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dsp73
4 years ago
9

Three blocks are arranged in a stack on a frictionless horizontal surface. The bottom block has a mass of 37.0 kg. A block of ma

ss 18.0 kg sits on top of it and a 16.0 kg block sits on top of the middle block. A downward vertical force of 170 N is applied to the top block. What is the magnitude of the normal force exerted by the bottom block on the middle block?
Physics
1 answer:
Alex777 [14]4 years ago
7 0

Answer:

N₂ = 503.8 N

Explanation:

given,

mass of bottom block = 37 Kg

mass of middle block = 18 Kg

mass of the top block = 16 Kg

force acting on the top block = 170 N

force on the block at top

N₁ be the normal force from block at middle

now,

N₁ = 170 + m g

N₁ = 170 + 16 x 9.8

now, force on block at middle

N₂ be the normal force exerted by the bottom block

N₂ = N₁ + m₂ g

N₂ = 326.8 + 18 x 9.8

N₂ = 503.8 N

hence, normal force by bottom block is equal to N₂ = 503.8 N

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The angular velocity of a process control motor is (13−12t2) rad/s, where t is in seconds. Part A At what time does the motor re
mihalych1998 [28]

Answer:

Explanation:

Given

\omega =13-\frac{1}{2}\cdot t^2

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13-0.5t^2=0

26=t^2

t=\sqrt{26}=5.099\approx 5.1 s

(b)

\frac{\mathrm{d} \theta }{\mathrm{d} t}=\omega

\int d\theta =\int_{0}^{5.1}\omega dt

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\theta =(13t-0.1667\times t^3)_0^{5.1}

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3 years ago
A test of the prototype of a new automobile shows that the minimum distance for a controlled stop from 95 km/h to zero is 55 m.
iris [78.8K]

Answer:

-0.64525g

Explanation:

t = Time taken for the car to stop

u = Initial velocity = 95 km/h

v = Final velocity = 0 km/h

s = Displacement

a = Acceleration

Equation of motion

v^2-u^2=2as\\\Rightarrow a=\frac{v^2-u^2}{2s}\\\Rightarrow a=\frac{0^2-95^2}{2\times 0.055}\\\Rightarrow a=-82045.45\ km/h^2

Converting to m/s²

a=82045.45=\frac{82045.45\times 1000}{3600\times 3600}=-6.33\ m/s^2

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Dividing both the accelerations, we get

\frac{a}{g}=\frac{-6.33}{9.81}=-0.64525\\\Rightarrow a=-0.64525g

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4 years ago
How much time does it take a pulse of light to travel through 150 m of water?
jenyasd209 [6]

Answer:

 6.65×10⁻⁷ s

Explanation:

Speed of light in water = Total distance traveled in water/time

c = d/t ....................... Equation 1

Where c = speed of the light pulse in water, d = distance traveled in water, t = time.

also

c = v/n ..................... Equation 2

Where v = speed of light in air, n = refractive index of water

Substituting equation 2 into equation 1

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making t the subject of the equation,

t = nd/v.................. Equation 3

Given: 150 m.

Constant: n = 1.33, v = 3.0×10⁸ m/s

Substitute into equation 3

t = 1.33(150)/3.0×10⁸

t = 6.65×10⁻⁷ seconds.

Hence the time =  6.65×10⁻⁷ s

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4 years ago
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