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Ratling [72]
3 years ago
15

If the force of gravity suddenly stopped acting on the planets, they would

Physics
2 answers:
stepladder [879]3 years ago
6 0
They would explode. lol.
Afina-wow [57]3 years ago
3 0
<span>If this were to happen, they would slowly spiral to the sun and continue to orbit the sun, thus moving in straight lines, tangent to their orbits, and slowly spiral away from the sun. </span>
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knowledge of first aid ... eg St John Ambulance, Red Cross etc. I think that everyone in a school should be taught First Aid.

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A 52 N sled is pulled across a cement sidewalk at constant speed. A horizontal force of 36 N is exerted. What is the coefficient
Andre45 [30]

Answer:

μ = 0.692

Explanation:

In order to solve this problem, we must make a free body diagram and include the respective forces acting on the body. Similarly, deduce the respective equations according to the conditions of the problem and the directions of the forces.

Attached is an image with the respective forces:

A summation of forces on the Y-axis is performed equal to zero, in order to determine the normal force N. this summation is equal to zero since there is no movement on the Y-axis.

Since the body moves at a constant speed, there is no acceleration so the sum of forces on the X-axis must be equal to zero.

The frictional force is defined as the product of the coefficient of friction by the normal force. In this way, we can calculate the coefficient of friction.

The process of solving this problem can be seen in the attached image.

5 0
3 years ago
What is the minimum v (in km/s) the rocket engines must provide to allow the craft to escape from the Earth?
coldgirl [10]

Answer:

11 kilometers (7 miles) per second, or over 40,000 kilometers per hour (25,000 miles per hour)

Explanation:

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3 years ago
Marine magnetic anomalies result from seafloor spreading in conjunction with
Artemon [7]
Earths magnetic field
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andreev551 [17]
The question is incomplete. I can help you by adding the information missing. They want you to calculate a) the radius of the cyclotron orbit for an electron with speed 1.0 * 10^6 m/s^2 and b) the radius of a cyclotron orbit for a proton with speed 5.0 * 10^4 m/s.

The two tasks involve combining the equations of the magnectic force and the centripetal force in a circular motion.

When you do that, you will obtain an expression to find the radius of the circular motion, which is the radius of the cyclotron that impulses the particles.

a)

Magentic force, F = q*v*B

q is the charge of the electron = 1.6 * 10^ -19 C
v is the speed = 1.0 * 10 ^ 6 m/s
B is the magentic field = 5.0 * 10 ^-5 T

Centripetal force, F = m*Ac = m * v^2 / R

where,

Ac = centripetal acceleration
m = mass of the electron = 9.11 * 10 ^-31 kg
R = the radius of the orbit

Now equal the two forces: q*v*B = m * v^2 / R => R =  m*v / (q*B)

=> R = (9.11 * 10^31 kg) (1.0*10^6m/s) / [ (1.6 * 10^-19C)* (5.0 * 10^-5T) ]

=> R = 0.114 m

b) The equations are the same, just now use the speed, charge and mass of the proton instead of those of the electron.

R = m*v / (qB) = (1.66*10^-27 kg)(5.0*10^4 m/s) / [(1.6*10^-19C)(5*10^-5T)]

=> R = 10.4 m

 

4 0
3 years ago
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