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marshall27 [118]
4 years ago
8

We push a 38.4-kg box across the floor at constant velocity. If we are pushing a horizontal force of 238 N, find the coefficient

of kinetic friction between the box and the floor.
Physics
1 answer:
zmey [24]4 years ago
5 0

Answer: 0.62

Explanation:

Coefficient of friction is defined as the ratio of the moving force (Fm) acting on a body to the normal reaction (R).

Note that the normal reaction acts vertically on the object and is equal to the objects weight (W) i.e W=R

Since W = mg, W = 38.4 ×10

W= 384N =R

Normal reaction = 384N

The horizontal force acting on the body will be the moving force which is 238N

Coefficient of friction = Fm/R

Coefficient of friction = 238/384

Coefficient of friction = 0.62

Therefore, coefficient of kinetic friction between the box and the floor is 0.62

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4. F

5. F

Explanation:

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6 0
3 years ago
A race car reaches the finish line and the driver slows down to come to a stop. If it takes the car 10 seconds to stop at a cons
boyakko [2]

Answer:

53.64 m/s

Explanation:

Applying,

a = (v-u)/t............. Equation 1

Where a = acceleration of the car, v = final velocity of the car, u = initial velocity of the car, t = time.

make u the subject of the equation

u = v-at............. Equation 2

From the question,

Given: a = -12 mph/s = -5.364 m/s², t = 10 seconds, v = 0 m/s (comes to stop)

Substitute these values into equation 2

u = 0-(-5.364×10)

u = 0+53.64

u = 53.64 m/s

6 0
3 years ago
To provide some perspective on the dimensions of atomic defects, consider a metal specimen that has a dislocation density of 105
-Dominant- [34]

Explanation:

LD₁ = 10⁵ mm⁻²

LD₂ = 10⁴mm⁻²

V = 1000 mm³

Distance = (LD)(V)

Distance₁ = (10⁵mm⁻²)(1000mm³) = 10×10⁷mm = 10×10⁴m

Distance₂ = (10⁹mm⁻²)(1000mm³) = 1×10¹² mm = 1×10⁹ m

Conversion to miles:

Distance₁ = 10×10⁴ m / 1609m = 62 miles

Distance₂ = 10×10⁹m / 1609 m = 621,504 miles.

7 0
3 years ago
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Explain the significance behind the universe having a flat geometry.
nirvana33 [79]
<span>The meaning of the Big Bang has been very often misunderstood. It is thought that something exploded somewhere and then the exploded part expanded to where we are currently.  I hope this did help If not I`m so sorry.</span>
5 0
3 years ago
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A long, thin straight wire with linear charge density λ runs down the center of a thin, hollow metal cylinder of radius R. The c
Delvig [45]

Answer:

E=\dfrac{\lambda }{2\pi \varepsilon _or}

Explanation:

Given that

For straight wire

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For hollow metal cylinder

Charge density=2 λ

We know that electric filed for wire given as

E_w=\dfrac{\lambda_{wire} }{2\pi \varepsilon _or}

E_w=\dfrac{\lambda }{2\pi \varepsilon _or}

Now the electric filed due to hollow metal cylinder

E_c=\dfrac{\lambda_{cylinder} }{2\pi \varepsilon _or}

E_c=\dfrac{2\lambda }{2\pi \varepsilon _or}

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E=\dfrac{\lambda }{2\pi \varepsilon _or}

5 0
4 years ago
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