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vova2212 [387]
3 years ago
7

When a 15.00 kg mass is attached to a vertical spring, the spring is stretched 2.0 m such that the mass is 6.0 m above the table

. The spring constant is 400.0N/m? What is the total potential energy associated with this system?
Physics
1 answer:
vlada-n [284]3 years ago
3 0
Add the two potential energy
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What do these letters stand for<br> P=mv
victus00 [196]

Answer:

The equation for momentum of a piece of matter.

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Explanation:

Hope This Helps

Have A Great Day

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Describe free fall and weightlessness ?<br>​
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A 25,000 kg traveling east collides with a 2,000 kg truck standing still on the tracks. After the collision the train and truck
Elis [28]

Answer:

24.084 m/s

Explanation:

From the law of conservation of linear momentum

Total momentum before collision equals to the total momentum after collision

Since momentum=mv where m is mass and v is velocity

M_{truck}V_{truck}=V_{common}*(M_{truck} +M_{standing}) where M_{truck} is the mass of the truck, V_{truck} is velocity of the truck, V_{common} is the common velocity of moving and standing truck after collision and M_{standing} is the mass of the standing truck

Making V_{truck} the subject we obtain

V_{truck}=\frac { V_{common}*(M_{truck} +M_{standing})}{M_{truck}}

Substituting M_{truck} as 25000 Kg, V_{common} as 22.3 m/s, M_{standing} as 2000 Kg we obtain

V_{truck}=\frac { 22.3 m/s *(25000 Kg +2000 Kg)}{25000}= 24.084 m/s

Therefore, assuming no friction and considering that after collision they still move eastwards hence common velocity and initial truck velocities are positive

The truck was moving at 24.084 m/s

3 0
3 years ago
A 1100 kg sports car accelerates from 0m/s to 32m/s in 10 seconds. What is the average power of the engine?
timurjin [86]

Answer:

The average power of the engine of the sports car is 56.32 kW

Explanation:

Given;

mass of the sports car, m = 1100 kg

initial velocity of the sports car, u = 0 m/s

final velocity of the sports car, v = 32 m/s

time of motion, t = 10 s

The kinetic energy of the car is given by;

K.E = ¹/₂m(v² - u²)

K.E = ¹/₂mv²

K.E = ¹/₂ x 1100 x 32²

K.E = 563200 J

The average power of the engine of the sports car is given by;

Pavg = Energy / time

Pavg = 563200 / 10

Pavg = 56320 W

Pavg = 56.32 kW

Therefore, the average power of the engine of the sports car is 56.32 kW

8 0
3 years ago
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