The mass of the planet Gallifrey is 8 times the mass of the Earth.
- let the gravitational field of Earth = g
- let the radius of the Earth = R
- gravitational field of Gallifrey = 2g
- radius of Gallifrey = 2R
<h3>What is gravitational potential energy?</h3>
- This is the work done in moving an object to a certain distance against gravitational field.
The gravitational field strength of the Earth is given as follows;

The gravitational field strength of the Planet Gallifrey is calculated as follows;


Thus, the mass of the planet Gallifrey is 8 times the mass of the Earth.
Learn more about gravitational field strength here: brainly.com/question/14080810
The correct answer is an ''electric current''.
Answer:

Explanation:
Kinetic energy is energy due to motion. The formula is half the product of mass and velocity squared.

The mass of the roller coaster car is 2000 kilograms and the car is moving 10 meters per second.
Substitute these values into the formula.

Solve the exponent.
- (10 m/s)²= 10 m/s * 10 m/s= 100 m²/s²

Multiply the first two numbers together.

Multiply again.

- 1 kilogram square meter per square second is equal to 1 Joule.
- Our answer of 100,000 kg*m²/s² is equal to 100,000 Joules.

The roller coaster car has <u>100,000 Joules</u> of kinetic energy.
1.5 m/s is the velocity.
9.3 m is the length of aisle, over which Distance will be covered.
Time is demanded in which the child will move the cart over the aisle with 1.5 m/s.
v=S/t
and,
t=S/v
Put values,
t=9.3/1.5=6.2 s
Answer:

Explanation:
First, we calculate the work done by this force after the box traveled 14 m, which is given by:
![W=\int\limits^{x_f}_{x_0} {F(x)} \, dx \\W=\int\limits^{14}_{0} ({18N-0.530\frac{N}{m}x}) \, dx\\W=[(18N)x-(0.530\frac{N}{m})\frac{x^2}{2}]^{14}_{0}\\W=(18N)14m-(0.530\frac{N}{m})\frac{(14m)^2}{2}-(18N)0+(0.530\frac{N}{m})\frac{0^2}{2}\\W=252N\cdot m-52N\cdot m\\W=200N\cdot m](https://tex.z-dn.net/?f=W%3D%5Cint%5Climits%5E%7Bx_f%7D_%7Bx_0%7D%20%7BF%28x%29%7D%20%5C%2C%20dx%20%5C%5CW%3D%5Cint%5Climits%5E%7B14%7D_%7B0%7D%20%28%7B18N-0.530%5Cfrac%7BN%7D%7Bm%7Dx%7D%29%20%5C%2C%20dx%5C%5CW%3D%5B%2818N%29x-%280.530%5Cfrac%7BN%7D%7Bm%7D%29%5Cfrac%7Bx%5E2%7D%7B2%7D%5D%5E%7B14%7D_%7B0%7D%5C%5CW%3D%2818N%2914m-%280.530%5Cfrac%7BN%7D%7Bm%7D%29%5Cfrac%7B%2814m%29%5E2%7D%7B2%7D-%2818N%290%2B%280.530%5Cfrac%7BN%7D%7Bm%7D%29%5Cfrac%7B0%5E2%7D%7B2%7D%5C%5CW%3D252N%5Ccdot%20m-52N%5Ccdot%20m%5C%5CW%3D200N%5Ccdot%20m)
Since we have a frictionless surface, according to the the work–energy principle, the work done by all forces acting on a particle equals the change in the kinetic energy of the particle, that is:

The box is initially at rest, so
. Solving for
:
