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3 years ago

What is the formula for finding gravitational potential energy??

Physics

2 answers:

allochka39001 [22]3 years ago

7 0

Gravitational potential energy = m * g* h

Explanation:

Gravitational potential is referred as the energy spent to move the mass from one point to another. It is represented in the unit as Joules per kilogram.

Gravitational potential energy = m * g * h where m denotes the mass of the object, g denotes the acceleration due to gravity which is 9.8 m / s 2, and h denotes the height in meters.

MrMuchimi3 years ago

3 0

The equation for gravitational potential energy is GPE = mgh, where m is the mass in kilograms, g is the acceleration due to gravity (9.8 on Earth), and h is the height above the ground in meters.

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A car travels 20 km South, then turns and travels 30 km East? What is the total displacement?

Ne4ueva [31]

Answer:

Explanation:36.05 km

Given

First car travels $r_1=20\ km$ South

then turns and travels $r_2=30\ km$ east

Suppose south as negative y axis and east as positive x axis

So, $r_1=-20\hat{j}$

$r_2=30\hat{i}$

Displacement is the shortest between initial and final point

Dispalcement $=r=r_1+r_2$

Displacement $=-20\hat{j}+30\hat{i}$

Displacement $=30\hat{i}-20\hat{j}$

Magnitude $=\sqrt{30^2+(-20)^2}$

Magnitude $=36.05\ km$

4 0

3 years ago

A racecar experiences a centripetal acceleration of 36.0 m/s2 as it travels at a constant speed of 27.0 m/s along a circular arc

alexdok [17]

Answer:

20.25 m

Explanation:

Centripetal acceleration is given by; the square of the velocity, divided by the radius of the circular path.

That is;

ac = v²/r

Where; ac = acceleration, centripetal, m/s², v is the velocity, m/s and r is the radius, m

Therefore;

r = v²/ac

= 27²/36

= 20.25 m

Hence the radius is 20.25 meters

5 0

3 years ago

Read 2 more answers

A distant planet with a mass of (7.2000x10^26) has a moon with a mass of (5.0000x10^23). The distance between the planet and the

BARSIC [14]

Answer:

Explanation:

This is a simple gravitational force problem using the equation:

$F_g=\frac{Gm_1m_2}{r^2}$ where F is the gravitational force, G is the universal gravitational constant, the m's are the masses of the2 objects, and r is the distance between the centers of the masses. I am going to state G to 3 sig fig's so that is the number of sig fig's we will have in our answer. If we are solving for the gravitational force, we can fill in everything else where it goes. Keep in mind that I am NOT rounding until the very end, even when I show some simplification before the final answer.

Filling in:

$F_g=\frac{(6.67*19^{-11})(7.2000*10^{26})(5.0000*10^{23})}{(6.10*10^{11})^2}$ I'm going to do the math on the top and then on the bottom and divide at the end.

$F_g=\frac{2.4012*10^{40}}{3.721*10^{23}}$ and now when I divide I will express my answer to the correct number of sig dig's:

$Fg=$ 6.45 × 10¹⁶ N

8 0

3 years ago

A fish swims 12.0 m in 5.0 s. It swims the first 4.0 m in 2.0 s, the next 3.0 m in 1.2 s, and the last 5.0 m in 1.8 s. What is t

xz_007 [3.2K]

Use the distance swan and the time elapsed in that interval.

Average velocity = distance / time

Average velocity = [4.0 m + 3.0m] / 3.2 s = 2.1875 m/s

7 0

3 years ago

Read 2 more answers

A hockey puck is sliding across a frozen pond with an initial speed of 9.3 m/s. It comes to rest after sliding a distance of 42.

kondaur [170]

Answer:

The coefficient of kinetic friction between the puck and the ice is 0.11

Explanation:

Given;

initial speed, u = 9.3 m/s

sliding distance, S = 42 m

From equation of motion we determine the acceleration;

v² = u² + 2as

0 = (9.3)² + (2x42)a

- 84a = 86.49

a = -86.49/84

|a| = 1.0296

$F_k = \mu_k N$ = ma

where;

Fk is the frictional force

μk is the coefficient of kinetic friction

N is the normal reaction = mg

μkmg = ma

μkg = a

μk = a/g

where;

g is the gravitational constant = 9.8 m/s²

μk = a/g

μk = 1.0296/9.8

μk = 0.11

Therefore, the coefficient of kinetic friction between the puck and the ice is 0.11

3 0

3 years ago