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Llana [10]
3 years ago
15

Help with this question plz and show how you got it :) tysm!

Physics
1 answer:
spayn [35]3 years ago
5 0
The hardest part of the job is to find the right formula to use, and write it down. You've already done that ! The rest is just turning the crank until an answer falls out.

You wrote. E = m g h.
Beautiful.
Now divide each side by (g h), and you'll have the formula for mass:

m = E / (g h).

You know all the numbers on the right side. Just pluggum in, do the arithmetic, and you'll have the mass.
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A block on a horizontal frictionless plane is attached to a spring, as shown below. The block oscillates along the x-axis with s
AleksandrR [38]

The question is about unclear since no picture provided. But from the question, it could be guessed that the box is moving back and forth on the frictionless plane at the amplitude of A in simple harmonic motion.

Answer:

D. At x=0, it's acceleration is at a maximum

Explanation:

As the box move forward, it reaches point A and than move backward. Theoretically, the box will move backwards, through its origin, to point -A and then going forward.

Point A is the maximum displacement of the box in this case. At this point, the box instantaneously stop to go backward. Therefore the velocity at that moment is zero.

From point -A, the box travel forward and keep building up speed due to the release in potential energy of the spring. And at point x=0, the velocity become maximum. After point x=0, the velocity of the box slows down due to the conversion of kinetic energy to potential energy of the spring. And as it reaches point A, it reaches zero velocity.

The same can be said as the box travels backward from point A to -A

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A wave is produced from a vibrating string on a violin. Two observers are standing 20 m and 40 m away from the musician. The obs
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Well the one that is closer can see and hear more

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Longitudinal waves transfer energy ___________ to the direction of the wave motion.
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Longitudinal waves transfer energy parallel to the direction of the wave motion
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A spherical asteroid of average density would have a mass of 8.7×1013kg if its radius were 2.0 km.A)If you and your spacesuit ha
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A) 0.189 N

The weight of the person on the asteroid is equal to the gravitational force exerted by the asteroid on the person, at a location on the surface of the asteroid:

F=\frac{GMm}{R^2}

where

G is the gravitational constant

8.7×10^13 kg is the mass of the asteroid

m = 130 kg is the mass of the man

R = 2.0 km = 2000 m is the radius of the asteroid

Substituting into the equation, we find

F=\frac{(6.67\cdot 10^{-11})(8.7\cdot 10^{13} kg)(130 kg)}{(2000 m)^2}0.189 N=

B) 2.41 m/s

In order to orbit just above the surface of the asteroid (r=R), the centripetal force that keeps the astronaut in orbit must be equal to the gravitational force acting on the astronaut:

\frac{GMm}{R^2}=\frac{mv^2}{R}

where

v is the speed of the astronaut

Solving the formula for v, we find the minimum speed at which the astronaut should launch himself and then orbit the asteroid just above the surface:

v=\sqrt{\frac{2GM}{R}}=\sqrt{\frac{2(6.67\cdot 10^{-11})(8.7\cdot 10^{13} kg)}{2000 m}}=2.41 m/s

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