A spring of constant k hangs from a ceiling with a mass m attached at the other end. The natural length of the hanging spring without the mass is L. a) Draw this scenario, including forces and known or calculable displacements. b) What is the equilibrium length of the spring with the mass attached? c) Suppose the spring is pulled down to create a displacement y from this equilibrium position and then released. Write an expression y (t) describing its position as a function of time t. Let "up" be the positive vertical direction. d) What is the period of oscillation for this simple harmonic oscillator? e) Suppose you cut the spring in half. What would be the spring constant of each half-spring? (The two are identical, so they will have the same spring constant.) To determine this, redraw your diagram as two springs of equal length connected "in series," i.e., top to bottom. Then indicate the forces acting on each of the three interesting bodies: the mass m and each halfspring of constant k’. Apply Hooke’s Law to each half-spring individually and then to the two springs together. f) Take the two half-springs from part e) and connect them "in parallel" (that is, side by side, hanging from the ceiling and each connected to the mass m). Draw the new diagram and repeat parts a) - d)

Answer:

C) It increases to 2.0 cm

Explanation:

In a double-slit diffraction experiment, the distance on the screen between two adjacent maxima is given by

[tex]\Delta y = \frac{\lambda D}{d}[/tex]

where

[tex]\lambda[/tex] is the wavelength of the wave

D is the distance of the screen from the slits

d is the separation between the slits

In this problem, the initial distance between adjacent maxima is 1.0 cm. Later, the slit separation is cut in a half, which means that the new slit separation is

[tex]d'=\frac{d}{2}[/tex]

Substituting into the equation, we find that the new separation between the maxima is

[tex]\Delta y' = \frac{\lambda D}{d/2}=2(\frac{\lambda D}{d})=2\Delta y[/tex]

So, the distance increases by a factor 2: therefore, the new separation between the maxima will be 2.0 cm.

In a double-slit experiment, when the slit separation is halved, the distance between adjacent maxima will double. Therefore, the correct answer is C) It increases to 2.0 cm.

In a double-slit experiment, the distance between adjacent maxima on a remote screen is determined by the separation of the slits and the wavelength of the light used. According to the formula for double-slit interference d sin θ = mλ (d - distance between slits, m - order of maxima, λ - wavelength of the light), when the slit separation (d) is cut in half, the distance between adjacent maxima will increase. Therefore, between the given options, C) It increases to 2.0 cm is the correct answer. This is because the maxima result from constructive interference which occurs when the path difference is an integral multiple of the wavelength.

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The answer is:

The first option, the force tending to lift Rover is equal to 14.5 N.

To calculate the force that is tending to lift Rover vertically, we need to calculate the vertical component force.

Since we know that the angle between the force and the ground is 29°, we can calculate the vertical component of the force using the following formula:

[tex]F_y=Force*Sin(29\°)[/tex]

We are given that the force is equal to 30.0 N, so, calculating we have:

[tex]F_y=Force*Sin(29\°)[/tex]

[tex]F_y=30N*Sin(29\°)=14.5N[/tex]

Also, we can calculate the horizontal component of the force using the following formula:

[tex]F_x=Force*Cos(29\°)[/tex]

[tex]F_x=30N*Cos(29\°)=26.24N[/tex]

Hence, we have that the correct option is the first option, the force tending to lift Rover is equal to 14.5 N.

Have a nice day!

Answer:

[tex]2.25\mu C[/tex]

Explanation:

At the beginning, we have:

V = 4.0 V potential difference across the capacitor

[tex]Q=9.0 \mu C=9.0\cdot 10^{-6}C[/tex] charge stored on the capacitor

Therefore, we can calculate the capacitance of the capacitor:

[tex]C=\frac{Q}{V}=\frac{9.0 \cdot 10^{-6} C}{4.0 V}=2.25\cdot 10^{-6} F[/tex]

Later, the battery is replaced with another battery whose voltage is

V = 5.0 V

Since the capacitance of the capacitor does not change, we can calculate the new charge stored:

[tex]Q=CV=(2.25\cdot 10^{-6} F)(5.0 V)=11.25 \cdot 10^{-6} C=11.25 \mu C[/tex]

Since the capacitor has been connected exactly as before, we have that the charge on the positive plate has increased from [tex]9.0 \mu C[/tex] to [tex]11.25 \mu C[/tex]. Therefore, the additional charge that moved to the positive plate is

[tex]\Delta Q = 11.25 \mu C-9.0 \mu C=2.25 \mu C[/tex]

In a uncharged capacitor when connected to 5.0 V battery, the additional charge flows to the positive plate,  is [tex]2.25\rm \mu C[/tex] .

The capacitance of capacitor is the ratio of the electric charge stored inside the capacitance to the potential difference. The capacitance of capacitor can be given as,

[tex]C=\dfrac{Q}{V}[/tex]

Here, (Q) is the electric charge and (V) is the potential difference.

Given information-

The potential difference of the first battery is 4.0 V.

The charge of the first battery is 9.0 μC.

The potential difference of the second battery is 5.0 V.

The capacitance of the first battery is,

[tex]C=\dfrac{9.0\times10^{-6}}{4}\\C=2.25\times10^{-6} \rm F[/tex]

Let the charge of the second battery is (q). Thus The capacitance of the first battery is,

[tex]C=\dfrac{q}{5}\\[/tex]

As the capacitance of the capacitor remain same. Thus put the value of C in the above equation as,

[tex]2.25\times10^{-6}=\dfrac{q}{5}\\q=11.25\rm \mu C[/tex]

The additional charge flows to the positive plate is the difference of the charge flows to the positive plate and second battery to the first battery. Thus,

[tex]\Delta q=11.25-9\\\Delta q=2.25\rm \mu C[/tex]

Thus the additional charge flows to the positive plate is [tex]2.25\rm \mu C[/tex] .

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Answer: Last option

They all have the same current.

