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kondor19780726 [428]
3 years ago
8

What is the effect of gravity pulling downward on an object

Physics
1 answer:
pishuonlain [190]3 years ago
3 0
The effect is that the object appears to have "weight".

I should mention that the question has some squishy 'rubber' in it.

Gravity ALWAYS pulls "down".  The reason is that WHEREVER
gravity pulls, we CALL that direction "down".  That's how we know
which way is 'down'.
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Answer:  5.05 m

Explanation:

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3 years ago
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Three observers watch a train pull away from a station toward the right of the platform. Observer A is in one of the train’s car
juin [17]

Observer A is moving inside the train

so here observer A will not be able to see the change in position of train as he is standing in the same reference frame

So here as per observer A the train will remain at rest and its not moving at all

Observer B is standing on the platform so here it is a stationary reference frame which is outside the moving body

So here observer B will see the actual motion of train which is moving in forward direction away from the platform

Observer C is inside other train which is moving in opposite direction on parallel track. So as per observer C the train is coming nearer to him at faster speed then the actual speed because they are moving in opposite direction

So the distance between them will decrease at faster rate

Now as per Newton's II law

F = ma

Now if train apply the brakes the net force on it will be opposite to its motion

So we can say

- F = ma

a = \frac{-F}{m}

so here acceleration negative will show that train will get slower and its distance with respect to us is now increasing with less rate

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5 0
4 years ago
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Answer:

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3 years ago
An electron is released from rest at the negative plate of a parallel plate capacitor. The charge per unit area on each plate is
Andreas93 [3]

Answer:

v=1.08\times 10^7\ m/s

Explanation:

Initial speed of the electron, u = 0

The charge per unit area of each plate, \dfrac{Q}{A}=1.69\times 10^{-7}\ C/m^2

Separation between the plates, d=1.75\times 10^{-2}\ m

An electron is released from rest, u = 0

Using equation of kinematics,

v^2-u^2=2ad..........(1)

Acceleration of the electron in electric field, a=\dfrac{qE}{m}............(2)

Electric field, E=\dfrac{\sigma}{\epsilon_o}............(3)

From equation (1), (2) and (3) :

v=\sqrt{\dfrac{2q\sigma d}{m\epsilon_o}}

v=\sqrt{\dfrac{2\times 1.6\times 10^{-19}\times 1.69\times 10^{-7}\times 1.75\times 10^{-2}}{9.1\times 10^{-31}\times 8.85\times 10^{-12}}}

v = 10840393.1799 m/s

or

v=1.08\times 10^7\ m/s

So, the electron is moving with a speed of 1.08\times 10^7\ m/s before it reaches the positive plate. Hence, this is the required solution.

3 0
4 years ago
A light wave has a wavelength of 0.015 m. What is the frequency of this light wave? (1 point)
Natalija [7]

Answer:

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Explanation:

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4 years ago
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