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WINSTONCH [101]
3 years ago
13

A 5kg rock is lifted 2m. Find the amount of work done. ​

Physics
1 answer:
lianna [129]3 years ago
5 0

Answer:

98J

Explanation:

Given parameters:

Mass of rock  = 5kg

Height = 2m

Unknown:

Work done  = ?

Solution:

The amount of work done is given as:

 Work done  = Force x distance

 Work done = Weight x height

  Work done  = mgH

Now insert the parameters and solve;

 Work done  = 5 x 9.8 x 2 = 98J

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The nebular hypothesis is the name of the theory where gravitational attraction caused bits of matter to collapse, eventually forming a solar system.
Hope this helps :)
7 0
3 years ago
Question 7 (1 point)
Temka [501]

Answer:

Forms over water, warm humid air mass, it's a polar air mass

Explanation: I think that's right sorry if it's not..

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4 0
3 years ago
A youngster having a mass of 50.0 kg steps off a 1.00 m high platform. If she keeps her legs fairly rigid and comes to rest in 1
attashe74 [19]

The average act on her during the deceleration is 4.47 meters per second.                                                                

<u>Explanation</u>:

<u>Given</u>:

youngster mass m = 50.0 kg

She steps off a 1.00 m high platform that is s = 1 meter

She comes to rest in the 10-meter second

<u>To Find</u>:

The average force and momentum

<u>Formulas</u>:

p = m * v

F * Δ t = Δ p

vf^2= vi^2+2as

<u>Solution</u>:

a = 9.8 m/s

vi = 0

vf^2= 0+2(9.8)(1)

vf^2 = 19.6

vf = 4.47 m/s .

Therefore the average force is 4.47 m/s.                                          

                                 

5 0
3 years ago
A 1200 kg car reaches the top of a 100 m high hill at A with a speed vA. What is the value of vA that will allow the car to coas
Pavel [41]

Complete question is:

A 1200 kg car reaches the top of a 100 m high hill at A with a speed vA. What is the value of vA that will allow the car to coast in neutral so as to just reach the top of the 150 m high hill at B with vB = 0 m/s. Neglect friction.

Answer:

(V_A) = 31.32 m/s

Explanation:

We are given;

car's mass, m = 1200 kg

h_A = 100 m

h_B = 150 m

v_B = 0 m/s

From law of conservation of energy,

the distance from point A to B is;

h = 150m - 100 m = 50 m

From Newton's equations of motion;

v² = u² + 2gh

Thus;

(V_B)² = (V_A)² + (-2gh)

(negative next to g because it's going against gravity)

Thus;

(V_B)² = (V_A)² - (2gh)

Plugging in the relevant values;

0² = (V_A)² - 2(9.81 × 50)

(V_A) = √981

(V_A) = 31.32 m/s

3 0
3 years ago
When performing the spike the ball should be hit with the ?
VMariaS [17]

Answer:

When you are performing spike it's most effective to strike the ball from the right or left side at a sharp downward angle. Whether you are spiking the ball from the right or left front position, position yourself behind the 10-foot line (attack line), which is the line that is about four steps away from the net.

5 0
3 years ago
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