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polet [3.4K]
3 years ago
8

A 2 kg picture frame sits on a shelf at a height of 0.5 m. how much gravitational potential energy is added to the picture frame

when it is lifted to a shelf of height 1.3 m? acceleration due to gravity is g=9.8 m/s^2
A. 15.68 J
B. 10.23 J
C. 25.48 J
D. 9.80 J
Physics
1 answer:
AysviL [449]3 years ago
7 0

The change in gravitational potential energy is A) 15.68 J

Explanation:

The gravitational potential energy of an object is the energy possessed by the object due to its position in the gravitational field.

The change in gravitational potential energy of an object is given by the equation:

\Delta U = mg\Delta h

where

m is the mass of the object

g=9.8 m/s^2 is the acceleration due to gravity

\Delta h is the change in height of the object

In this problem, we have

m = 2 kg is the mass of the frame

\Delta h = 1.3 - 0.5 = 0.8 m is the change in height

Substituting,  we find:

\Delta U = (2)(9.8)(0.8)=15.68 J

Learn more about potential energy:

brainly.com/question/1198647

brainly.com/question/10770261

#LearnwithBrainly

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what evidence can you cite that the interstellar medium contains both gas and dust? (select all that apply.)
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(a) The gas of interstellar medium can be detected from the radiations of photons of  wavelength 21 cm.

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4 0
2 years ago
7. A toy car of mass 1.2 kg is driving vertical circles inside a hollow cylinder of radius 2.0m. It is moving at a constant spee
wlad13 [49]

Answer:

a)

N_{top}=9.8N\\N_{bottom}=33.4N

b) v_{min}=4.4m/s

Explanation:

The net force on the car must produce the centripetal acceleration necessary to make this circle, which is a_{cp}=\frac{v^2}{R}. At the top of the circle, the normal force and the weight point downwards (like the centripetal force should), while at the bottom the normal force points upwards (like the centripetal force should) and the weight downwards, so we have (taking the upwards direction as positive):

-m\frac{v^2}{R}=-N_{top}-mg\\m\frac{v^2}{R}=N_{bottom}-mg

Which means:

N_{top}=m\frac{v^2}{R}-mg=(1.2kg)\frac{(6m/s)^2}{2m}-(1.2kg)(9.8m/s^2)=9.8N\\N_{bottom}=m\frac{v^2}{R}+mg=(1.2kg)\frac{(6m/s)^2}{2m}+(1.2kg)(9.8m/s^2)=33.4N

The limit for falling off would be N_{top}=0, so the minimum speed would be:

0=m\frac{v_{min}^2}{R}-mg\\v_{min}=\sqrt{Rg}=\sqrt{(2m)(9.8m/s^2)}=4.4m/s

3 0
3 years ago
shelly starts from rest on her bicycle at the top of a hill. After 6.0s she has reached a final velocity of 14m/s. what is shell
earnstyle [38]
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8 0
3 years ago
Read 2 more answers
a particle with a charge of 5.5 x 10^-8 c is 3.5 cm from a particle with a charge of -2.3 x10^-8 c. the potential energy of this
Yuri [45]

Answer:

-32.5 * 10^-5 J

Explanation:

The potential energy of this system of charges is;

Ue = kq1q2/r

Where;

k is the Coulumb's constant

q1 and q2 are the magnitudes of the charges

r is the distance of separation between the charges

Substituting values;

Ue = 9.0×10^9 N⋅m2/C2 * 5.5 x 10^-8 C *( -2.3 x10^-8) C/(3.5 * 10^-2)

Ue= -32.5 * 10^-5 J

4 0
3 years ago
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