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Studentka2010 [4]
3 years ago
7

Suppose a particle is accelerated through space by a constant 10-N force. Suddenly the particle encounters a second force of 10-

N in a direction opposite to that of the first force. The particle
Physics
1 answer:
Harlamova29_29 [7]3 years ago
3 0

Answer:

The particle will continue moving at constant velocity

Explanation:

When the particle encounters the second force of 10 N, the net force acting on the particle becomes zero, because the two forces are equal in magnitude but opposite in direction:

F_{net}=10 N-10 N=0

For Newton's second Law, the acceleration of the particle is proportional to the net force acting on it:

F_{net}=ma

Therefore, since F_{net}=0, the acceleration is zero (a=0) and so the particle will keep a constant velocity.

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A small block has constant acceleration as it slides down a frictionless incline. The block is released from rest at the top of
ehidna [41]

Answer:

Explanation:

Given

Speed of block at bottom is v=3.8\ m/s

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2 years ago
Read 2 more answers
. If a pendulum-driven clock gains 5.00 s/day, what fractional change in pendulum length must be made for it to keep perfect tim
inn [45]

Answer:

The appropriate response will be "Length must be increased by 0.012%".

Explanation:

The given values is:

ΔT = 5 s/day

Now,

⇒ \frac{\Delta T}{T} =\frac{5}{24\times 60\times 60}

On multiplying both sides by  "100", we get

⇒ \frac{\Delta T}{T}\times 100 =\frac{500}{24\times 60\times 60}

⇒ \frac{\Delta T}{T}\times 100=0.005787 (%)

∵  T=2\pi\sqrt{\frac{l}{g} }

On substituting the values, we get

⇒ \frac{\Delta T}{T}% = \frac{1}{2}\times \frac{\Delta l}{l}%

On applying cross multiplication, we get

⇒ \frac{\Delta l}{l}% = 2\times \frac{\Delta T}{T}%

⇒        = 2\times 0.05787

⇒        = 0.011574

⇒        = 0.012%

6 0
2 years ago
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