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jonny [76]
3 years ago
11

Scientists observe an approaching asteroid that is on a collision course with

Physics
1 answer:
nasty-shy [4]3 years ago
3 0

Answer:

The approximate velocity the rocket must have to stop the asteroid completely after the collision is;

C. -324 m/s

Explanation:

The parameters of the asteroid and the rocket are;

The mass of the asteroid, m₁ = 11,000 kg

The initial velocity with which the asteroid is approaching Earth, v₁ = 50 m/s

The mass of the rocket, m₂ = 1700 kg

The initial velocity of the rocket = v₂

The final velocity of the combined asteroid and rocket after the collision, v₃ = 0 m/s

By the law of conservation of linear momentum, we have;

The total initial momentum = The total final momentum

m₁·v₁ + m₂·v₂ = (m₁ + m₂)·v₃

Substituting the known values, we get;

11,000 kg × 50 m/s + 1,700 kg × v₂ = (11,000 kg + 1,700 kg) × 0 m/s

11,000 kg × 50 m/s + 1,700 kg × v₂ = 0

∴ 1,700 kg × v₂ = -11,000 kg × 50 m/s

v₂ = (-11,000 kg × 50 m/s)/(1,700 kg) = -323.529412 m/s ≈ -324 m/s

The approximate initial velocity the jet must have to completely stop the asteroid after the collision is -324 m/s.

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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)
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  • τ  = 294 Nm

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

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

5 0
2 years ago
If a car starts at rest and accelerates at 3.4 m/s^2 over a distance of 200 meters, how fast will it be travelling at the end?
CaHeK987 [17]

Answer:

36.9 m/s

Explanation:

From;

v^2 = u^2 + 2as

Where;

v = final velocity =?

a = acceleration = 3.4 m/s^2

u = initial velocity = 0 m/s

s = distance covered = 200 meters

v^2 = 0^2 + 2 * 3.4 * 200

v^2 = 1360

v = √1360

v = 36.9 m/s

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Had to look for the options and here is my answer.
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A boy is pulling a 150 kg crate with a force of 1150 N. If the crate experiences a frictional force of 490 N, find:
valentina_108 [34]
<h3><u>For the aceleration:</u></h3>

First, let's find the resultant, and <u>applicate 2nd law of Newton</u> using the resultant, so:

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Replacing according our data:

1150 N - 490 N = 150 kg * a

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<h3><u>For the normal force:</u></h3>

The normal force IS NOT the resultant force, the normal force's the force between the ground and the object, in another words, is the weight of the object, and for the weight:

w = mg

w = 150 kg * 10 m/s²

w = 1500 N   ← Normal force between object and ground.

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