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nalin [4]
2 years ago
5

A bomb falls with velocity v(t)=C(1−e−kt), where C and k are constants. What is the terminal velocity of the bomb? That is, what

is the maximum valocity the bomb can reach regardless of the height of the bomber?
Physics
1 answer:
irinina [24]2 years ago
6 0

Answer:

The terminal velocity is equal to C.

Explanation:

Making the assumption that k and t are positive, we then have that -kt is negative. The value of e^(-kt) will be equal to 1/(e^(kt))

If t increases, e^(kt) will increase exponentially and its reciprocal 1/(e^(kt)) will approach zero.

So, we have:

v(t) = C*(1-0)

v(t) = C*(1)

v(t) = C

Therefore, C is the terminal velocity.

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6 0
3 years ago
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lisov135 [29]

The net torque on the seesaw is 294 Nm.

<h3>What is torque?</h3>

Torque is the force that tends to rotate the body to which it was applied.

To calculate the net torque, we use the formula below

Formula:

  • τ = mgd-m'gd'........... Equation 1

Where:

  • τ = Net torque
  • m = Jenny's mass
  • m' = Tom's mass
  • d = Jenny's distance from the pivot
  • d' = Tom's distance from the pivot
  • g = acceleration due to gravity

From the question,

Given:

  • m = 40 kg
  • m' = 30 kg
  • d = 1.5 m
  • d' = 1 m
  • g = 9.8 m/s²

Substitute these values into equation 1

  • τ = (40×1.5×9.8)-(30×1×9.8)
  • τ  = 588-294
  • τ  = 294 Nm

Hence, The net torque on the seesaw is 294 Nm.

Learn more about torque here: brainly.com/question/14839816

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