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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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A mass is tied to a string and swung in a horizontal circle with a constant angular speed. show answer No Attempt If this speed
Liono4ka [1.6K]

Answer:

The tension in the string is quadrupled i.e. increased by a factor of 4.

Explanation:

The tension in the string is the centripetal force. This force is given by

F = \dfrac{mv^2}{r}

m is the mass, v is the velocity and r is the radius.

It follows that F \propto v^2, provided m and r are constant.

When v is doubled, the new force, F_1, is

F_1 = \dfrac{m(2v)^2}{r} = \dfrac{4mv^2}{r} = 4\dfrac{mv^2}{r} = 4F

Hence, the tension in the string is quadrupled.

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2 years ago
Stu wanted to calculate the resistance of a light bulb connected to a 4.0 V battery, with a resulting current of 0.5 A. He used
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His answer was incorrect because according to ohm's law the formula used should have been R=V/I instead of multiplying and the answer should be 8ohms
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2 years ago
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A baseball player runs 27.4 meters from the
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6.0 m longer because the player ran 3 and came back 3 at the very end, which looks like he went nowhere but in reality he ran 6.
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) What is the electric potential due to the nucleus of hydrogen at a distance of 7.50× 10-11 m? Assume the potential is equal to
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V = k q / r
where k is the Coulombs law constant = 9 x 10^9 N
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2 years ago
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A uniform sphere with mass M and radius R is rotating with angular speed ω1 about a frictionless axle along a diameter of the sp
liq [111]

Answer:

W_2=\sqrt{\frac{3}{5} }W_1

Explanation:

For the first ball, the moment of inertia and the kinetic energy is:

I_1 =\frac{2}{5}MR^2

K_1 = \frac{1}{2}IW_1^2

So, replacing, we get that:

K_1 = \frac{1}{2}(\frac{2}{5}MR^2)W_1^2

At the same way, the moment of inertia and kinetic energy for second ball is:

I_2 =\frac{2}{3}MR^2

K_2 = \frac{1}{2}IW_2^2

So:

K_2 = \frac{1}{2}(\frac{2}{3}MR^2)W_2^2

Then, K_2 is equal to K_1, so:

K_2 = K_1

\frac{1}{2}(\frac{2}{3}MR^2)W_2^2 = \frac{1}{2}(\frac{2}{5}MR^2)W_1^2

\frac{1}{3}MR^2W_2^2 = \frac{1}{5}MR^2W_1^2

\frac{1}{3}W_2^2 = \frac{1}{5}W_1^2

Finally, solving for W_2, we get:

W_2=\sqrt{\frac{3}{5} }W_1

5 0
2 years ago
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