Explanation:

A connection of three elements in series is represented as follows:

--------[4Ω]-------[8Ω]--------[12Ω]----------

→ I

Note that the three elements share the same current line I .

By definition when the resistors or other electrical components are connected in series then the same current passes through them. Therefore in this case the magnitude of the resistance does not influence the magnitude of the current.

The answer is the last option

Answer:

C.) Fleming shows that the Penicillium mood could kill bacteria and observed that it was very effective when taken orally or intravenously

Out of the given options, “Fleming shows that the Penicillium mold could kill bacteria and observed that it was very effective when taken orally or intravenously” is most accurately describes the development of the antibiotic penicillin

Answer: Option C

Explanation:

It was in 1928, when Alexander Fleming discovered the most effective antibacterial drug to knock down severe bacteria and hence, saved lives. It all began with the development of mold on a staphylococcus culture plate while Fleming was experimenting with the influenza virus.

Penicillin has helped numerous people to get aided from severe illness in the 20th century and especially during the World War II. The use of penicillin has helped in treating bacterial endocarditis, pneumococcal pneumonia, bacterial meningitis etc.

Answer:

5.4 cm³

Explanation:

Ideal gas law:

PV = nRT

where P is absolute pressure, V is volume, n is number of moles, R is ideal gas constant, and T is absolute temperature.

Since n and R are constant, we can say:

PV / T = constant

At the bottom of the lake, the pressure is:

P = ρgh + Patm

P = (1000 kg/m³) (9.8 m/s²) (40.5 m) + 101,325 Pa

P = 498,225 Pa

And the temperature is:

T = 2.3 + 273.15 K

T = 275.45 K

At the top of the lake, the pressure is:

P = Patm

P = 101,325 Pa

And the temperature is:

T = 28.1 + 273.15 K

T = 301.25 K

Therefore:

PV / T = PV / T

(498225 Pa) (1.00 cm³) / (275.45 K) = (101325 Pa) V / (301.25 K)

V = 5.4 cm³

Final answer:

The radius of the air bubble just before it reaches the surface is calculated using Charles's Law and the volume of a sphere. It is approximately 6.20 mm once we consider the change in temperature from the bottom to the top of the lake.

Explanation:

To determine the radius of the air bubble just before it reaches the surface of the lake, we must consider the effects of pressure and temperature changes on the volume of the bubble. Due to the nature of the question involving thermal expansion and the principles of gas laws, Physics is the subject, and the content is suitable for High School level.

First, we acknowledge some basic relations for the behavior of gases. Assuming that the amount of gas and the pressure remain constant (since it's mentioned that we can ignore the pressure change), we can use Charles's Law (V1/T1 = V2/T2), where V1 and T1 are the original volume and temperature, and V2 and T2 are the final volume and temperature respectively. Temperatures should be in Kelvin for gas laws, where Kelvin is the Celsius temperature plus 273.15.

V1 = 1.00 cm3 = 1.00 x 10⁻⁶ m³
T1 = 2.3 °C + 273.15 = 275.45 K
T2 = 28.1 °C + 273.15 = 301.25 K

Applying Charles's Law:

V2 = V1 * (T2/T1) = (1.00 x 10⁻⁶ m³) * (301.25 K / 275.45 K)

Now let's calculate V2.

V2 = (1.00 x 10⁻⁶ m³) * (301.25 / 275.45)
V2 ≈ 1.0937 x 10⁻⁶ m3

The volume of a sphere is given by  (4/3)πr³. To find the radius, we solve for r:

V = (4/3)πr³
r = ∛(3V / 4π)

Substituting the value of V2:

r ≈ ∛(3 * 1.0937 x 10⁻⁶ m3) / (4 * π)
r ≈ 6.20 x 10⁻³ m

So, the radius of the bubble just before it reaches the surface is approximately 6.20 mm.

a) [tex]1.55\cdot 10^{-3} T[/tex]

The magnetic force exerted on a charged particle in motion is given by:

[tex]F=qvB sin \theta[/tex]

where

q is the charge of the particle

v is the velocity of the particle

B is the magnetic field strength

[tex]\theta[/tex] is the angle between the direction of v and B

The expression can be rewritten as

[tex]B=\frac{F}{qv sin \theta}[/tex]

We see that the minimum magnetic field needed is the one for which [tex]sin \theta=1[/tex], so with [tex]\theta=90^{\circ}[/tex]. In this problem, we have:

[tex]q=1.6\cdot 10^{-19} C[/tex] (charge of the electron)

[tex]f=5.7 fN=5.7\cdot 10^{-15}N[/tex] is the force

[tex]v=23 Mm/s = 23\cdot 10^6 m/s[/tex] is the electron's velocity

Substituting, we find

[tex]B=\frac{5.7\cdot 10^{-15}N}{(1.6\cdot 10^{-19} C)(23\cdot 10^6 m/s) sin 90^{\circ}}=1.55\cdot 10^{-3} T[/tex]

b) [tex]2.19\cdot 10^{-3} T[/tex]

In this case, the field is at 45 degrees to the electron's velocity, so we have

[tex]\theta=45^{\circ}[/tex]

Therefore, the field strength required to obtain a force of

[tex]f=5.7 fN=5.7\cdot 10^{-15}N[/tex] is the force

will be equal to

[tex]B=\frac{F}{qv sin \theta}=\frac{5.7\cdot 10^{-15}N}{(1.6\cdot 10^{-19} C)(23\cdot 10^6 m/s) sin 45^{\circ}}=2.19\cdot 10^{-3} T[/tex]

The minimum magnetic field needed to exert the given force is [tex]1.55 \times 10^{-3} \ T[/tex]

When the field is 45 degrees to the electron's speed, the magnetic field strength is [tex]2.19\times 10^{-3} \ T[/tex]

The given parameters;

The minimum magnetic field is calculated as follows;

[tex]F = qvB \\\\B = \frac{F}{qv} \\\\B = \frac{5.7\times 10^{-15}}{(1.602\times 10^{-19})(23\times 10^6)} \\\\B = 1.55 \times 10^{-3} \ T[/tex]

