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3 years ago

It takes 40 J to push a large box 4 m across a floor.

Physics

1 answer:

astra-53 [7]3 years ago

5 0

Answer: D

Explanation:

Use equation for work:

W=F*s

W=40J

s=4m

-------------

W=F*s

F=W/s

F=40J/4m

F=10N

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What does balanced and unbalanced forces have to do with newtons laws of motion

Ivahew [28]

Newton's second law of motion involves the effect of force acting on a mass.
It has to be understood that this actually means the combination of any and
all individual forces acting on it, with their individual strengths and directions
all added together.

When all the individual forces acting on an object add up to zero, then the
whole group of forces is said to be 'balanced'.

When all the individual forces acting on an object don't add up to zero, then
the whole group of forces is said to be 'unbalanced'.

Notice that there's no such thing as 'a balanced force' or 'an unbalanced force'.
It's a <u>group</u> of two or more forces that's balanced or unbalanced.

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3 years ago

A vehicle of mass 100kg has a kinetic energy of 5000 J at an instant. The velocity at that instant is

snow_lady [41]

Answer:

$\bf\pink{10\:m/s}$

Explanation:

<h2><u>Given</u> :-</h2>

$\sf\red{Mass \ of \ vehicle \ (m) = 100 \ kg}$
$\sf\orange{Kinetic \ energy \ (K.E.) = 5000 \ J}$

$\sf\green{Velocity\: of\: the \:vehicle \:at \:the \:instant}$

<h2><u>Formula to be used</u> :-</h2>

$\sf\blue{K.E. = \dfrac{1}{2} mv^{2}}$

Where,

K.E. = Kinetic energy possessed by the body
M = Mass of the body
V = Velocity of the body

<h2><u>Solution</u> :-</h2>

$\to\:\:\sf\red{K.E. = \dfrac{1}{2}mv^{2}}$

$\to\:\:\sf\orange{5000 = \dfrac{1}{2} \times 100 \times v^{2}}$

$\to\:\:\sf\green{5000 = 50 \times v^{2}}$

$\to\:\:\sf\blue{\dfrac{5000}{50} = v^{2}}$

$\to\:\:\sf\purple{100 = v^{2}}$

$\to\:\:\sf\red{\sqrt{100} = v}$

$\to\:\:\sf\orange{ 10 = v}$

$\to\:\:\bf\pink{v = 10\:m/s}$

Velocity of the vehicle at the instant is $\bf\green{10\:m/s}$

7 0

3 years ago

Read 2 more answers

Water drops fall from the edge of a roof at a steady rate. a fifth drop starts to fall just as the first drop hits the ground. a

Alchen [17]

The height of the roof is <u>3.57m</u>

Let the drops fall at a rate of 1 drop per t seconds. The first drop takes 5t seconds to reach the ground. The second drop takes 4t seconds to reach the bottom of the 1.00 m window, while the 3rd drop takes 3t s to reach the top of the window.

Calculate the distances traveled by the second and the third drops s₂ and s₃, which start from rest from the roof of the building.

$s_2=\frac{1}{2} g(4t)^2=8gt^2\\ s_3=\frac{1}{2} g(3t)^2=(4.5)gt^2$