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marissa [1.9K]
1 year ago
8

A 0.15-m-diameter pulley turns a belt rotating the driveshaft of a power plant pump. The torque applied by the belt on the pulle

y is 200 N ∙ m, and the power transmitted is 7 kW. Determine the net force applied by the belt on the pulley, in kN, and the rotational speed of the driveshaft, in RPM.
Physics
1 answer:
Sveta_85 [38]1 year ago
6 0

Answer:

F= 2666.66 N

N=334.22 RPM          

Explanation:

Given that

Diameter of pulley ,d= 0.15 m

Radius ,r= 0.075 m

Torque ,T= 200 N.m

Power ,P = 7 kW

Lets take force = F

T = F .r

F=\dfrac{T}{r}

Now by putting the values

F=\dfrac{200}{0.075}\ N

F= 2666.66 N

Lets take rotational speed = N RPM

We know that  

P=\dfrac{2\pi N\ T}{60}

N=\dfrac{60\times P}{2\pi \ T}

Now by putting the values in the above equation

N=\dfrac{60\times 7000}{2\pi \times 200}\ RPM

N=334.22 RPM

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a 250 mH coil of negligible resistance is connected to an AC circuit in which as effective current of 5 mA is flowing. if the fr
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A wheel rotates without friction about a stationary horizontal axis at the center of the wheel. A constant tangential force equa
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Answer:

I = 16 kg*m²

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τ = I * α   Formula  (1)

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I :  it is the moment of inertia of the body with respect to the axis of rotation (kg*m²)

α : It is angular acceleration. (rad/s²)

Kinematics of the wheel

Equation of circular motion uniformly accelerated :  

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Where:  

α : Angular acceleration (rad/s²)  

ω₀ : Initial angular speed ( rad/s)  

ωf : Final angular speed ( rad

t : time interval (rad)

Data  

ω₀ = 0

ωf = 1.2 rad/s

t = 2 s

Angular acceleration of the wheel  

We replace data in the formula (2):  

ωf = ω₀+ α*t

1.2= 0+ α*(2)

α*(2) = 1.2

α = 1.2 / 2

α = 0.6 rad/s²

Magnitude of the net torque (τ )

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Where:

F  = tangential force (N)

R  = radio (m)

τ = 80 N *0.12 m

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Rotational inertia of the wheel

We replace data in the formula (1):

τ = I * α

9.6 = I *(0.6 )

I = 9.6 / (0.6 )

I = 16 kg*m²

8 0
1 year ago
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