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2 years ago

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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.

1 answer:

Sveta_85 [38]2 years 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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Answer:

2.39 revolutions

Explanation:

As she jumps off the platform horizontally at a speed of 10m/s, the gravity is the only thing that affects her motion vertically. Let g = 10m/s2, the time it takes for her to fall 10m to water is

$h = gt^2/2$

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$t^2 = 2$

$t = \sqrt{2} = 1.414 s$

Knowing the time it takes to fall to the pool, we calculate the angular distance that she would make at a constant acceleration of 15 rad/s2:

$\theta = \alpha t^2/2$

$\theta = 15 * 2/2 = 15 rad$

As each revolution is 2π, the total number of revolution that she could make is: 15 / 2π = 2.39 rev

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These are the answers:

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Julius competes in the hammer throw event. The hammer has a mass of 7.26 kg and is 1.215 m long. What is the centripetal force o

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In the circular motion of the hammer, the centripetal force is given by
$F=m \frac{v^2}{r}$
where m is the mass of the hammer, v its tangential speed and r is the distance from the center of the motion, i.e. the length of the hammer.
Using the data of the problem, we find:
$F=m \frac{v^2}{r}=(7.26 kg) \frac{(31.95 m/s)^2}{1.215 m}=6100 N$

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A 5.0-kg crate is resting on a horizontal plank. The coefficient of static friction is 0.50 and the coefficient of kinetic frict

Harlamova29_29 [7]

Answer:

The mass of the crate is 5kg.

We know that the force of friction can be obtained by:

F = N*k

where k is the coefficient of friction, where we use the static one if the object is at rest, and the kinetic one if the object os moving. N is the normal force

If we tilt the base making an angle of 30° with the horizontal, now the normal force against the plank will be equal to the fraction of the weight in the direction normal to the surface of the plank.

Knowing that the angle is 30°, then the fraction of the weight that pushes against the normal is Cos(30°)*W = cos(30°)*5kg*9.8m/s^2 = 42.4N

The fraction of the force in the parallel direction to the plank (the force that would accelerate the crate downwards) is:

F = sin(30°)*5k*9,8m/s = 24.5N

now, the statical friction force is:

Fs = 42.4N*0.5 = 21.2N

The statical force is less than the 24.5N, so the crate will move downwards, then the force that acts on the crate is the kinetic force of friction:

Fk = 42.4N*0.4 = 16.96N

Then, the total force that acts on the crate is:

total force = F - Fk = 24.5N - 16.69N = 7.54N and the direction of this force points downside along the parallel direction of the plank.

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2 years ago

A thin, horizontal, 18-cm-diameter copper plate is charged to -3.8 nC. Assume that the electrons are uniformly distributed on th

son4ous [18]

Answer:

Part a)

$E = 8436.7 N/C$

Part b)

$E = 8436.7 N/C$

Explanation:

Part a)

Electric field due to large sheet is given as

$E = \frac{\sigma}{2\epsilon_0}$

$\sigma = \frac{Q}{A}$

$Q = -3.8 nC$

$A = \pi(0.09)^2$

$A = 0.025 m^2$

$\sigma = \frac{-3.8\times 10^{-9}}{0.025}$

$\sigma = -1.5 \times 10^{-7} C/m^2$

now the electric field is given as

$E = \frac{-1.5 \times 10^{-7}}{2(8.85 \times 10^{-12})}$

$E = 8436.7 N/C$

Part b)

Now since the electric field is required at same distance on other side

so the field will remain same on other side of the plate

$E = 8436.7 N/C$

5 0

2 years ago