When the field is 45 degrees to the electron's speed, the magnetic field strength is calculated as follows;

[tex]B = \frac{F}{q\times v \times sin(45)} = \frac{5.7 \times 10^{-15}}{1.602 \times 10^{-19} \times 23 \times 10^6 \times sin(45)} = 2.19 \times 10^{-3} \ T[/tex]

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Answer:

C) 2.44 × 106 N/C

Explanation:

The electric flux through a circular loop of wire is given by

[tex]\Phi = EA cos \theta[/tex]

where

E is the electric field

A is the cross-sectional area

[tex]\theta[/tex] is the angle between the direction of the electric field and the normal to A

The flux is maximum when [tex]\theta=0^{\circ}[/tex], so we are in this situation and therefore [tex]cos \theta =1[/tex], so we can write

[tex]\Phi = EA[/tex]

Here we have:

[tex]\Phi = 7.50\cdot 10^5 N/m^2 C[/tex] is the flux

d = 0.626 m is the diameter of the coil, so the radius is

r = 0.313 m

and so the area is

[tex]A=\pi r^2 = \pi (0.313 m)^2=0.308 m^2[/tex]

And so, we can find the magnitude of the electric field:

[tex]E=\frac{\Phi}{A}=\frac{7.50\cdot 10^5 Nm^2/C}{0.308 m^2}=2.44\cdot 10^6 N/C[/tex]

Answer:

2.38 m/s, 4.31 m/s, lower

Explanation:

a)

Initial energy = final energy

½ m v₀² + ½ I ω₀² = mgh + ½ m v₁² + ½ I ω₁²

Since the ball is rolling without slipping, ω = v / r.

For a hollow sphere, I = ⅔ m r².

½ m v₀² + ½ (⅔ m r²) (v₀ / r)² = mgh + ½ m v₁² + ½ (⅔ m r²) (v₁ / r)²

½ m v₀² + ⅓ m v₀² = mgh + ½ m v₁² + ⅓ m v₁²

⅚ m v₀² = mgh + ⅚ m v₁²

⅚ v₀² = gh + ⅚ v₁²

v₀² = 1.2gh + v₁²

v₁ = √(v₀² − 1.2gh)

Given v₀ = 4.03 m/s, g = 9.80 m/s, h = 0.900 m:

v₁ = √((4.03)² − 1.2 (9.80) (0.900))

v₁ ≈ 2.38 m/s

At the top of the loop, the sum of the forces in the radial direction is:

∑F = ma

W + N = m v² / R

N = m v² / R - mg

N = m (v² / R - g)

Given v = 2.38 m/s, R = 0.450 m, and g = 9.80 m/s²:

N = m ((2.38)² / 0.450 - 9.80)

N = 2.77m

N ≥ 0, so the ball stays on the track.

b)

Initial energy = final energy

Borrowing from part a):

v₂ = √(v₀² − 1.2gh)

This time, h = -0.200 m:

v₂ = √((4.03)² − 1.2 (9.80) (-0.200))

v₂ ≈ 4.31 m/s

c)

Without the rotational energy:

½ m v₀² = mgh + ½ m v₁²

½ v₀² = gh + ½ v₁²

v₀² = 2gh + v₁²

v₁ = √(v₀² - 2gh)

This is less than v₁ we calculated earlier.

Refer the below Solution for better understanding.

Given :

Speed = 4.03 m/sec

Vertical circular loop of 90 cm diameter.

Solution :

a)

Initial energy = Final Energy

[tex]\rm \dfrac{1}{2}mv_0^2 + \dfrac{1}{2}I\omega_0^2 = mgh + \dfrac{1}{2}mv_1^2 + \dfrac{1}{2}I\omega_1^2[/tex]

here,

[tex]\rm \omega = \dfrac{v}{r}[/tex]

for a hollow sphere,

[tex]\rm I = \dfrac{2}{3}mr^2[/tex]

[tex]\rm \dfrac{1}{2}mv_0^2 + \dfrac{1}{2}\times\dfrac{2}{3}mr^2(\dfrac{v_0}{r})^2 = mgh + \dfrac{1}{2}mv_1^2 + \dfrac{1}{2}\times\dfrac{2}{3}mr^2(\dfrac{v_1}{r})^2[/tex]

by further solving above equation,

[tex]\rm v_1=\sqrt{v_0^2-1.2gh}[/tex]  --- (1)

Now put the values of [tex]\rm v_0 , \;g\;and\;h[/tex] in equation (1),

[tex]\rm v_1 = \sqrt{4.03^2-1.2(9.8)(0.9)}[/tex]

[tex]\rm v_1 = 2.38 \; m/sec[/tex]

Now,

F = ma

[tex]\rm mg + N = \dfrac{mv_1^2}{R}[/tex]

[tex]\rm N = m(\dfrac{v_1^2}{R}-g)[/tex]   --- (2)

Now put the values of R, g, m and [tex]\rm v_1[/tex] in equation (2) we get,

N = 2.77m

[tex]\rm N\geq 0[/tex]

ball stays on the track.

b) To find the speed of the ball as it leaves the track,

[tex]\rm v_2=\sqrt{v_0^2-1.2gh}[/tex] ---- (3)

put h = -0.2m in equation (3)

[tex]\rm v_2=\sqrt{4.03^2-1.2(9.8)(-0.2)}[/tex]

[tex]\rm v_2=4.31\;m/sec[/tex]

c) Again, but without rotational energy

Initial energy = Final energy

[tex]\rm \dfrac{1}{2}mv_0^2 = mgh + \dfrac{1}{2}mv_1^2[/tex]

by further solving the above equation we get,

[tex]\rm v_1 = \sqrt{v_0^2-2gh}[/tex] and this is less than [tex]\rm v_1[/tex] we calculated earlier.

