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Sidana [21]
3 years ago
13

A 50 kg box is at rest on a horizontal surface. When an 80 N force is applied to the box, it slides on the surface with a consta

nt velocity. How will the force of friction change if an identical box is placed on top of the 50 kg box
Physics
1 answer:
vovikov84 [41]3 years ago
3 0
As the box is moving with a constant velocity, the two forces acting on the box are canceling each other.

Then   friction force = 80 Newtons              but in the opposite direction.

Friction force =  Mu  * Normal force exerted by ground  =  Mu * weight of box
So we find Mu.
Mu = coefficient of friction between box and horizontal surface
           = Force of friction / weight  =  80 / 50 * 9.81 = 0.163

When an identical box is placed on top, the force of friction is
       = Mu * total weight = 0.163 * (50+50) * 9.81 = 159.9 Newtons

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valentina_108 [34]

Answer:

  94 kg

Explanation:

The mass registered by the scale is based on the assumption that the force applied is due entirely to gravity. If the force is greater, then the indicated mass will be greater.

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<h3>how many g's</h3>

As a fraction of the acceleration of gravity, the elevator's acceleration is ...

  (1.2 m/s²)/(9.8 m/s²) ≈ 6/49

<h3>net force</h3>

The force required to produce a given acceleration is found by the formula ...

  F = ma . . . . . . . force on mass m to produce acceleration 'a'

When the man is stationary on the scale, the upward force it supplies is balanced by the downward force on the man due to gravity. The force and the mass are proportional, and the constant of proportionality (the acceleration due to gravity) is used to calibrate the scale. More force is thus translated to a higher mass reading.

Since the man's net acceleration is upward at the rate of 6/49×g, the total force applied by the scale is (1 +6/49) = 55/49 times as great as when the man is stationary. This greater force gets translated to a greater mass reading.

The force is equivalent to what would be required to support a stationary man with a mass of ...

  (84 kg)(55/49) = 94 2/7 kg

The scale would read about 94 kg during the upward acceleration period.

3 0
2 years ago
The object will feel minimum force of gravity at the
Alika [10]

Answer:

space , small amount of gravity is found in the space ,infact we can say that there is no gravity in the space

4 0
3 years ago
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Three moles of an ideal gas are taken around the eyele acb shown in Fig.For this gas Co-29.2J/mol-K.Process ac is at constant pr
Scorpion4ik [409]

For three moles of an ideal gas are taken around the eye, the total work for the cycle is mathematically given as

Wt=1945.475J

<h3>What is the total work F for the cycle.(R=8.31J/molK)?</h3>

Generally, the equation for work  is mathematically given as

Wt=wac+Wc+Wba

Therefore

Wac=Pa(Vc-Va)=nR(Tc-Ta)

Wac=3(8.314*192)

Wac=4788.864J

Wcb=P1v1-p2v2/v-1

Wcb-3*8.34*108/0.4

Wcb=-6734.34J

Wab=0

In conclusion,

Wt=wac+Wc+Wba

Wt=4788.864J+0+(-6734.34J)

Wt=+1945.475J

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8 0
2 years ago
Which of these would NOT produce visible light? A) flint B) lightning C) microwave D) sun
ioda

Answer:

A microwave

Explanation

The human retina can only detect incident light that falls in waves 400 to 720 nanometers long, so we can't see microwave or ultraviolet wavelengths. This also applies to infrared lights which has wavelengths longer than visible and shorter than microwaves, thus being invisible to the human eye.

6 0
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liberstina [14]

Answer:

d: 39W

Explanation:

Power(P) = w/t = F*d/t = F*v = m*a*v= 2*9.8*2 = 39.2W

6 0
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