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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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In an "atom smasher," two particles collide head on at relativistic speeds. If the velocity of the first particle is 0.741c to t
USPshnik [31]

Answer:

W_x = 0.9156\ c

Explanation:

given,

velocity of particle 1 = 0.741 c to left

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relative velocity between the particle = ?

for the relative velocity calculation we have formula

W_x = \dfrac{|u_x - v_x|}{1-\dfrac{u_xv_x}{c^2}}

u_x = 0.543 c

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W_x = \dfrac{0.543 c - (-0.741 c)}{1-\dfrac{(0.543 c)(-0.741 c)}{c^2}}

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Which of the following is true about a planet orbiting a star in uniform circular
Mnenie [13.5K]

The velocity vector of the planet points toward the center of the  circle is the following is true about a planet orbiting a star in uniform circular  motion.

A. The velocity vector of the planet points toward the center of the  circle.

<u>Explanation:</u>

Motion of the planet around the star is mentioned to be uniform and around a circular path. Objects in uniform circular motion motion has constant angular speed but the velocity of the object will not remain constant. Since the planet is in circular motion the direction of velocity vector at a particular point is tangential to the circular path at that particular point.

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Answer:

D

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6 0
3 years ago
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