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Answer:

III quadrant

Explanation:

First of all, let's write the components of each vector along the two directions. We have:

[tex]B_x = B cos 33^\circ = (5.6) cos 33^\circ=-4.7\\B_y = B sin 33^\circ = (5.6) sin 33^\circ=-3.0[/tex]

where we put a negative sign on both components, since they are in the negative x- and y- direction.

Similarly,

[tex]C_x = C sin 22^\circ = (4.8) sin 22^\circ=1.8\\C_y = C cos 22^\circ = (4.8) cos 22^\circ=-4.5[/tex]

where we put a negative sign on the y-direction, since it is along the negative direction.

Now we sum the components along each axis:

[tex]B_x + C_x = -4.7+1.8=-2.9\\B_y + C_y = -3.0 + (-4.5)=-7.5[/tex]

We see that both components of (B+C) are negative, so this vector must be in the 3rd quadrant.

Explanation:

The heat (thermal energy) needed in to raise the temperature in a process can be found using the following equation:

[tex]Q=m.C.\Delta T[/tex]   (1)

Where:

[tex]Q[/tex] is the heat

[tex]m=8 g[/tex] is the mass of the element (water in this case)

[tex]C[/tex] is the specific heat capacity of the material. In the case of water is [tex]C=4.184\frac{J}{g\°C}[/tex]

[tex]\Delta T=15\°C[/tex] is the variation in temperature  (which is increased in this case)

Knowing this, let's rewrite (1) with these values:

[tex]Q=(8 g)(4.184\frac{J}{g\°C})(15\°C)[/tex]  (2)

Finally:

[tex]Q=502.08 J[/tex]  

(a) The diffraction decreases

The formula for the diffraction pattern from a single slit is given by:

[tex]sin \theta = \frac{n \lambda}{a}[/tex]

where

[tex]\theta[/tex] is the angle corresponding to nth-minimum in the diffraction pattern, measured from the centre of the pattern

n is the order of the minimum

[tex]\lambda[/tex] is the wavelength

a is the width of the opening

As we see from the formula, the longer the wavelength, the larger the diffraction pattern (because [tex]\theta[/tex] increases). In this problem, since the wavelength of the signal has been decreased from 54 cm to 13 mm, the diffraction of the signal has decreased.

(b) [tex]10.8^{\circ}[/tex]

The angular spread of the central diffraction maximum is equal to twice the distance between the centre of the pattern and the first minimum, with n=1. Therefore:

[tex]sin \theta = \frac{(1) \lambda}{a}[/tex]

in this case we have

[tex]\lambda=54 cm = 0.54 m[/tex] is the wavelength

[tex]a=5.7 m[/tex] is the width of the opening

Solving the equation, we find

[tex]\theta = sin^{-1} (\frac{\lambda}{a})=sin^{-1} (\frac{0.54 m}{5.7 m})=5.4^{\circ}[/tex]

So the angular spread of the central diffraction maximum is twice this angle:

[tex]\theta = 2 \cdot 5.4^{\circ}=10.8^{\circ}[/tex]

(c) [tex]0.26^{\circ}[/tex]

Here we can apply the same formula used before, but this time the wavelength of the signal is

[tex]\lambda=13 mm=0.013 m[/tex]

so the angle corresponding to the first minimum is

[tex]\theta = sin^{-1} (\frac{\lambda}{a})=sin^{-1} (\frac{0.013 m}{5.7 m})=0.13^{\circ}[/tex]

So the angular spread of the central diffraction maximum is twice this angle:

[tex]\theta = 2 \cdot 0.13^{\circ}=0.26^{\circ}[/tex]

The change from a 54 cm wavelength to a 13 mm wavelength will decrease the diffraction of signals. The angular spread for a 54 cm wavelength is 0.1154 degrees and for a 13 mm wavelength, it is 0.00279 degrees.

The shift from conventional television signals with a wavelength of about 54 cm to future digital television signals with a wavelength of about 13 mm will decrease the diffraction of the signals into shadow regions of obstacles. This is because diffraction is inversely proportional to wavelength.

The angular spread of the central diffraction maximum can be calculated using diffraction spreading for a single slit given by the equation θ = 1.22 λ / D, where θ is the angle, λ is the wavelength, and D is the slit width. For a wavelength (λ) of 54 cm and a slit width (D) of 5.7 m, the angular spread θ is:

θ = 1.22 * (0.54 m) / 5.7 m = 0.1154 °

For a wavelength of 13 mm, the angular spread θ is:

θ = 1.22 * (0.013 m) / 5.7 m = 0.00279 °

Answer : The products of the combustion of methanol are, carbon dioxide [tex](CO_2)[/tex] and water [tex](H_2O)[/tex]

Explanation :

Combustion reaction : It is a type of reaction in which the hydrocarbon react with the oxygen gas to give carbon dioxide and water as products.

The balanced chemical reaction of combustion of methanol [tex](CH_3OH)[/tex] is :

[tex]2CH_3OH(l)+3O_2(g)\rightarrow 2CO_2(g)+4H_2O(g)[/tex]

By the stoichiometry we can say that, 2 mole of methanol react with 3 moles of oxygen gas to give 2 mole of carbon dioxide and 4 moles of water as products.

In this reaction, methanol and oxygen gas are the reactants and carbon dioxide and water are the products.

Therefore, the products of the combustion of methanol are, carbon dioxide [tex](CO_2)[/tex] and water [tex](H_2O)[/tex]

The combustion of methanol produces carbon dioxide and water in the presence of oxygen, as represented by the reaction equation CH3OH(l) + 1.5 O2(g) -> CO2(g) + 2 H2O(l).

In a combustion reaction, a substance combines with oxygen to produce heat and light. For methanol, CH3OH(l), the reactants are methanol and oxygen (O2), and the products are carbon dioxide (CO2) and water (H2O). The balanced chemical equation for this reaction is CH3OH(l) + 1.5 O2(g) -> CO2(g) + 2 H2O(l). This indicates that one molecule of methanol reacts with 1.5 molecules of oxygen to produce one molecule of carbon dioxide and two molecules of water.

