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vitfil [10]
2 years ago
5

Gravitational potential energy (U) depends on three things: the object’s mass (m), its height (h), and gravitational acceleratio

n (g), which is 9.81 m/s2 on Earth’s surface: U = mgh Energy is measured in joules (J). One joule is equal to one 1 kg•m2/s2. When calculating the energy of an object, it is helpful to convert the mass and height to kilograms and meters.
A. What is the mass of the 50-gram car, in kilograms?

B. Set Hill 1 to 75 cm and the other hills to 0 cm. What is the height in meters?

C. What is the potential energy of the car, in joules?
Physics
2 answers:
Viefleur [7K]2 years ago
8 0

A. 500N

B. 0.75 m

C. 7865432J


adoni [48]2 years ago
7 0

Explanation:

A. Mass of the car ,m = 50 g

1 g = 0.001 kg

50 g = 0.001 × 50 kg = 0.05 kg

B. Height of hill ,h= 75 cm = 0.75 m

1 cm = 0.01 m

Height of other hill ,h' = 0 cm = 0 m

C. Gravitational potential energy (U):

U=mgh

Mass of car = m = 0.05 kg

Height at which car is = h = 0.75 m

U= 0.05 kg\times 0.75 m\times 9.8 m/s^2=0.3675 kg/m^2/s^2=0.3675 J

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lisov135 [29]

You asked the question twice I answered it on the last one

7 0
3 years ago
A spring has a spring constant of 330 n/m. how far is the spring compressed when 150 newtons of force are used?
pishuonlain [190]

Answer:

I think its B

Explanation:

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How much work is done when you lift a 5kg bag of groceries 1.5 m?
olga55 [171]

Answer:

7.5J

Explanation:

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8 0
3 years ago
An aluminum bar 600mm long, with diameter 40mm long has a hole drilled in the center of the bar.The hole is 30mm in diameter and
Svetradugi [14.3K]

Answer:

Total contraction on the Bar  = 1.22786 mm

Explanation:

Given that:

Total Length for aluminum bar = 600 mm  

Diameter for aluminum bar  = 40 mm

Hole diameter  = 30 mm

Hole length = 100 mm

elasticity for the aluminum is 85GN/m² = 85 × 10³ N/mm²

compressive load P = 180 KN = 180  × 10³ N

Calculate the total contraction on the bar = ???

The relation used in  calculating the contraction on the bar is:

\delta L = \dfrac{P *L }{A*E}

The relation used in  calculating the total contraction on the bar can be expressed as :

Total contraction in the Bar = (contraction in part of bar without hole + contraction in part of bar with hole)

i.e

Total contraction on the Bar = \dfrac{P *L_1 }{A_1*E} +  \dfrac{P *L_2 }{A_2 *E}

Let's find the area of cross section without the hole and with the hole

Area of cross section without the hole is :

Using A = πd²/4

A = π (40)²/4

A = 1256.64 mm²

Area of cross section with the hole is :

A = π (40²-30²)/4

A = 549.78 mm²

Total contraction on the Bar = \dfrac{P *L_1 }{A_1*E} +  \dfrac{P *L_2 }{A_2 *E}

Total contraction on the Bar  = \dfrac{180 *10^3 \N  }{85*10^3 \ N/mm^2} [\dfrac{500}{1256.64}+ \dfrac{100}{549.78}]

Total contraction on the Bar  = 2.117( 0.398 + 0.182)

Total contraction on the Bar  = 2.117*(0.58)

Total contraction on the Bar  = 1.22786 mm

5 0
3 years ago
Help
photoshop1234 [79]

Answer:

It conserves both energy and momentum in the collision at the same time. By design, when the balls collide the strings that hold them up are vertical (assuming balls are only swung from one side).

Explanation:

Hope This Helps!!

7 0
2 years ago
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