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3 years ago

The force shown in the figure(Figure 1) moves an object from x = 0 to x = 0.75 m.

Physics

1 answer:

Norma-Jean [14]3 years ago

3 0

Answer:

Explanation:

Work is force times distance in the direction of that force.

W = 0.6(0.25 - 0.00) + 0.4(0.50 - 0.25) + 0.8(0.75 - 0.50)

W = 0.45 J

W = 0.6(0.25 - 0.20) + 0.4(0.50 - 0.25) + 0.8(0.55 - 0.50)

W = 0.17 J

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A loudspeaker in a parked car is producing sound whose frequency is 20 510 Hz. A healthy young person with normal hearing is sta

Vsevolod [243]

Answer:

The car must be moving away from the person.

Explanation:

From Doppler's Effect, we know that when a sound source moves towards a stationary observer, the apparent frequency of that sound increases. While the apparent frequency decreases if the source moves away from the stationary observer.

The audible range of frequencies for a human ear is 20 Hz to 20000 Hz. Therefore, in order for the sound of a loud speaker to be audible for the person, the frequency must decrease below 20000 Hz.

<u>Due to this reason, the car must be moving away from the person.</u>

4 0

3 years ago

In this experiment, you need to examine the idea of thermal energy transfer. Using a controlled experiment, what might a good qu

Mamont248 [21]

We'll look at two properties:

1. The variation in temperature

2. The material's heat transfer coefficient

By taking an example;

Use a circular rod made of a certain material (for example, steel) that is insulated all the way around.

One end of the rod is immersed in a huge reservoir of 100°C water, while the other is immersed in water at 40°C. The cold water is kept in an insulated cylinder on both sides. The temp of the chilly water is measured using a meter as a time - dependent.

Conclusion of experiment;

Heat is transferred from a hot location to a cooler region.
Whenever heat is applied to a body, its thermal power rises, and its temperature rises.

Learn more:

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4 0

3 years ago

14. The design for a rotating spacecraft below consists of two rings. The outer ring with a radius of 30 m holds the living quar

Zolol [24]

Answer:

$T= 11.0003$ s

Explanation:

From the question we are told that

The outer ring with a radius of 30 m

inner Gravity Approximately 9.80 m/s'

Outer Gravity Approximately 5.35 m/s.

Generally the equation for centripetal force is given mathematically as

Centripetal acceleration enables Rotation therefore?

$\omega ^2 r =Angular\ acc$

Considering the outer ring,

$\omega ^2 r = 9.8$

$\omega ^2= \frac{9.8}{30}$

$\omega = \sqrt{\frac{9.8}{30}}$

$\omega= 0.571 rad/s$

Therefore solving for Period T

Generally the equation for solving Period T is mathematically given as

$T= \frac{2\pi}{\omega}$

$T= \frac{2\pi}{0.571 rad/s}$

$T= 11.0003$ s

5 0

3 years ago

child slides down a snow‑covered slope on a sled. At the top of the slope, her mother gives her a push to start her off with a s

Strike441 [17]

Answer:

θ = 13.7º

Explanation:

According to the work-energy theorem, the change in the kinetic energy of the combined mass of the child and the sled, is equal to the total work done on the object by external forces.
The external forces capable to do work on the combination of child +sled, are the friction force (opposing to the displacement), and the component of the weight parallel to the slide.
As this last work is just equal to the change in the gravitational potential energy (with opposite sign) , we can write the following equation:

$\Delta K + \Delta U = W_{nc} (1)$

ΔK, is the change in kinetic energy, as follows:

$\Delta K = \frac{1}{2}* m* (v_{f} ^{2} - v_{0} ^{2}) (2)$

ΔU, is the change in the gravitational potential energy.
If we choose as our zero reference level, the bottom of the slope, the change in gravitational potential energy will be as follows:

$\Delta U = 0 - m*g*h = -m*g*d* sin\theta (3)$

Finally, the work done for non-conservative forces, is the work done by the friction force, along the slope, as follows:

$W_{nc} = F_{f} * d * cos 180\º \\\\ = 0.2*m*g*d* cos 180\º = -0.2*m*g*d (4)$

Replacing (2), (3), and (4) in (1), simplifying common terms, and rearranging, we have:

$\frac{1}{2}* (v_{f} ^{2} - v_{0} ^{2}) = g*d* sin\theta -0.2*g*d$

Replacing by the givens and the knowns, we can solve for sin θ, as follows: $\frac{1}{2}*( (4.30 m/s) ^{2} - (0.75 m/s)^{2}) = 9.8 m/s2*25.5m* sin\theta -0.2*9.8m/s2*25.5m\\ \\ 8.56 (m/s)2 = 250(m/s)2* sin \theta -50 (m/s)2\\ \\ sin \theta = \frac{58.6 (m/s)2}{250 (m/s)2} = 0.236$ ⇒ θ = sin⁻¹ (0.236) = 13.7º

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3 years ago

Explain how the pressure at the bottom of a container depends on the container shape and the fluid height

3241004551 [841]

The pressure at the bottom of a column of fluid in a container
depends only on the depth of the fluid, not on the shape of the
container. The pressure is simply the result of the weight of the
fluid resting on the bottom.

3 0

3 years ago