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ahrayia [7]
3 years ago
5

Calculate the Torque that a child must apply to swing a rock of mass 0.4 kg in a radius 0.91 meters. From rest to a final linear

velocity of 6 m/s in 3 seconds.
Physics
1 answer:
antoniya [11.8K]3 years ago
6 0

Answer:

who you ganna throw that rock at child

Explanation:

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How can seismographs be used to predict hurricane intensity?
nikdorinn [45]

Answer: The earth is a noisy place. Seismometers, which measure ground movements to detect earthquakes, volcanic eruptions, and manmade explosives, are constantly recording smaller vibrations caused by ocean waves, rushing rivers, and industrial activity.

Explanation:

6 0
3 years ago
If each of the charges is increased by two times and the distance between them is also increased by two times, the electromagnet
PIT_PIT [208]

Answer: The force does not change.

Explanation:

The force between two charges q₁ and q₂ is:

F = k*(q₁*q₂)/r^2

where:

k is a constant.

r is the distance between the charges.

Now, if we increase the charge of each particle two times, then the new charges will be: 2*q₁ and 2*q₂.

If we also increase the distance between the charges two times, the new distance will be 2*r

Then the new force between them is:

F = k*(2*q₁*2*q₂)/(2*r)^2 = k*(4*q₁*q₂)/(4*r^2) = (4/4)*k*(q₁*q₂)/r^2 = k*(q₁*q₂)/r^2

This is exactly the same as we had at the beginning, then we can conclude that if we increase each of the charges two times and the distance between the charges two times, the force between the charges does not change.

8 0
3 years ago
already did the work, i just need someone to see if i did the tangent line right? the line has to touch the point 0.6! thank you
Dmitrij [34]

The tangent looks good.

The curve is a bit crooked, at the 0.9 and 1.

But overall, cool graph.

5 0
3 years ago
How can a moving coil galvanometer can be made into a dc ammeter?
Dennis_Churaev [7]
I am absolutely sure that the way how can a moving coil galvanometer can be made into a dc ammeter is of course by connecting a. low resistance across the meter. You should remember that you must connect <span>a shunt resistor straight across the galvanometer. Do hope this answer will help you! Regards.</span>
7 0
3 years ago
A factory worker pushes a 30.0-kg crate a distance of 4.5 m along a level floor at constant velocity by pushing horizontally on
Olin [163]
The crate moves at constant velocity, this means that its acceleration is zero, so the net force acting on the crate is zero (Newton's second law). 

There are only two forces acting on the crate: the force F applied by the worker and the frictional force, acting in the opposite direction: \mu m g, where \mu=0.25 is the coefficient of friction and m=30.0 kg is the mass of the crate. Since the net force should be equal to zero, the two forces must have same magnitude, so we have:
F=\mu m g=(0.25)(30.0 kg)(9.81 m/s^2)=73.8 N
And so, this is the force that the worker must apply to the crate.
5 0
3 years ago
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