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Answer:

5120 m/s

Explanation:

The acceleration due to gravity is:

g = MG / r²

where M is the mass of the earth, G is the universal constant of gravitation, and r is the distance from the earth's center to the object's center.

Here, r = h + R, where h is the height of the chest above the surface and R is the radius of the earth.

g = MG / (h + R)²

Acceleration is the derivative of velocity:

dv/dt = MG / (h + R)²

Using chain rule, we can say:

(dv/dh) (dh/dt) = MG / (h + R)²

(dv/dh) v = MG / (h + R)²

Separate the variables:

v dv = MG / (h + R)² dh

Integrating:

∫₀ᵛ v dv = MG ∫₀ʰ dh / (h + R)²

½ v² |₀ᵛ = -MG / (h + R) |₀ʰ

½ (v² − 0²) = -MG / (h + R) − -MG / (0 + R)

½ v² = -MG / (h + R) + MG / R

½ v² = MGh / (R(h + R))

v² = 2MGh / (R(h + R))

Given:

M = 5.98×10²⁴ kg

R = 6.37×10⁶ m

h = 1.69×10⁶ m

G = 6.67×10⁻¹¹ m³/kg/s²

Plugging in:

v² = 2 (5.98×10²⁴) (6.67×10⁻¹¹) (1.69×10⁶) / ((6.37×10⁶) (1.69×10⁶ + 6.37×10⁶))

v² = 2 (5.98) (6.67) (1.69) / ((6.37) (1.69 + 6.37)) × 10⁷

v ≈ 5120 m/s

Notice that if we had approximated g as a constant 9.8 m/s², we would have gotten an answer of:

v² = v₀² + 2a(x - x₀)

v² = (0 m/s)² + 2 (9.8 m/s²) (1.69×10⁶ m - 0 m)

v ≈ 5760 m/s

So we know that our calculated velocity of 5120 m/s is a reasonable answer.

The Newton's second law  and the law of universal gravitation allows to find the result for the speed of the chest when reaching the ocean is:

The law of universal gravitation is stable that the gravitational force between bodies is attractive and is proportional to the mass of the bodies and inversely proportional to the square of the distance.

           [tex]F= - G \frac{Mm}{r^2}[/tex]

Where M and m are the mass of the two bodies and r is the distance.

Indicate the height from where the chest falls h= 1690 km = 1,690 10⁶ m, they also give the radius and the mass of the earth.

Newton's second law establishes a relationship between force, mass and the acceleration of bodies.

           F = m a  

          [tex]- G \frac{Mm}{r^2} = m a \\a= - G \frac{M}{r^2}[/tex]

The distance of the body from the center of the planet is

          r = R + h

The acceleration is defined as the variation of velocity with time.

          [tex]a= \frac{dv}{dt} \\ \frac{dv}{dt} = - G \frac{M}{(h+R)^2}[/tex]

Let's use the chain rule

         [tex]\frac{dv}{dh}\ \frac{dh}{dt} = - GM \frac{1}{(h+R)^2 }[/tex]  

The velocity is the derivative of the position with respect to the time.

         [tex]v=\frac{dh}{dt} \\ v \ \frac{dv}{dh} = - GM \ \frac{1}{(h+R)^2}[/tex]    

To solve we use the method of separation of variables and we integrate.

       [tex]\int\limits^v_0 {v} \, dv = -GM \int\limits^0_h {\frac{1}{(h+R)^2} } \, dh \\\frac{1}{2} ( v^2 - 0) = -GM (-1) [ \frac{1}{(0+R)} - \frac{1}{(h+R)}] \\\frac{1}{2} v^2 = GM \ \frac{h}{(h+R)R}[/tex]

      [tex]v^2 = 2GM \ \frac{h}{(h+R) R }[/tex]  

Let's calculate

      v² = [tex]2 \ 6.67 \ 10^{-11} \ 5.98 \ 10^{24}} \ \frac{1.690 }{(1.609 + 6.370) 6.370} \ 10^{-6}[/tex]

      v = 5120 m/s

In conclusion we use Newton's second law and the universal gravitation's law we can find the result for the speed of the chest when reaching the ocean is;

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(a) 23.4

The fiber-to-matrix load ratio is given by

[tex]\frac{F_f}{F_m}=\frac{E_f V_f}{E_m V_m}[/tex]

where

[tex]E_f = 131 GPa[/tex] is the fiber elasticity module

[tex]E_m = 2.4 GPa[/tex] is the matrix elasticity module

[tex]V_f=0.3[/tex] is the fraction of volume of the fiber

[tex]V_m=0.7[/tex] is the fraction of volume of the matrix

Substituting,

[tex]\frac{F_f}{F_m}=\frac{(131 GPa)(0.3)}{(2.4 GPa)(0.7)}=23.4[/tex] (1)

(b) 44,594 N

The longitudinal load is

F = 46500 N

And it is sum of the loads carried by the fiber phase and the matrix phase:

[tex]F=F_f + F_m[/tex] (2)

We can rewrite (1) as

[tex]F_m = \frac{F_f}{23.4}[/tex]

And inserting this into (2):

[tex]F=F_f + \frac{F_f}{23.4}[/tex]

Solving the equation, we find the actual load carried by the fiber phase:

[tex]F=F_f (1+\frac{1}{23.4})\\F_f = \frac{F}{1+\frac{1}{23.4}}=\frac{46500 N}{1+\frac{1}{23.4}}=44,594 N[/tex]

(c) 1,906 N

Since we know that the longitudinal load is the sum of the loads carried by the fiber phase and the matrix phase:

[tex]F=F_f + F_m[/tex] (2)

Using

F = 46500 N

[tex]F_f = 44594 N[/tex]

We can immediately find the actual load carried by the matrix phase:

[tex]F_m = F-F_f = 46,500 N - 44,594 N=1,906 N[/tex]

