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sweet [91]
3 years ago
10

Two people carry a heavy electric motor by placing it on a light board 1.80 m long. One person lifts at one end with a force of

350 N, and the other lifts the opposite end with a force of 620 N.
What is the weight of the motor?
Physics
1 answer:
bixtya [17]3 years ago
6 0

Answer:

W=970\ N

Explanation:

When two people carry a weight placing it on a very light board then we neglect the weight of the board and the board acts like a beam with two mass placed over it and two end point supports are here provided by men.

Then balancing the forces in the vertical direction, weight of the motor is:

W=350+620

W=970\ N

These men are able to carry the motor in equilibrium condition because they have placed the motor (considering it a point mass) at a position such that the moment due to the lifting forces are equal and opposite.

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In a science museum, a 140 kg brass pendulum bob swings at the end of a 16.8 m -long wire.
Mice21 [21]

The period of the pendulum is 8.2 s

Explanation:

The period of a simple pendulum is given by the equation:

T=2\pi \sqrt{\frac{L}{g}}

where

L is the length of the pendulum

g is the acceleration of gravity

T is the period

We notice that the period of a pendulum does not depend at all on its mass, but only on its length.

For the pendulum in this problem, we have

L = 16.8 m

and

g=9.8 m/s^2 (acceleration of gravity)

Therefore the period of this pendulum is

T=2\pi \sqrt{\frac{16.8}{9.8}}=8.2 s

#LearnWithBrainly

3 0
3 years ago
75 kg + 1352 g = ______________g?
JulsSmile [24]
Answer is 76,352 just look it up
8 0
3 years ago
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A racquetball strikes a wall with a speed of 30 m/s and rebounds in the opposite direction with a speed of 26 m/s. The collision
Fudgin [204]

Answer:

The average acceleration of the ball during the collision with the wall is a=2,800m/s^{2}

Explanation:

<u>Known Data</u>

We will asume initial speed has a negative direction, v_{i}=-30m/s, final speed has a positive direction, v_{f}=26m/s, \Delta t=20ms=0.020s and mass m_{b}.

<u>Initial momentum</u>

p_{i}=mv_{i}=(-30m/s)(m_{b})=-30m_{b}\ m/s

<u>final momentum</u>

p_{f}=mv_{f}=(26m/s)(m_{b})=26m_{b}\ m/s

<u>Impulse</u>

I=\Delta p=p_{f}-p_{i}=26m_{b}\ m/s-(-30m_{b}\ m/s)=56m_{b}\ m/s

<u>Average Force</u>

F=\frac{\Delta p}{\Delta t} =\frac{56m_{b}\ m/s}{0.020s} =2800m_{b} \ m/s^{2}

<u>Average acceleration</u>

F=ma, so a=\frac{F}{m_{b}}.

Therefore, a=\frac{2800m_{b} \ m/s^{2}}{m_{b}} =2800m/s^{2}

8 0
3 years ago
Describe the interaction with the sun and producers
bagirrra123 [75]

Answer: the sun's rays is one of the raw Materials recquired by plants to make food

Explanation:plants trap light used in splitting water into hydrogen ions and oxygen molecules

4 0
3 years ago
A 51.0 kg crate, starting from rest, is pulled across a floor with a constant horizontal force of 225 N. For the first 10.0 m th
shepuryov [24]

Answer:

The final speed of the crate is 12.07 m/s.

Explanation:

For the first 10.0 meters, the only force acting on the crate is 225 N, so we can calculate the acceleration as follows:

F = ma

a = \frac{F}{m} = \frac{225 N}{51.0 kg} = 4.41 m/s^{2}

Now, we can calculate the final speed of the crate at the end of 10.0 m:

v_{f}^{2} = v_{0}^{2} + 2ad_{1}                  

v_{f} = \sqrt{0 + 2*4.41 m/s^{2}*10.0 m} = 9.39 m/s    

For the next 10.5 meters we have frictional force:

F - F_{\mu} = ma

F - \mu mg = ma

So, the acceleration is:

a = \frac{F - \mu mg}{m} = \frac{225 N - 0.17*51.0 kg*9.81 m/s^{2}}{51.0 kg} = 2.74 m/s^{2}

The final speed of the crate at the end of 10.0 m will be the initial speed of the following 10.5 meters, so:

v_{f}^{2} = v_{0}^{2} + 2ad_{2}  

v_{f} = \sqrt{(9.39 m/s)^{2} + 2*2.74 m/s^{2}*10.5 m} = 12.07 m/s  

Therefore, the final speed of the crate after being pulled these 20.5 meters is 12.07 m/s.  

I hope it helps you!                              

7 0
3 years ago
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