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IRISSAK [1]
3 years ago
15

8. The moon has about 1/6th the gravity of the Earth. How would the potential energy of an object be different on the moon if it

was the same height above the ground? (5 pts)
Physics
1 answer:
topjm [15]3 years ago
7 0

The potential energy on the moon would be 1/6 of the potential energy on Earth

Explanation:

The potential energy of an object is the energy possessed by the object due to its position in a gravitational field, and it is given by

PE=mgh

where

m is the mass of the object

g is the strength of the gravitational field

h is the height of the object above the ground

In this problem, the potential energy of the object on the Earth is

PE=mgh

On the Moon, the gravity is

g'=\frac{1}{6}g

Therefore, if the same object (same mass) is placed at the same height h on the Moon, its potential energy would be

PE' = m(\frac{1}{6}g)h = \frac{1}{6}(mgh) = \frac{1}{6}PE'

Learn more about potential energy:

brainly.com/question/1198647

brainly.com/question/10770261

#LearnwithBrainly

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Car a has a mass of 5 kg and 10 m/s .cart B has a mass of 10 kg and 5 m/s .which of the following statements best compares the i
weeeeeb [17]

The two cars have same momentum

Explanation:

The momentum of an object is given by the equation

p=mv

where

m is the mass of the object

v is its velocity

For the car A in this problem,

m = 5 kg

v = 10 m/s

So its momentum is

p_A = (5 kg)(10 m/s) = 50 kg m/s

For car B we have,

m = 10 kg

v = 5 m/s

So its momentum is

p_B = (10 kg)(5 m/s)=50 kg m/s

Therefore, the two cars have same momentum.

Learn more about momentum:

brainly.com/question/7973509

brainly.com/question/6573742

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7 0
2 years ago
You are driving to the grocery store at 14 m/s. You are 115 m from an intersection when the traffic light turns red. Assume that
kogti [31]

Answer:

111.5 m

Explanation:

Given that You are driving to the grocery store at 14 m/s. You are 115 m from an intersection when the traffic light turns red. Assume that your reaction time is 0.50 s and that your car brakes with constant acceleration. 

Use first equation of motion

V = U - at

Since the car is going to rest, V = 0 and a = negative

0 = 14 - a × 0.5

0.5a = 14

a = 14 /0.5

a = 28 m/s^2

Let us use second equation of motion

S = Ut - 1/2at^2

S = 14 × 0.5 - 0.5 × 28 × 0.5^2

S = 7 - 3.5

S = 3.5 m

115 - 3.5 = 111.5

Therefore, you are 111.5 metres from the intersection (in m) when you begin to apply the brakes.

6 0
3 years ago
An object with a mass of 2.0 kg is accelerated at 5.0 m/s/s. the net force acting on the mass is
Vitek1552 [10]
First we need to know the equation:
F= Mass times acceleration
F = 2.0 kg times 5.0 m/s^2 
multiply them to get the net force!
F= 10 N 
N is newton 
Hope this heps
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3 years ago
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solmaris [256]
The answer would be hypothesis testing
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2 years ago
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