(d) 437 MPa

The cross-sectional area of the fiber phase is

[tex]A_f = A V_f[/tex]

where

[tex]A=340 mm^2=340\cdot 10^{-6}m^2[/tex] is the total cross-sectional area

Substituting [tex]V_f=0.3[/tex], we have

[tex]A_f = (340\cdot 10^{-6} m^2)(0.3)=102\cdot 10^{-6} m^2[/tex]

And the magnitude of the stress on the fiber phase is

[tex]\sigma_f = \frac{F_f}{A_f}=\frac{44594 N}{102\cdot 10^{-6} m^2}=4.37\cdot 10^8 Pa = 437 MPa[/tex]

(e) 8.0 MPa

The cross-sectional area of the matrix phase is

[tex]A_m = A V_m[/tex]

where

[tex]A=340 mm^2=340\cdot 10^{-6}m^2[/tex] is the total cross-sectional area

Substituting [tex]V_m=0.7[/tex], we have

[tex]A_m = (340\cdot 10^{-6} m^2)(0.7)=238\cdot 10^{-6} m^2[/tex]

And the magnitude of the stress on the matrix phase is

[tex]\sigma_m = \frac{F_m}{A_m}=\frac{1906 N}{238\cdot 10^{-6} m^2}=8.0\cdot 10^6 Pa = 8.0 MPa[/tex]

(f) [tex]3.34\cdot 10^{-3}[/tex]

The longitudinal modulus of elasticity is

[tex]E = E_f V_f + E_m V_m = (131 GPa)(0.3)+(2.4 GPa)(0.7)=41.0 Gpa[/tex]

While the total stress experienced by the composite is

[tex]\sigma = \frac{F}{A}=\frac{46500 N}{340\cdot 10^{-6}m^2}=1.37\cdot 10^8 Pa = 0.137 GPa[/tex]

So, the strain experienced by the composite is

[tex]\epsilon=\frac{\sigma}{E}=\frac{0.137 GPa}{41.0 GPa}=3.34\cdot 10^{-3}[/tex]

The fiber-matrix load ratio, actual load carried by the fiber and matrix phase, stress on the fiber and matrix phase, and the strain in the composite can be calculated using the given volume fractions and elastic moduli.

Given the load P = 46500N, volume fraction of fiber (Vf) = 0.3, volume fraction of matrix (Vm) = 0.7, elastic modulus of fiber (Ef) = 131 GPa = 131 * 10⁹ Pa, and elastic modulus of matrix (Em) = 2.4 Gpa = 2.4 * 10⁹ Pa, the fiber-matrix load ratio is given by the ratio of the product of the volume fraction and elastic modulus of fiber to the product of the volume fraction and elastic modulus of matrix, i.e., (Vf * Ef)/(Vm * Em). (b) The actual load carried by the fiber phase can be calculated by multiplying load P with Vf and the ratio of Ef to the sum of Vf * Ef and Vm * Em. (c) The actual load carried by the matrix phase can be calculated by subtracting the load carried by fiber from total load P. (d) The stress on the fiber phase can be calculated as the load carried by the fiber phase divided by the cross-sectional area. (e) The stress on the matrix phase can be calculated as the load carried by the matrix phase divided by the cross-sectional area. (f) The strain in the composite can be calculated by dividing the total applied stress by the equivalent stiffness of the composite material (i.e., (Vf * Ef + Vm * Em)).

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Solution here,

mass of cylinder(m)=100 g

radius of cylinder(r)= 4 cm

momoent of inertia(I)=?

we have,

I=mr^2=100×4^2=1600 g/cm^2

Answer:

1.78 J

Explanation:

Find the spring coefficient using Hooke's law:

F = k Δx

23 N = k (0.20 m − 0.14 m)

k = 383.33 N/m

The work is the change in energy:

W = PE₂ − PE₁

W = ½ kx₂² − ½ kx₁²

W = ½ k (x₂² − x₁²)

W = ½ (383.33 N/m) ((0.17 m)² − (0.14 m)²)

W = 1.78 J

B) The velocity of the car reduced from 50k/m over one min.

Answer:

Answer b

Explanation:

The answer is:

The third option, the approximate magnitude of the given vector is 9.9 units.

[tex]|A|=9.9units[/tex]

To calculate the magnitude (length) of a vector, we need to apply the following formula:

[tex]|A|=\sqrt{(A_x)^{2} +(A_y)^{2} }[/tex]

So, we are given the vector:

[tex]A=(7.6,-6.4)[/tex]

Then,  substituting and calculating the magnitude of the vector, we have:

[tex]|A|=\sqrt{(7.6)^{2} +(-6.4)^{2}}=\sqrt{57.76+40.96}=\sqrt{98.72}\\\\|A|=\sqrt{98.72}=9.9units[/tex]

Hence, we have that the correct option is the third option, the approximate magnitude of the given vector is 9.9 units.

[tex]|A|=9.9units[/tex]

Have a nice day!

The answer is:

The time took for the rock to reach its maximum height is 0.110 seconds.

In order to calculate the time needed for the rock to reach its maximum height, we need to calculate the initial vertical speed.

From the statement we know that the rock was launched at an initial speed of 2.1 m/s at an angle of 30° above the horizontal, so, calculating we have:

[tex]V_x=2.1\frac{m}{s} *cos(30\°)=1.82\frac{m}{s} \\\\V_y=2.1\frac{m}{s} *sin(30\°)=1.05\frac{m}{s}[/tex]

Also, we know that at the maximum height, the speeds tends to 0. So, using the following equation and substituting "v" equal to 0, we have:

[tex]V=V_o-g*t\\\\0=V_o-g*t\\\\g*t=V_o\\\\t=\frac{V_o}{g}[/tex]

Where,

t is the time in seconds.

V is the initial speed

g is the gravity acceleration.

