All categories

3 years ago

How much work is done when a 5 N force moves a block 4 m?

Physics

1 answer:

katen-ka-za [31]3 years ago

4 0

20N•m or 20J. Work is equal to force•distance, and 5N•4m is 20N•m, or J

You might be interested in

The law of conservation of energy states

igor_vitrenko [27]

That the total energy of an isolated system remains constant and is said to be conserved over time

3 0

3 years ago

Could someone please please help meee

Kobotan [32]

Explanation:

noo she has potencial i think kinetic is just if shes moving and shes just about to jump so i think is no

4 0

2 years ago

What information do you need to describe an object's location

stira [4]

longitude and latitude<span />

4 0

3 years ago

Can you answer this math homework? Please!

steposvetlana [31]

Using the count data and observational data you acquired, calculate the number of CFUs in the original sample

5 0

2 years ago

A 600kg car is moving at 5 m/s to the right and elastically collided with a stationary 900 kg car. What is the velocity of the 9

11111nata11111 [884]

Answer:

$\mathrm{v}_{2} \text { velocity after the collision is } 3.3 \mathrm{m} / \mathrm{s}$

Explanation:

It says “Momentum before the collision is equal to momentum after the collision.” Elastic Collision formula is applied to calculate the mass or velocity of the elastic bodies.

$m_{1} v_{1}=m_{2} v_{2}$

$\mathrm{m}_{1} \text { and } \mathrm{m}_{2} \text { are masses of the object }$

$\mathrm{v}_{1} \text { velocity before the collision }$

$\mathrm{v}_{2} \text { velocity after the collision }$

$\mathrm{m}_{1}=600 \mathrm{kg}$

$\mathrm{m}_{2}=900 \mathrm{kg}$

$\text { Velocity before the collision } v_{1}=5 \mathrm{m} / \mathrm{s}$

$600 \times 5=900 \times v_{2}$

$3000=900 \times v_{2}$

$\mathrm{v}_{2}=\frac{3000}{900}$

$\mathrm{v}_{2}=3.3 \mathrm{m} / \mathrm{s}$

$\mathrm{v}_{2} \text { velocity after the collision is } 3.3 \mathrm{m} / \mathrm{s}$

5 0

4 years ago