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miv72 [106K]
3 years ago
14

You want to push a 71-kg box up a 16° ramp. The coefficient of kinetic friction between the ramp and the box is 0.23. With what

magnitude force parallel to the ramp should you push on the box so that it moves up the ramp at a constant speed?
Physics
1 answer:
ycow [4]3 years ago
8 0

Answer:

<em>345.60N</em>

Explanation:

The magnitude of the force parallel to the ramp that will push the box up the ramp are the frictional force(Ff) and the moving force Fm

Total force = Ff + Fm

Ff = ηR

Ff = ηmgcosθ

Fm = Wsinθ

Fm = mgsinθ

Total force = ηmgcosθ + mgsinθ

Total force = mg(ηcosθ + sinθ)

Given

mass m = 71kg

acceleration due to gravity g = 9.8m/s²

angle of inclination θ = 16°

coefficient of friction η = 0.23

Substitute into the formula;

Total force = 71(9.8)(0.23cos16 + sin16)

Total force = 695.8(0.2211+0.2756)

Total force = 695.8(0.4967)

Total force = 345.60N

<em>Hence the magnitude force parallel to the ramp that you should push on the box so that it moves up the ramp at a constant speed is 345.60N</em>

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Explanation:

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mechanical energy.

Explanation:

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Read 2 more answers
Normal atmospheric pressure is 1.013 105 Pa. The approach of a storm causes the height of a mercury barometer to drop by 27.1 mm
Burka [1]

Answer:

The atmospheric pressure is 0.97622\times10^{5}\ Pa.

Explanation:

Given that,

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drop height h'= 27.1 mm

Density of mercury \rho= 13.59 g/cm^3

We need to calculate the height

Using formula of pressure

p = \rho g h

Put the value into the formula

1.013\times10^{5}=13.59\times10^{3}\times9.8\times h

h =\dfrac{1.013\times10^{5}}{13.59\times10^{3}\times9.8}

h=0.76\ m

We need to calculate the new height

h''=h - h'

h''=0.76-27.1\times10^{-3}

h''=0.76-0.027

h''=0.733\ m

We need to calculate the atmospheric pressure

Using formula of atmospheric pressure

P=\rho g h

Put the value into the formula

P= 13.59\times10^{3}\times9.8\times0.733

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7 0
3 years ago
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Kruka [31]

Explanation:

Given that,

Distance, s = 47 m

Time taken, t = 8.6 s

Final speed of the truck, v = 2.3 m/s

Let u is the initial speed of the truck and a is its acceleration such that :

a=\dfrac{v-u}{t}.............(1)

Now, the second equation of motion is :

s=ut+\dfrac{1}{2}at^2

Put the value of a in above equation as :

s=ut+\dfrac{1}{2}\times \dfrac{v-u}{t}\times t^2

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u=\dfrac{2\times 47}{8.6}-2.3

u = 8.63 m/s

So, the original speed of the truck is 8.63 m/s. Hence, this is the required solution.

8 0
3 years ago
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