Using gravity acceleration equal to [tex]9.81\frac{m}{s^{2} }[/tex]  we have:

[tex]V=V_o-g*t\\\\0=V_o-g*t\\\\g*t=V_o\\\\t=\frac{V_o}{g}[/tex]

[tex]t=\frac{V_o}{g}[/tex]

[tex]t=\frac{1.05\frac{m}{s}}{9.81\frac{m}{s^{2}}}=0.1070seconds=0.110seconds[/tex]

Hence, the correct option is the last option, the time took for the rock to reach its maximum height is 0.110 seconds.

Have a nice day!

(a) 2250 Hz, 0.152 m

In this situation, both the ambulance and observer are stationary.

This means that there is no shift in frequency/wavelength due to the Doppler effect. So, the frequency heard by the observer is exactly identical to the frequency emitted by the ambulance:

f = 2250 Hz

While the wavelength is given by the formula:

[tex]\lambda=\frac{v}{f}[/tex]

where

v = 343 m/s is the speed of sound

f = 2250 Hz is the frequency of the sound

Substituting, we find

[tex]\lambda=\frac{343 m/s}{2250 Hz}=0.152 m[/tex]

(b) 2439.2 Hz, 0.141 m

The Doppler effect formula for a moving source is

[tex]f'=(\frac{v}{v+v_s})f[/tex]

where

f' is the apparent frequency

f is the original frequency

v is the speed of sound

[tex]v_s[/tex] is the velocity of the source (the ambulance), which is positive if the source is moving away from the observer, negative otherwise

Here the ambulance is moving toward the observer, so

[tex]v_s = -26.6 m/s[/tex]

Substituting into the formula, we find the frequency heard by the observer:

[tex]f'=(\frac{343 m/s}{343 m/s-26.6 m/s})(2250 Hz)=2439.2 Hz[/tex]

while the wavelength seen by the observer will be:

[tex]\lambda' = \frac{v}{f'}=\frac{343 m/s}{2439.2 Hz}=0.141 m[/tex]

(c) 2517.4 Hz, 0.136 m

In this situation, we must use the most general formula for the Doppler effect, which is

[tex]f'=(\frac{v+v_r}{v+v_s})f[/tex]

where

[tex]v_r[/tex] is the velocity of the observer, which is positive if the observer is moving toward the source, negative otherwise

[tex]v_s[/tex] is the velocity of the source (the ambulance), which is positive if the source is moving away from the observer, negative otherwise

In this situation,

[tex]v_s = -26.6 m/s[/tex]

[tex]v_r = +11.0 m/s[/tex]

Therefore, the frequency heard by the observer is

[tex]f'=(\frac{343 m/s+11.0 m/s}{343 m/s-26.6 m/s})(2250 Hz)=2517.4 Hz[/tex]

while the wavelength seen by the observer will be:

[tex]\lambda' = \frac{v}{f'}=\frac{343 m/s}{2517.4 Hz}=0.136 m[/tex]

A mass suspended from a spring is oscillating up and down, (as stated but not indicated).

A). At some point during the oscillation the mass has zero velocity but its acceleration is non-zero (can be either positive or negative).  Yes.  This statement is true at the top and bottom ends of the motion.

B). At some point during the oscillation the mass has zero velocity and zero acceleration.  No.  If the mass is bouncing, this is never true.  It only happens if the mass is hanging motionless on the spring.

C). At some point during the oscillation the mass has non-zero velocity (can be either positive or negative) but has zero acceleration.  Yes.  This is true as the bouncing mass passes through the "zero point" ... the point where the upward force of the stretched spring is equal to the weight of the mass.  At that instant, the vertical forces on the mass are balanced, and the net vertical force is zero ... so there's no acceleration at that instant, because (as Newton informed us), A = F/m .  

D). At all points during the oscillation the mass has non-zero velocity and has nonzero acceleration (either can be positive or negative).  No.  This can only happen if the mass is hanging lifeless from the spring.  If it's bouncing, then It has zero velocity at the top and bottom extremes ... where acceleration is maximum ... and maximum velocity at the center of the swing ... where acceleration is zero.  

In a bouncing spring, the mass will have zero velocity and non-zero acceleration at the top and bottom end of motion.

It is defined as the rate of change in velocity or change speed/direction of the object.

An object can have zero velocity but non-zero acceleration. It can be understood by 2 examples,

Therefore, in a bouncing spring, the mass will have zero velocity and non-zero acceleration at the top and bottom end of motion.

Learn more about Oscillation:

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Answer:

7 units, north

Explanation:

This is a problem of vector subtraction. We have:

- Vector A: magnitude 2 units, direction to the north

- Vector B: magnitude 5 units, direction to the south

If we take the north as positive direction, we can write

[tex]A = +2\\B=-5[/tex]

Since we want to find [tex]A-B[/tex] (vector subtraction), we have to change the sign of B, so we find:

[tex]A-B=+2-(-5))=+2+5=+7[/tex]

And the positive sign means the direction is north.

(a) [tex]-3.48\cdot 10^{-7} J[/tex]

The gravitational potential energy of the two-sphere system is given by

[tex]U=-\frac{Gm_A m_B}{r}[/tex] (1)

where

G is the gravitational constant

[tex]m_A = 94 kg[/tex] is the mass of sphere A

[tex]m_B = 100 kg[/tex] is the mass of sphere B

r = 1.8 m is the distance between the two spheres

Substitutign data in the formula, we find

[tex]U=-\frac{(6.67\cdot 10^{-11})(94 kg)(100 kg)}{1.8 m}=-3.48\cdot 10^{-7} J[/tex]

and the sign is negative since gravity is an attractive force.

(b) [tex]1.74\cdot 10^{-7}J[/tex]

According to the law of conservation of energy, the kinetic energy gained by sphere B will be equal to the change in gravitational potential energy of the system:

[tex]K_f = U_i - U_f[/tex] (2)

where

[tex]U_i=-3.48\cdot 10^{-7} J[/tex] is the initial potential energy

The final potential energy can be found by substituting

r = 1.80 m -0.60 m=1.20 m

inside the equation (1):

U=-\frac{(6.67\cdot 10^{-11})(94 kg)(100 kg)}{1.2 m}=-5.22\cdot 10^{-7} J

So now we can use eq.(2) to find the kinetic energy of sphere B:

[tex]K_f = -3.48\cdot 10^{-7}J-(-5.22\cdot 10^{-7} J)=1.74\cdot 10^{-7}J[/tex]

"A concave lens is thinner at the center than it is at the edges."

