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Vladimir79 [104]
3 years ago
10

So we were working on some pulley problems but this one has kinda left me scratching my head, please help! My question is for pa

rt C :' )

Physics
1 answer:
USPshnik [31]3 years ago
8 0

Explanation:

(c) I assume we're looking for mA.

Sum of forces on B in the -y direction:

∑F = ma

mBg − T = mBa

Sum of forces on A in the +x direction:

∑F = ma

T = mAa

Substitute:

mBg − mAa = mBa

mBg − mBa = mAa

mA = mB (g − a) / a

Plug in values:

mA = (5 kg) (10 m/s² − 0.01 (10 m/s²)) / (0.01 (10 m/s²))

mA = 495 kg

The answer key seems to have a mistake.  It's possible they meant mB = 1 kg, or they changed mB to 5 kg but forgot to change the answer.

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A small object carrying a charge of -2.50 nc is acted upon by a downward force of 18.0 nn when placed at a certain point in an e
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Missing question in the text:
"A.What are the magnitude and direction of the electric field at the point in question?

B.<span>What would be the magnitude and direction of the force acting on a proton placed at this same point in the electric field?"</span>

<span>Solution:

A) A charge q </span>under an electric field of intensity E will experience a force F  equal to:

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In our problem we have q=-2.5 nC=-2.5\cdot 10^{-9} C and F=18 nN = 18 \cdot 10^{-9} N, so we can find the magnitude of the electric field:

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