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2 years ago

If gravity between the Sun and Earth suddenly vanished, Earth would continue moving in

Physics

1 answer:

Ksenya-84 [330]2 years ago

6 0

Answer:

Earth would continue moving by uniform motion, with constant velocity, in a straight line

Explanation:

The question can be answered by using Newton's first law of motion, also known as law of inertia, which states that:

"an object keeps its state of rest or of uniform motion in a straight line unless acted upon by an external net force different from zero"

This means that if there are no forces acting on an object, the object stays at rest (if it was not moving previously) or it continues moving with same velocity (if it was already moving) in a straight line.

In this problem, the Earth is initially moving around the Sun, with a certain tangential velocity v. When the Sun disappears, the force of gravity that was keeping the Earth in circular motion disappears too: therefore, there are no more forces acting on the Earth, and so by the 1st law of Newton, the Earth will continue moving with same velocity v in a straight line.

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On a dry day, just after washing your hair to remove natural oils and drying it thoroughly, run a plastic comb through it. Small

Hitman42 [59]

When the surface of the comb rubs on your hair, the comb is electrically charged. When the comb comes close to the paper, the charge on the comb causes charge separation on the paper bits. Since paper is neutral, positive and negative charges are equivalent. The charge on the comb charges the area of the bit of paper nearest the comb to the opposite. Thus, the bits of paper become attracted to the comb.

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2 years ago

Lauren wants to know which location in her apartment is best for growing African violets. She has three African violets. She put

Harrizon [31]

First, before determining which variable is which, we go over the definition of each.
The independent variable is the one which is intentionally changed in order to investigate its effect on the dependent variable.
The dependent variable is monitored and changes occur in it due to the changing conditions of the independent variable.

In this case, the location of the African violets is the independent variable as it is intentionally changed, while the rate of growth of the African violets is the dependent variable as it is being measured.

5 0

2 years ago

(4%) Problem 10: The standard frequency of the musical pitch called "middle C" is 261.6 Hz. show answer No Attempt How fast, in

bearhunter [10]

Answer:

16.77 m/s

Explanation:

Given that

Frequency of middle pitch, Fo = 261.6 Hz

Frequency of C sharp, f = 277.2 Hz

Velocity of sound in air, v = 298 m/s

Speed of sound from the source, Vs = ? m/s

Using the formula

f = Fo•(V + Vr)/(V + Vs)

← Doppler

Vr would be +ve if the receiver is moving toward source;

Vs would be -ve if source is moving toward the receiver

277.2 Hz = 261.6Hz * (298 + 0) / (298 - Vs)

277.2 = 77956.8 / (298 - Vs)

298 - Vs = 77956.8 / 277.2

298 - Vs = 281.23

Vs = 298 - 281.23

Vs = 16.77 m/s

Thus, the speed needed is 16.77 m/s

6 0

2 years ago

Workers do 8000 J of work on a 2000-N crate to push it up a ramp. If the ramp is 2 m high, what is the efficiency of the ramp?

IRISSAK [1]

Answer:

50%

Explanation:

Efficiency = work out / work in

e = Fd / W

e = (2000 N) (2 m) / (8000 J)

e = 0.5

7 0

2 years ago

Read 2 more answers

A long-distance swimmer is able to swim through still water at 4.0 km/h. She wishes to try to swim from Port Angeles, Washington

Roman55 [17]

Let $\theta$ be the direction the swimmer must swim relative to east. Then her velocity relative to the water is

$\vec v_{S/W}=\left(4.0\dfrac{\rm km}{\rm h}\right)(\cos\theta\,\vec\imath+\sin\theta\,\vec\jmath)$

The current has velocity vector (relative to the Earth)

$\vec v_{W/E}=\left(3.0\dfrac{\rm km}{\rm h}\right)\,\vec\imath$

The swimmer's resultant velocity (her velocity relative to the Earth) is then

$\vec v_{S/E}=\vec v_{S/W}+\vec v_{W/E}$

$\vec v_{S/E}=\left(\left(4.0\dfrac{\rm km}{\rm h}\right)\cos\theta+3.0\dfrac{\rm km}{\rm h}\right)\,\vec\imath+\left(4.0\dfrac{\rm km}{\rm h}\right)\sin\theta\,\vec\jmath$

We want the resultant vector to be pointing straight north, which means its horizontal component must be 0:

$\left(4.0\dfrac{\rm km}{\rm h}\right)\cos\theta+3.0\dfrac{\rm km}{\rm h}=0\implies\cos\theta=-\dfrac{3.0}{4.0}\implies\theta\approx138.59^\circ$

which is approximately 41º west of north.

6 0

2 years ago