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Anastaziya [24]
2 years ago
5

You are in a spacecraft moving at a constant velocity. The front thruster rocket fires incorrectly, causing the craft to slow do

wn. You try to shut it off but fail. Instead, you fire the rear thruster, which exerts a force equal in magnitude but opposite in direction to the front thruster. How does the craft respond
Physics
1 answer:
Alchen [17]2 years ago
8 0

Answer:

It continue to move forward at a constant velocity which will be slower than before the front thruster was fired.

Explanation:

Before the front thruster was fired, the spacecraft was already moving at a particular velocity.

After the malfunction, the front thruster is fired and then the force exerted by that front thruster slows the spacecraft down, as we are told.

By using the rear thruster to exert a force equal to that from the front thrusters, a force equal in magnitude to that of the front thrusters is added, cancelling out the effect of the front thrusters. Because the spacecraft is already moving at a slower speed at this point compared to the beginning, it continues to move at that speed.

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B

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A body is projected upward at an angle of 30 degree to the horizontal at an initial speed of 200ms-.In how many seconds will it
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Answer:

20.41 s

3534.80 m

Explanation:

<em><u>In how many seconds will it reach the ground?</u></em>

We are given the initial velocity of the body, which is 200 m/s at a 30° angle.

We know the acceleration in the vertical direction is -9.8 m/s², assuming that the upwards/right direction is positive and the downwards/left direction is negative.

Since we are using acceleration in the y-direction, let's use the vertical component of the initial velocity.

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Let's use the fact that at the top of its trajectory, the body will have a final velocity of 0 m/s.

Now we have one missing variable that we are trying to solve for: time t.

Find the constant acceleration equation that contains v₀, v, a, and t.

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Substitute known values into the equation.

  • 0 = 200 · sin(30) + (-9.8)t
  • -200 · sin(30) = -9.8t
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Recall that this is only half of the body's trajectory, so we need to double the time value we found to find the total time the body is in the air.

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The body will reach the ground in 20.41 seconds.

<em><u>How far from the point of projection would it strike? </u></em>

We want to find the displacement in the x-direction for the body.

Let's find the constant acceleration equation that contains time t, that we just found, and displacement (Δx).

  • Δx = v₀t + 1/2at²

Substitute known values into the equation. Remember that we want to use the horizontal component of the initial velocity and that the acceleration in the x-direction is 0 m/s².

  • Δx = (200 · cos(30) · 20.40816327) + 1/2(0)(20.40816327)²
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4 0
1 year ago
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Answer:

5.1*10^3 J/m^3

Explanation:

Using E = q/A*eo

And

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A = 0.25

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Energy density = 1/2*eo*(E^2) = 1/2*eo*(q/A*eo)^2 = [q^2] / [2*(A^2)*eo]

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You are in a hot-air balloon that, relative to the ground, has a veloc- ity of 6.0 m/s in a direction due east. You see a hawk m
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