In 'coin on card' experiment a smooth card is used.

Physics

2 answers:

liq [111]3 years ago

8 0

In three words I can sum up everything I’ve learned about life: it goes on. ― Robert Frost

KIM [24]3 years ago

5 0

Answer:

In coin card experiment smooth card is used so that the card can slide easily from glass

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The planet Gallifrey has 2 times the gravitational field strength and 2 times the radius of the Earth.

Alexeev081 [22]

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;

$g = \frac{GM}{r^2} \\\\G = \frac{gr^2}{M}$

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

$g_2 = \frac{GM_2}{r_2^2}$

$G = \frac{g_2r_2^2}{M_2} \\\\\frac{g_2r_2^2}{M_2} = \frac{gr^2}{M}\\\\M_2gr^2 = Mg_2r_2^2\\\\M_2 = \frac{Mg_2r_2^2}{gr^2} \\\\M_2 = \frac{M \times 2g \times (2r)^2}{gr^2} \\\\M_2 = \frac{M \times 2g \times 4r^2}{gr^2} \\\\M_2 = 8M$

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

4 0

2 years ago

If you have to stop on an incline, your stopping distance will be __________ on a flat surface.

Nataliya [291]

Not as far on a flat surface

5 0

3 years ago

A woman takes her dog Rover for a walk on a leash. She pulls on the leash with a force of 30.0 N at an angle of 29° above the ho

ValentinkaMS [17]

Answer:

The force parallel to the horizontal is 26.24 N

Explanation:

She pulls on the leash with a force F = 30 N, this force, since its at an angle of 29° (i will cal this angle $\theta$ ), it has a force component on x (the horizontal, i will call this force $F_{x}$ ) and a force component on y (the vertical, i will call this $F_{y}$ ).

This can be seen in the attached picture.

Since we are asked about the force parallel to the horizontal, we need to find the component of the force $F_{x}$ , since $F_{x}$ is the adjacent angle, we need to use cosine:

$F_{x}=Fcos \theta$