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zhuklara [117]
4 years ago
11

3. You are to determine the power rating of the motor in an elevator system. The elevator (with a full load) weights 1700 kg and

is required to move upward 3 m/sec at constant speed. The lifting mechanism is 90% efficient (i.e., 10% of the power generated by the motor is wasted in the mechanism, e.g., for overcoming friction). What is the minimum output power rating in HP of the motor that meets the requirement?
Physics
1 answer:
Talja [164]4 years ago
3 0

Answer:

3. You are to determine the power rating of the motor in an elevator system. The elevator (with a full load) weights 1700 kg and is required to move upward 3 m/sec at constant speed. The lifting mechanism is 90% efficient (i.e., 10% of the power generated by the motor is wasted in the mechanism, e.g., for overcoming friction). What is the minimum output power rating in HP of the motor that meets the requirement?The Elevator (with A Full Load) Weights 2000 Kg And Is Required To Move Upward 3 M/sec At Constant Speed. The Lifting Mechanism Is 85% Efficient (i.e., 15% Of The Power Generated By The Motor Is Wasted In The Mechanism, E.g., For Overcoming Friction).

Explanation:

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A skateboarder with mass ms = 54 kg is standing at the top of a ramp which is hy = 3.3 m above the ground. The skateboarder then
Elan Coil [88]

Answer:

A) W_{ff} =-744.12J

B) F_f=-W_{ff}*sin\theta /hy = 112.75N

C) F_{f2}=207.58N

Explanation:

This question is incomplete. The full question was:

<em>A skateboarder with mass ms = 54 kg is standing at the top of a ramp which is hy = 3.3 m above the ground. The skateboarder then jumps on his skateboard and descends down the ramp. His speed at the bottom of the ramp is vf = 6.2 m/s.  </em>

<em>Part (a) Write an expression for the work, Wf, done by the friction force between the ramp and the skateboarder in terms of the variables given in the problem statement.  </em>

<em>Part (b) The ramp makes an angle θ with the ground, where θ = 30°. Write an expression for the magnitude of the friction force, fr, between the ramp and the skateboarder.  </em>

<em>Part (c) When the skateboarder reaches the bottom of the ramp, he continues moving with the speed vf onto a flat surface covered with grass. The friction between the grass and the skateboarder brings him to a complete stop after 5.00 m. Calculate the magnitude of the friction force, Fgrass in newtons, between the skateboarder and the grass.</em>

For part A), we make a balance of energy to calculate the work done by the friction force:

W_{ff}=\Delta E

W_{ff}=1/2*m*vf^2-m*g*hy

W_{ff}=-744.12J

For part B), we use our previous value for the work:

W_{ff}=-F_f*(hy/sin\theta)   Solving for friction force:

F_f=-W_{ff}*sin\theta /hy

F_f=112.75N

For part C), we first calculate the acceleration by kinematics and then calculate the module of friction force by dynamics:

Vf^2=Vo^2+2*a*d

Solving for a:

a=-3.844m/s^2

Now, by dynamics:

|F_f|=|m*a|

|F_f|=207.58N

8 0
3 years ago
A stone is thrown straight up from the ground with an initial speed of 44 m/s.At the same instant, a stone is dropped from a hei
Lubov Fominskaja [6]

Answer:

394.26 m

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration due to gravity = 9.8 m/s²

v=u+at\\\Rightarrow 0=44-9.8\times t\\\Rightarrow \frac{-44}{-9.8}=t\\\Rightarrow t=4.49 \s

s=ut+\frac{1}{2}at^2\\\Rightarrow s=44\times 4.49+\frac{1}{2}\times -9.8\times 4.49^2\\\Rightarrow s=98.77\ m

s=ut+\frac{1}{2}at^2\\\Rightarrow 98.77=0t+\frac{1}{2}\times 9.8\times t^2\\\Rightarrow t=\sqrt{\frac{98.77\times 2}{9.8}}\\\Rightarrow t=4.49\ s

Total time taken by the stone to reach the ground is 4.49+4.49 = 8.97 seconds

s=ut+\frac{1}{2}at^2\\\Rightarrow h=0t+\frac{1}{2}\times 9.81\times t^2\\\Rightarrow t=\sqrt{\frac{h\times 2}{9.8}}

The times are equal so,

8.97=\sqrt{\frac{h\times 2}{9.8}}\\\Rightarrow h=\frac{8.97^2\times 9.8}{2}\\\Rightarrow h=394.26\ m

The height is 394.26 m

6 0
3 years ago
1) Calculate the torque required to accelerate the Earth in 5 days from rest to its present angular speed about its axis. 2) Cal
disa [49]

Answer:

a) τ = 4.47746 * 10^25 N-m

b) E = 2.06301 * 10^13 J

c) P = 3.25511*10^21 W

Explanation:

Given:

- The radius of earth r = 6.3781×10^6 m

- The angular speed of earth w = 7.27*10^-5 rad/s

- The time taken to reach above speed t = 5 yrs = 1.57784760 * 10^8 s

- The mass of earth m = 5.972 × 10^24 kg

- The inertia of sphere I = 2/5 * m* r^2

Solution:

- The angular acceleration of the earth from rest to w is given by α:

                                α = w / t

                                α = (7.27*10^-5) / (1.57784760 * 10^8)

                                α = 4.60754*10^-13 rad/s^2

- The required torque τ is given by:

                                τ = I*α

                                τ = 2/5 * m* r^2 * α

                           τ = 2/5 *(5.972 × 10^24) * (6.3781×10^6)^2 * (4.60754*10^-13)

                            τ = 4.47746 * 10^25 N-m

- The power required P to turn the earth to the speed w is:

                           P = τ*w

                           P = (4.47746 * 10^25)*(7.27*10^-5)

                           P = 3.25511*10^21 W

- The energy E required is :

                           E = P / t

                           E = (3.25511*10^21) / (1.57784760 * 10^8)

                           E = 2.06301 * 10^13 J

4 0
3 years ago
When two objects collide and stick together, what will happen to their speed, assuming momentum is conserved?
sladkih [1.3K]

Answer:

When two objects collide and stick together, what will happen to their speed, assuming momentum is conserved? They will move at the same velocity as whichever object was fastest initially. They will move at the same velocity of whichever object was slowest initially.

Explanation:

7 0
3 years ago
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Can someone help me on this please?
vladimir2022 [97]
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7 0
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