If this isn't on the list of choices, that's tough.  We can't help you choose the best one if we don't know what any of them is.

Answer:

they did it

Explanation:

Answer:

56.0 m

Explanation:

We know that after 0.510 s, the ball is level with Bob again.  We can use this to find the vertical component of the initial velocity.

y = y₀ + v₀ᵧ t + ½ gt²

h+2 = h+2 + v₀ᵧ (0.510) + ½ (-9.8) (0.510)²

v₀ᵧ = 2.50 m/s

Since the magnitude is 34.1 m/s, we can now find the horizontal component:

v₀² = v₀ₓ² + v₀ᵧ²

(34.1)² = v₀ₓ² + (2.50)²

v₀ₓ = 34.0 m/s

And since we know the ball lands 126 m from the base of the cliff, we can find the time it takes to land:

x = x₀ + v₀ₓ t + ½ at²

126 = 0 + (34.0) t + ½ (0) t²

t = 3.71 s

Finally, we can now find the height of the cliff:

y = y₀ + v₀ᵧ t + ½ gt²

0 = h+2 + (2.50) (3.71) + ½ (-9.8) (3.71)²

h = 56.0 m

Bob climbed approximately 4.4 meters.

To determine the height that Bob climbed, we can use the kinematic equation for vertical motion:

Y = Yo + Voy*t -1/2gt^2

where:

Y = final height (unknown)

Yo = initial height (2m)

Voy = initial vertical velocity (unknown)

t = time taken to reach the final height (0.510s)

g = acceleration due to gravity (9.8m/s^2)

We need to find the initial vertical velocity. When the ball reaches the final height, it has the same vertical velocity as it did when it was thrown. Using the equation for vertical velocity:

Vy = Voy - gt

where:

Vy = vertical velocity (0 m/s)

Substituting the known values:

0 = Voy - (9.8)(0.510)

Solving for Voy, we get:

Voy ≈ 5.0 m/s

Now we can calculate the final height:

Y = Yo + Voy*t - 1/2gt^2

Y = 2 + (5.0)(0.510) - 1/2(9.8)(0.510)^2

Y ≈ 2 + 2.55 - 0.126

Y ≈ 4.425m

Therefore, Bob climbed approximately 4.4 meters.

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Answer:

A transverse wave is one in which the disturbance is parallel to the direction of travel

Explanation:

There are two types of waves:

- Transverse waves: in transverse waves, the disturbance (oscillation) occurs in a plane perpendicular to the direction of propagation of the wave

- Longitudinal waves: in longitudinal waves, the disturbance (oscillation) occurs parallel to the direction of propagation of the wave

Therefore, we immediately see that the statement:

"A transverse wave is one in which the disturbance is parallel to the direction of travel"

is wrong, because it is actually the opposite: in a transverse wave, the disturbance is perpendicular to the direction of travel.

The false statement is that a transverse wave has a disturbance parallel to its direction of travel. A transverse wave has the disturbance perpendicular to its direction of travel, not parallel. An example of this would be the waves on a stringed instrument, contrasted with sound waves which are longitudinal (disturbance and wave travel in the same direction).

The false statement among the given options is: "A transverse wave is one in which the disturbance is parallel to the direction of travel." This is incorrect because in a transverse wave, the disturbance or oscillation is perpendicular to the direction of wave propagation. For example, the movement of a stringed instrument creates transverse waves, where the strings vibrate upwards and downwards while the wave travels horizontally along the string.

On the other hand, in a longitudinal wave, the disturbance is parallel to the direction of wave propagation. An example would be sound waves, where the air particles vibrate back and forth, in the same direction that the wave propagates.

It's also correct that waves can have both transverse and longitudinal components, such as seismic waves from earthquakes. Furthermore, a wave does carry energy from one location to another but does not result in the bulk movement of the material medium.

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Answer: Third option

More than w and equal to F

Explanation:

If the horse presses the floor with a force F, then it follows that by Newton's third law, the floor makes a force -F on the horse.

Then if this force F is greater than the weight w of the horse, then the horse accelerates upwards and rises from the ground.

We know that after pressing the floor with a force F, the horse gets up off the floor. Then it is true that:

The force exerted by the floor on the horse is equal to F and greater than w

The answer is the third option more than w and equal to F

To become airborne, the force exerted by the ground on the horse must be greater than both the horse's weight and the force applied by the horse.

When a horse becomes airborne after pressing down on the ground, the magnitude of the force that the ground exerted on the horse must have been more than w (weight of the horse) and more than F (force applied by the horse on the ground). This is because for the horse to become airborne, the ground must exert a force greater than the horse's weight, which is the normal reaction force, to overcome gravitational pull. Additionally, the applied force F would be a combination of the horse's weight (mg) and the additional force caused by the horse's muscles to initiate the jump.

Answer:

Explanation:

6.7 amps

The total current of the circuit is 6.7 amps.

In a parallel circuit, the total current is the sum of the individual currents flowing through each resistor. To find the total current, we can use Ohm's law, which states that I = V/R, where I is the current, V is the voltage, and R is the resistance.

For the given circuit, the total resistance can be calculated by adding the reciprocals of the individual resistances: 1/Rt = 1/R1 + 1/R2 + 1/R3. Plugging in the resistance values, we get 1/Rt = 1/4.0 + 1/6.0 + 1/8.0.

After calculating the reciprocal of the sum, we find that Rt = 1.714 ohms. Finally, using Ohm's law, we can find the total current: I = V/Rt = 12 volts / 1.714 ohms = 6.7 amps.