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Dennis_Churaev [7]
4 years ago
10

Two workers are sliding 290 kg crate across the floor. One worker pushes forward on the crate with a force of 430 N while the ot

her pulls in the same direction with a force of 360 N using a rope connected to the crate. Both forces are horizontal, and the crate slides with a constant speed. What is the crate's coefficient of kinetic friction on the floor?
Physics
1 answer:
m_a_m_a [10]4 years ago
6 0

Answer:0.27

Explanation:

Given

One worker Pushes with force F_1=430 N

other Pulls it with a rope of rope F_2=360 N

mass of crate m=290 kg

both forces are horizontal and crate slides with a constant speed

Both forces are in the same direction so Friction will oppose the forces and will be equal in magnitude of sum of two forces because crate is moving with constant speed i.e. net force is zero on it

f_r=F_1+F_2

where f_r is the friction force

f_r=430+360

f_r=790 N

f_r=\mu N

where \muis the coefficient of static friction

N=mg

790=\mu 29\cdot 9.8

\mu =0.27

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The black ball moving to the right after being struck by the white ball

Explanation:

4 0
3 years ago
A 25.0 kg block is initially at rest on a horizontal surface. A horizontal force of 75.0 N is required to set the block in motio
QveST [7]

Answer:

(a) 0.31

(b) 0.245

Explanation:

(a)

F' = μ'mg.................... Equation 1

Where F' = Horizontal Force required to set the block in motion, μ' = coefficient of static friction, m = mass of the block, g = acceleration due to gravity.

make μ' the subject of the equation above

μ' = F'/mg............. Equation 2

Given: F' = 75 N, m = 25 kg

constant: g = 9.8 m/s²

Substitute these values into equation 2

μ' = 75/(25×9.8)

μ' = 75/245

μ' = 0.31.

(b) Similarly,

F = μmg.................. Equation 3

Where F = Horizontal force that is required to keep the block moving with constant speed, μ = coefficient of kinetic friction.

make μ the subject of the equation

μ = F/mg.............. Equation 4

Given: F = 60 N, m = 25 kg, g = 9.8 m/s²

Substitute these values into equation 4

μ  = 60/(25×9.8)

μ = 60/245

μ = 0.245

3 0
3 years ago
Write the equation for eccentricity
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<span>e=ca{\displaystyle e={\frac {c}{a}}}.</span>

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4 years ago
A ball dropping from 70 cm took 0.6 seconds to travel a horizontal distance of 34 cm.
Alenkasestr [34]

Answer:

vₓ = 0.566 m / s,   W_total = 9.1 J

Explanation:

This exercise is a parabolic type movement, for the x axis where there is no acceleration

         x = v t

         vₓ = x / t

         vₓ = 0.34 / 0.6

         vₓ = 0.566 m / s

the work done is

X axis

In this axis there is no acceleration, therefore the sum of the forces is zero and since the work is the force times the distance, we conclude that the lock in this axis is zero.

        W₁ = 0

Y axis  

in this axis the force that exists is the force of gravity, that is, the weight of the body

         W₂ = Fy y

         W₂ = mg and

          W₂ = m 9.8 0.70

          W₂ = m 9.1

the work is a scalar for which we have to add the quantities obtained

          W_total = W₁ + W₂

          W_total = 0 + 9.1 m

           W_total = 9.1 m

In order to complete the calculation, the mass of the body is needed if we assume that the mass is m = 1

           W_total = 9.1 J

4 0
3 years ago
A student carries a backpack for one mile , another student carries the same back pack for two miles . Compared to the first stu
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Answer:

he did twice as much since the first student carried the backpack for 1 mile whilst the second one carried it for 2 miles

Explanation:

8 0
3 years ago
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