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Genrish500 [490]
2 years ago
9

The relationship between distance from the sun and orbital period is that as the distance from the sun increases, the orbital pe

riod increases.
We can see this because as we go through the planets in our solar system, the farther away they are from the Sun, the longer their orbital period becomes. For example, Mercury is about 67 million km away from the Sun and has an orbital period of 88 days, where as Neptune is about 4.5 billion km away from the Sun and has an orbital period of 165 years.

This tells us that as the distance from the sun increases, gravity _________________. Because gravity _______________, the orbital period is __________________.
Physics
1 answer:
iren [92.7K]2 years ago
4 0
Second law is it:))))))))
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A point charge with charge q1 = 4.00 μC is held stationary at the origin. A second point charge with charge q2 = -4.40 μC moves
Bezzdna [24]

Answer:

W=0.94J

Explanation:

Electrostatic potential energy is the energy that results from the position of a charge in an electric field. Therefore, the work done to move a charge from point 1 to point 2 will be the change in electrostatic potential energy between point 1 and point 2.

This energy is given by:

U=\frac{K\left |q_1 \right |\left |q_2 \right |}{r}\\

So, the work done to move the chargue is:

W=U_1-U_2\\W=\frac{K\left |q_1 \right |\left |q_2 \right |}{r_1}-\frac{K\left |q_1 \right |\left |q_2 \right |}{r_2}\\r_1=\sqrt{((0.155 m)^2+0 m)^2}=0.115m\\r_2=\sqrt{((0.245 m)^2+(0.270 m)^2}=0.365m\\W=K\left |q_1 \right |\left |q_2 \right |(\frac{1}{r_1}-\frac{1}{r_2})\\W=8.99*10^9\frac{Nm^2}{c^2}(4.00*10^{-6}C)(4.40*10^{-6}C)(\frac{1}{0.115m}-\frac{1}{0.365})\\W=0.94J

The work is positive since the potential energy in 1 is greater than 2.

5 0
3 years ago
Please answer both questions and not just one. Thanks!
Margarita [4]
It would be d and c hoped i helped!
8 0
3 years ago
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
WITCHER [35]

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

3 0
3 years ago
Two tugboats are moving a barge. Tugboat A pushes on the barge with a force of 3000n. Tugboat B pulls with a force of 5000 Newto
kenny6666 [7]

The net force on the barge is 8000 N

Explanation:

In order to find the net force on the badge, we have to use the rules of vector addition, since force is a vector quantity.

In this problem, we have two forces:

  • The force of tugboat A, F_A = 3000 N, acting in a certain direction
  • The force of tugboat B, F_B = 5000 N, also acting in the same direction

Since the two forces act in the same direction, this means that we can simply add their magnitudes to find the net combined force on the barge. Therefore, we get

F=F_A+F_B = 3000 + 5000 = 8000 N

and the direction is the same as the direction of the two forces.

Learn more about forces:

brainly.com/question/11179347

brainly.com/question/6268248

#LearnwithBrainly

5 0
3 years ago
A boat travels north across a river at a velocity of 22 meters/second with respect to the water. The river's velocity is 2.2 met
MArishka [77]
The magnitude of the resultant is

         √ (22² + 2.2²)  =  √ (484 + 4.84)  =  √488.84  =  22.11 m/s .

The direction of the resultant is

         tan⁻¹(22N / 2.2E)  =  tan⁻¹(10)  =  5.71° east of north .
7 0
3 years ago
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