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sashaice [31]
2 years ago
6

How much work is done when 100 N of force is applied to a rock to move it 20 m

Physics
1 answer:
qaws [65]2 years ago
4 0

Answer: 2000 J

Explanation: work W = F s

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Which idea would most likely be dangerous for a student to think while entering a lab?
aleksley [76]
Your answer should be a i can figure out how the tools work as i go along
3 0
2 years ago
Please help me with the following question:
pishuonlain [190]

Answer: a. 667N

b. 665N

c. 54.5N

Explanation:

a) on the surface of the earth

W = mg

W = 68 × 9.81

= 667N

b) at the top of Everest (8848 m above sea level).

W =mg × R²/(R + H)²

W = 667 × [6378²/(6378 + 8.848)²

W = 665N

c) has 2 1/2 times the radius of the earth

W = mg × R²/(R + H)²

W = 667 × R²/(R + 2.5R)²

W = 54.5N

3 0
2 years ago
A satellite in a circular orbit of radius R around planet X has an orbital period T. If Planet X had one-fourth as much mass, th
Iteru [2.4K]
<h2>Answer: 2T</h2>

According to the Third Kepler’s Law of Planetary motion <em>“The square of the orbital period of a planet is proportional to the cube of the semi-major axis (size) of its orbit”.</em>

In other words, this law states a relation between the orbital period T of a body (moon, planet, satellite) orbiting a greater body in space with the size R of its orbit.

This Law is originally expressed as follows (in the case of planet X and assuming we have a circular orbit):

T^{2}=\frac{4\pi^{2}}{GM}R^{3}    (1)

Where:

G is the Gravitational Constant

M=1.9(10)^{27}kg is the mass of planet X

R  is the radius of the orbit of the satellite around planet X

If we want to find the period, we have to express equation (1) as written below and substitute all the values:

T=2\pi\sqrt{\frac{R^{3}}{GM}}   (2)

Now, we are asked to find the period when tha mass of the planet is \frac{1}{4}M. In order to do this, we have to rewrite equation (2) with this new value:

T=2\pi\sqrt{\frac{R^{3}}{G(\frac{1}{4}M)}}  (3)

Solving:

T=4\pi\sqrt{\frac{R^{3}}{G(\frac{1}{4}M)}}   (4)

On the other hand, if we multiply both sides of equation (2) by 2, we have:

2T=4\pi\sqrt{\frac{R^{3}}{GM}}    (5)

As we can see, (5) is equal to (4). This means the orbital period is twice the orignal period.

Hence, the answer is:

If Planet X had <u>one-fourth </u>as much mass, the <u>orbital period</u> of this satellite in an orbit of the same radius would be <u>2T.</u>

3 0
3 years ago
What is difference between non uniform and uniform circular motion?
statuscvo [17]

Answer:

The object in a uniform motion covers same distances in an equal time period. Objects in a non-uniform motion cover dissimilar distances in an equal time period.

Explanation:

The speed of the object traveling in uniform motion is constant, the actual speed and the average speed of the moving body is same.

6 0
3 years ago
When a wire with a current is placed in a magnetic field,
Jet001 [13]
A. electrical energy is transformed into mechanical energy.
6 0
3 years ago
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