular velocity to perform the respective conversion with the given values. To find the velocity in the upper part of the tire we will use the mathematical relation that expresses that it is twice the linear velocity. Let's start
PART A)
[tex]\omega = \frac{v}{r}[/tex]
[tex]\omega = \frac{29}{0.3}[/tex]
[tex]\omega = 96.66 rad/s[/tex]
Now we now that [tex]2\pi rad = 1 rev[/tex], then
[tex]\omega = 96.66rad/s (\frac{1 rev}{2\pi rad})[/tex]
[tex]\omega = 15.38rev/s[/tex]
PART B)
[tex]v = 2v_0[/tex]
[tex]v = 2(29)[/tex]
[tex]v = 58m/s[/tex]
(II) At t =0, an 885-9 mass at rest on the end of a horizontal Spring (K: 184 N/m) is struck by a hammer which gives it aninitial speed of 226 m/s. Determine (a) the period and frequency ofthe motion, (b) the amplitude, (c) the maximumacceleration, (d) the total energy, and (e) the kinetic energy when x =O.4O A where A is the amplitude.
Answer:
A. T = 0.4358s and f = 2.29hz
B. A = 15.67m
C. amax = 3258.71m/s
D. amax = 22601J
E. Ek = 3616.16J
Explanation:
A. The period of the motion, T = 2pi*(sqrt(m/k))
Where m is the mass of the body in motion = 885g = 0.885kg
k = the spring constant = 184N/m2
T = 2pi*(sqrt(0.885/184))
= 0.4358s
Frequency of the motion, f = 1/T
T = 0.4358s
f = 2.2949hz
B. Maximum speed, Vmax = A*(sqrt(k/m))
Where A = amplitude of the motion
Making amplitude subject of formula,
A = Vmax(sqrt(m/k))
= 226*(sqrt(0.885/184))
= 15.6739m
C. Maximum acceleration, amax = A*(k/m)
= 15.6739*(184/0.885)
= 3258.71m/s
D. Total energy, Etotal = 1/2*(m * Vmax)2
= 1/2 * 0.885 * (226)2
= 22601J
E. Kinetic energy, Ek = Etotal - mechanical energy
Ek = 1/2*(k*A2) - 1/2*(k*x2)
Where x = 0.40A
Ek = 1/2*((k*A2) - (k*0.40A)2)
= 1/2*k*A2*(1 - 0.16)
= 1/2*k*A2*0.16
But 1/2*k*A2 = 22601J
Therefore, Ek = 22601*0.16
= 3616.16J