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3 years ago

When force is measured in newtons and distance is measured in meters, the work done is measured in

Physics

2 answers:

zalisa [80]3 years ago

7 0

Newton meters, which is called Joules

RUDIKE [14]3 years ago

3 0

Jewls my spelling is off but, in Physics it would be jewls. so Work=fxdxcos0= 900J. Is what it'd look like for an answer.

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A stone is thrown vertically upward with an initial velocity of 25m/s , calculate time taken to return to the thrower

Nonamiya [84]

Answer:

can you give me abrainlest??

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3 years ago

A dart is loaded into a spring-loaded toy dart gun by pushing the spring in by a  distance d. For the next loading?the spring is

Mrac [35]

Answer:

$W'=\dfrac{W}{9}$

Explanation:

The potential energy of the spring or the work done by the spring is given by :

$W=\dfrac{1}{2}kd^2$ ............(1)

k is the spring constant

d is the compression

When the spring is compressed a distance d' = d/3, let W' is the work is required to load the second dart. Then the work done is given by :

$W'=\dfrac{1}{2}kd'^2$

$W'=\dfrac{1}{2}k(d/3)^2$ .............(2)

Dividing equation (1) and (2) :

$\dfrac{W}{W'}=\dfrac{1/2kd^2}{1/2k(d/3)^2}$

$\dfrac{W}{W'}=9$

$W'=\dfrac{W}{9}$

So, the work required to load the second dart compared to that required to load the first is one-Ninth as much. Therefore, the correct option is (E).

7 0

3 years ago

What is the number one and most important rule to remember in soccer?

AlekseyPX

To score in the correct goal

8 0

3 years ago

What will happen if a car experiences a 300 N force to the right from the engine and a separate 150 N force due to friction and

gayaneshka [121]

Force applied on the car due to engine is given as

$F_1 = 300 N$ towards right

Also there is a force on the car towards left due to air drag

$F_2 = 150 N$ towards left

now the net force on the car will be given as

$\vec F_{net} = \vec F_1 + \vec F_2$

now we can say that since the two forces are here opposite in direction so here the vector sum of two forces will be the algebraic difference of the two forces.

So we can say

$F_{net} = F_1 - F_2$