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morpeh [17]
3 years ago
7

A box is sitting on the ground and weighs 100 kg and the coefficient of friction is 0.23. Is it easier to push by applying the f

orce?

Physics
1 answer:
Hitman42 [59]3 years ago
7 0

Answer:

-No

Explanation:

Given that

Mass of box = 100 kg

Coefficient of friction ,μ= 0.23

We know that friction force depends on the normal force acts on the box

Fr= μ N

When we pull the box then :

Normal force N= mg

Friction force Fr=  μ mg                  

When we push the box :

Lets take pushing force = F

θ =Angle make by pushing force from the vertical line

Normal force

N = mg + F cosθ

Fr= μ ( mg + F cosθ )

The friction force is more when we push the box.That is why this is not easier to push the box.

Therefore answer is ---No

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Explanation:

Projectile motion is characterized as the motion that an object undergoes when it is thrown into the air and it is only exposed to acceleration due to gravity.

As per the question, 'any change in the initial velocity of the projectile(object having gravity as the only force) would lead to a change in the range as well as the maximum height of the projectile.' To illustrate numerically:

Horizontal range: As per expression:

R= (u^{2}*sin2θ)/g

the range depending on the square of the initial velocity.

Maximum height: As per expression:

H= (u^{2} * sin^{2}θ )/2g

the maximum distance also depends upon square of the initial velocity.

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When only several competitors in an industry maintain most of the sales of those items, it is considered a(n) Industry of Social
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"The burning of fossil fuels and ___________ from nuclear power provide about 87% of the energy used in the world."
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6 0
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Physics question, help please!
vova2212 [387]

The change in potential energy when the block falls to ground is -480J.

The maximum change in kinetic energy of the ball is 480 J.

The initial kinetic energy of the ball is 0 J.

The final  kinetic energy of the ball is 0.148J.

The initial potential energy of the ball is 0.187 J.

The final  potential energy of the ball is 0 J.

The work done by the air resistance is 0.039 J.

<h3>Change in potential energy when the block falls to ground</h3>

ΔP.E = -mgh

ΔP.E = -Wh

ΔP.E = - 40 x 12

ΔP.E = -480 J

<h3>Maximum change in kinetic energy of the ball</h3>

ΔK.E = - ΔP.E

ΔK.E = - (-480 J)

ΔK.E = 480 J

<h3>Initial kinetic energy of the ball</h3>

K.Ei = 0.5mv²

where;

  • v is zero since it is initially at rest

K.Ei = 0.5m(0) = 0

<h3>Final kinetic energy</h3>

K.Ef =  0.5mv²

K.Ef = 0.5(0.0091)(5.7)²

K.Ef = 0.148 J

<h3>Initial potential energy of the ball</h3>

P.Ei = mghi

P.Ei = 0.0091 x 9.8 x 2.1

P.Ei = 0.187 J

<h3>Final potential energy</h3>

P.Ef = mghf

P.Ef = 0.0091 x 9.8 x 0

P.Ef = 0

<h3>Work done by the air resistance</h3>

W = ΔE

W = P.E - K.E

W = 0.187 J - 0.148 J

W = 0.039 J

Learn more about potential energy here: brainly.com/question/1242059

#SPJ1

<h3 />
7 0
1 year ago
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