All categories

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

<u>Explanation:</u>

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.

You might be interested in

ken, 0.75 kg, moves toward a wall (his path normal to the wall) at 52 m/s. 13.0 ms after he touches the wall he pushes himself o

shtirl [24]

Q: ken, 0.75 kg, moves toward a wall (his path normal to the wall) at 52 m/s. 13.0 ms after he touches the wall he pushes himself off in the opposite direction at 60 m/s. What is the magnitude of the average force the wall exerts on Ken during this rapid maneuver

Answer:

-6461.54 N

Explanation:

From Newton's Fundamental equation,

F = m(v-u)/t.................... Equation 1

Where F = Force exerted in sonic, m = mass of ken, v = final velocity, u = initial velocity, t = time.

Given: m = 0.75 kg, v = - 60 m/s (opposite direction), u = 52 m/s, t = 13 ms = 0.013 s

Substitute into equation 1

F = 0.75(-60-52)/0.013

F = 0.75(-112)/0.013

F = -84/0.013

F = -6461.54 N

Note: The negative sign tells that the force act in opposite direction to the initial motion of ken.

Hence the magnitude of the average force of the wall = -6461.54 N

4 0

3 years ago

Un hamster esta sentado sobre un tocadisco cuya rapidez angular es constante si el hamster se mueve a un punto localizado al dob

kotegsom [21]

Answer:

b) se duplica

Explanation:

The disk is moving with constant angular velocity, let's call it $\omega$ .

The linear velocity of a point on the disk is given by

$v=\omega r$

where r is the distance of the point from the axis of rotation.

In this problem, the object is moved at a distance twice as far as the initial point, so

$r' = 2r$

Therefore, the new linear velocity is

$v'=\omega r' = \omega (2r) = 2 \omega r = 2 v$

So, the velocity has doubled, and the correct answer is

b) se duplica

8 0

2 years ago

An automobile with a tangential speed of 54.1 km/h follows a circular road that has a radius of 41.6 m. The automobile has a mas

Viefleur [7K]

1) Available force of friction: 6174 N

2) No

Explanation:

The magnitude of the frictional force between the car's tires and the pavement of the road is given by

$F_f=\mu mg$

where

$\mu$ is the coefficient of friction

m is the mass of the car

g is the acceleration of gravity

For the car in this problem, we have:

$\mu=0.500$ (coefficient of friction)

m = 1260 kg (mass of the car)

$g=9.8 m/s^2$

Therefore, the force of friction is

$F_f=(0.500)(1260)(9.8)=6174 N$

In order to mantain the car in circular motion, the force of friction must be at least equal to the centripetal force.

The centripetal force is given by

$F=m\frac{v^2}{r}$

where

m is the mass of the car

v is the tangential speed

r is the radius of the curve

In this problem, we have

m = 1260 kg

$v=54.1 km/h =15.0 m/s$ is the tangential speed

r = 41.6 m is the radius of the curve

Therefore, the centripetal force is

$F=(1260)\frac{15.0^2}{41.6}=6814 N$

Therefore, the force of friction is not enough to keep the car in the curve, since $F_f$

4 0

3 years ago

A truck is traveling at 27 m/s down the interstate highway where you are changing a flat tire. frequency of 185 Hz.

Reil [10]

Answer:

(a) the observed frequency is 200 Hz

(b) the observed frequency is 188 Hz.

Explanation:

speed of the truck, Vs = 27 m/s

frequency of the truck as it approaches, Fs = 185 Hz

(a) Apply Doppler effect to determine the frequency you will hear.

As the truck approaches you, the observed frequency will be higher than the source frequency because of decrease in distance.

$F_s = F_o [\frac{V}{V_S + V} ]$

Where;

Fo is the observed frequency which is the frequency you will hear.

V is speed of sound in air

$F_s = F_o [\frac{V}{V_S + V} ]\\\\185 = F_o [\frac{340}{27 + 340} ]\\\\185 = F_o (0.926)\\\\F_o = \frac{185}{0.926}\\\\F_o = 199.78 \ Hz$

$F_o = 200 \ Hz$

(b) Apply the following formula for a moving observer and a moving source;

$F_o = F_s[\frac{V-V_o}{V} ](\frac{V}{V-V_S} )$

The observed frequency is negative since you are driving away from the truck and the source frequency is also negative since it is driving towards you.

$F_o = F_s[\frac{V-V_o}{V} ](\frac{V}{V-V_S} )\\\\F_o = 185[\frac{340-22}{340} ](\frac{340}{340-27} )\\\\F_o = 185(0.9353)(1.0863)\\\\F_o = 188 \ Hz$