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Alex Ar [27]
3 years ago
12

The diagram shows a ramp with a toy car at the bottom. A string attached to the front of the car and the string goes over a pull

ey at the top of the ramp. A weight is attached to the other end of the string. What would be most responsible for the toy being pulled up the ramp ?
A. Friction
B. Gravity
C. Kinetic energy
D. momentum
Physics
1 answer:
Annette [7]3 years ago
3 0

The answer would be Gravity.

Gravity is pulling the weight down, which is pulling the car up the ramp.

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Answer: Mid-ocean ridges are geologically important because they occur along the kind of plate boundary where new ocean floor is created as the plates spread apart. Thus the mid-ocean ridge is also known as a "spreading center" or a "divergent plate boundary." The plates spread apart at rates of 1 cm to 20 cm per year.

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The speed of sound in air is 345 m/s. A tuning fork vibrates above the open end of a sound resonance tube. If sound waves have w
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To develop this problem it is necessary to apply the concept of Frequency based on speed and wavelength.

According to the definition the frequency can be expressed as

f = \frac{v}{\lambda}

Where,

v = Velocity

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Consider an insulating sphere of radius 6 cm surrounded by a conducting sphere of inner radius 18 cm and outer radius 26 cm. Fur
dimulka [17.4K]

Answer:

-1.7908787542\times 10^{-9}\ C

3.4260289211\times 10^{-9}\ C

1.7908787542\times 10^{-9}\ C

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Explanation:

r = Radius

k = Coulomb constant = 8.99\times 10^{9}\ Nm^2/C^2

Electric field is given by

E=-\dfrac{kq}{r^2}\\\Rightarrow q=-\dfrac{Er^2}{k}\\\Rightarrow q=-\dfrac{1610\times 0.1^2}{8.99\times 10^9}\\\Rightarrow q=-1.7908787542\times 10^{-9}\ C

The charge is -1.7908787542\times 10^{-9}\ C

Q+q=\dfrac{Er^2}{k}\\\Rightarrow Q=\dfrac{Er^2}{k}-q\\\Rightarrow Q=\dfrac{120\times 0.35^2}{8.99\times 10^9}-(-1.7908787542\times 10^{-9})\\\Rightarrow Q=3.4260289211\times 10^{-9}\ C

The charge is 3.4260289211\times 10^{-9}\ C

The charge inside will have the polarity changed

q=+1.7908787542\times 10^{-9}\ C

Outside the charge will be

3.4260289211\times 10^{-9}-1.7908787542\times 10^{-9}\\ =1.6351501669\times 10^{-9}\ C

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