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3 years ago

What causes the apparent motions of stars across the sky each night

Physics

1 answer:

densk [106]3 years ago

3 0

Answer:

Explanation:

for the motion every night, it is because the Earth spins on its axis.

for the changes over a year, this is because the Earth rotates around the sun.

for the changes over hundreds, thousands or millions of years, this is because the whole solar system (including the Earth, of course) rotates with all the other stars in our galaxy around the center of the galaxy. and each star system has its own orbit (similar to the planets in our solar system).

it is very rare to see objects outside of our galaxy without a telescope, but they change too over a long period of time, because our galaxy not only rotates by also moves through the universe, and these other objects move on their own too.

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A 1000-kg car is moving at 30 m/s around a horizontal unbanked curve whose diameter is 0.20 km. What is the magnitude of the fri

omeli [17]

Answer:

4500 N

Explanation:

When a body is moving in a circular motion it will feel an acceleration directed towards the center of the circle, this acceleration is:

a = v^2/r

where v is the velocity of the body and r is the radius of the circumference:

Therefore, a body with mass m, will feel a force f:

f = m v^2/r

Therefore we need another force to keep the body(car) from sliding, this will be given by friction, remember that friction force is given a the normal times a constant of friction mu, that is:

fs = μN = μmg

The car will not slide if f = fs, i.e.

fs = μmg = m v^2/r

That is, the magnitude of the friction force must be (at least) equal to the force due to the centripetal acceleration

fs = (1000 kg) * (30m/s)^2 / (200 m) = 4500 N

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Assume that the loop is initially positioned at θ=30∘θ=30∘ and the current flowing into the loop is 0.500 AA . If the magnitude

labwork [276]

Answer: $\tau=1.03\times 10^{-4}\ N-m$

Torque,

Explanation:

Given that,

The loop is positioned at an angle of 30 degrees.

Current in the loop, I = 0.5 A

The magnitude of the magnetic field is 0.300 T, B = 0.3 T

We need to find the net torque about the vertical axis of the current loop due to the interaction of the current with the magnetic field. We know that the torque is given by :

$\tau=NIAB\ \sin\theta$

Let us assume that, $A=0.0008\ m^2$

$\theta$ is the angle between normal and the magnetic field, $\theta=90^{\circ}-30^{\circ}=60^{\circ}$

Torque is given by :

$\tau=1\times 0.5\ A\times 0.0008\ m^2\times 0.3\ T\ \sin(60)\\\\\tau=1.03\times 10^{-4}\ N-m$

So, the net torque about the vertical axis is $1.03\times 10^{-4}\ N-m$ . Hence, this is the required solution.

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