Urgent please physic

How long does it take mercury to orbit the sun?

Oliga [24]

That is called the "Revolution Period of a planet" and one year on mercury equals to 88 days as compared to Earth

In short, Your Answer would be 88

Hope this helps!

3 0

2 years ago

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A proton and an electron are released from rest, with only the electrostatic force acting. Which of the following statements mus

Anit [1.1K]

Answer:

Their kinetic energy will increase but potential energy will decrease.

Explanation:

Given that

Initial velocities of electron and proton is zero.

We know that ,electron have negative charge and proton have positive charge it means that they will attract to each other.We know that opposite charge attract to each other and same charge repels to each other.

It means that the velocities of proton and electron will increase and that leads to increase in the kinetic energy of proton and electron.

We know that potential energy U

$U\alpha -\dfrac{1}{r}$

So when r will decrease then U will increase but in negative direction it means that U will decrease.

So we can say that their kinetic energy will increase but potential energy will decrease.

5 0

3 years ago

A 75-g bullet is fired from a rifle having a barrel 0.540 m long. choose the origin to be at the location where the bullet begin

lyudmila [28]

Part a) The work done by the gas on the bullet is the integral of the force in dx, where x is the distance covered by the bullet inside the barrel with respect to the origin:
$W= \int\limits^{0.540m}_{0} {F} \, dx = \int\limits^{0.540m}_{0} {(16000+10000x-26000x^2)} \, dx =$
$=16000x+10000 \frac{x^2}{2} - 26000 \frac{x^3}{3}$
By substituting the length of the barrel, L=0.540 m, we find the total work done by the gas on the bullet:
$W=16000(0.540m)+10000 \frac{(0.540m)^2}{2} - 26000 \frac{(0.540m)^3}{3} =$
$=8733 J=8.73 kJ$

part b) The resolution of the problem is the same, we just have to use the new length of the barrel (L=0.95 m) inside the final formula, and we find the new value of the work:
$W=16000(0.95m)+10000 \frac{(0.95m)^2}{2} - 26000 \frac{(0.95m)^3}{3} =$
$=12280 J=12.28 kJ$

5 0

2 years ago

What is the speed over the ground of a mosquito flying 2 m/s relative to the air against a 2 m/s headwind?

salantis [7]

We have that the speed over the ground of a mosquito flying 2 m/s relative to the air against a 2 m/s headwind is

X=0m/s

From the question we are told that

mosquito flying 2 m/s

against a 2 m/s headwind

Generally

The speed over the ground is the Flight Speed minus resistance speed

Generally the equation for the speed over the ground is mathematically given as

X=Flight Speed-resistance speed

Therefore

X=2-2

X=0m/s

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8 0

2 years ago

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A submarine is 2.84 102 m horizontally from shore and 1.00 102 m beneath the surface of the water. A laser beam is sent from the

xxMikexx [17]

Answer:

468 m

Explanation:

So the building and the point where the laser hit the water surface make a right triangle. Let's call this triangle ABC where A is at the base of the building, B is at the top of the building, and C is where the laser hits the water surface. Similarly, the submarine, the projected submarine on the surface and the point where the laser hit the surface makes a another right triangle CDE. Let D be the submarine and E is the other point.

The length CE is length AE - length AC = 284 - 234 = 50 m

We can calculate the angle ECD:

$tan(\hat{ECD}) = \frac{ED}{EC} = \frac{100}{50} = 2$