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nevsk [136]
3 years ago
12

A motor running at 2600 rev/min is suddenly switched off and decelerates uniformly to rest after 10 s. Find the angular decelera

tion and the number of rotations to come to rest?
Physics
1 answer:
Mama L [17]3 years ago
6 0
Angular acceleration = (change in angular speed) / (time for the change)

change in angular speed = (zero - 2,600 RPM)  =  -2,600 RPM

time for the change = 10 sec

Angular acceleration = -2600 RPM / 10 sec = -260 rev / min-sec

(-260 rev/min-sec) x (1 min / 60 sec) = <em>-(4 1/3)  rev / sec²</em>

Since the acceleration is negative, the motor is slowing down.
You might call that a 'deceleration' of (4 1/3) rev/sec² .

The average speed is  1/2(2,600 + 0) = 1,300 rev/min = (21 2/3) rev/sec.

Number of revs = (average speed) x (time) = (21 2/3) x (10sec) = <em>(216 2/3) revs</em>
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They generate energy through hydrogen fusion in their core.

Explanation:

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2 years ago
Technician A says that a power-assisted brake system reduces stopping distances compared to a nonpower-assisted brake system. Te
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Answer:

Technician B

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The power-assisted brake system does not reduce the distance of stopping. What it does is it reduces the force to be applied by the driver. Thus, making the drive more comfortable.

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3 years ago
Calculate the radius of each ball to the nearest tenth. Use the formula r = d/2 where r is the radius and d is the diameter. Rec
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Answer:

What is the radius of the table tennis ball?

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Explanation:

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8 0
3 years ago
Read 2 more answers
One of the waste products of a nuclear reactor is plutonium-239 . This nucleus is radioactive and decays by splitting into a hel
Gekata [30.6K]

Answer:

a) v_{U-235} = 2.68 \cdot 10^{5} m/s

v_{He-4} = -1.57 \cdot 10^{7} m/s  

b) E_{He-4} = 8.23 \cdot 10^{-13} J

E_{U-235} = 1.41 \cdot 10^{-14} J

 

Explanation:

Searching the missed information we have:                                        

E: is the energy emitted in the plutonium decay = 8.40x10⁻¹³ J

m(⁴He): is the mass of the helium nucleus = 6.68x10⁻²⁷ kg  

m(²³⁵U): is the mass of the helium U-235 nucleus = 3.92x10⁻²⁵ kg            

a) We can find the velocities of the two nuclei by conservation of linear momentum and kinetic energy:

Linear momentum:

p_{i} = p_{f}

m_{Pu-239}v_{Pu-239} = m_{He-4}v_{He-4} + m_{U-235}v_{U-235}

Since the plutonium nucleus is originally at rest, v_{Pu-239} = 0:

0 = m_{He-4}v_{He-4} + m_{U-235}v_{U-235}  

v_{He-4} = -\frac{m_{U-235}v_{U-235}}{m_{He-4}}    (1)

Kinetic Energy:

E_{Pu-239} = \frac{1}{2}m_{He-4}v_{He-4}^{2} + \frac{1}{2}m_{U-235}v_{U-235}^{2}

2*8.40 \cdot 10^{-13} J = m_{He-4}v_{He-4}^{2} + m_{U-235}v_{U-235}^{2}    

1.68\cdot 10^{-12} J = m_{He-4}v_{He-4}^{2} + m_{U-235}v_{U-235}^{2}   (2)    

By entering equation (1) into (2) we have:

1.68\cdot 10^{-12} J = m_{He-4}(-\frac{m_{U-235}v_{U-235}}{m_{He-4}})^{2} + m_{U-235}v_{U-235}^{2}  

1.68\cdot 10^{-12} J = 6.68 \cdot 10^{-27} kg*(-\frac{3.92 \cdot 10^{-25} kg*v_{U-235}}{6.68 \cdot 10^{-27} kg})^{2} +3.92 \cdot 10^{-25} kg*v_{U-235}^{2}  

Solving the above equation for v_{U-235} we have:

v_{U-235} = 2.68 \cdot 10^{5} m/s

And by entering that value into equation (1):

v_{He-4} = -\frac{3.92 \cdot 10^{-25} kg*2.68 \cdot 10^{5} m/s}{6.68 \cdot 10^{-27} kg} = -1.57 \cdot 10^{7} m/s                        

The minus sign means that the helium-4 nucleus is moving in the opposite direction to the uranium-235 nucleus.

b) Now, the kinetic energy of each nucleus is:

For He-4:

E_{He-4} = \frac{1}{2}m_{He-4}v_{He-4}^{2} = \frac{1}{2} 6.68 \cdot 10^{-27} kg*(-1.57 \cdot 10^{7} m/s)^{2} = 8.23 \cdot 10^{-13} J

For U-235:

E_{U-235} = \frac{1}{2}m_{U-235}v_{U-235}^{2} = \frac{1}{2} 3.92 \cdot 10^{-25} kg*(2.68 \cdot 10^{5} m/s)^{2} = 1.41 \cdot 10^{-14} J

 

I hope it helps you!                                                                                    

3 0
3 years ago
“Is it correct to say that a radio wave can be considered a low-frequency light wave?
klio [65]

It's weird but technically correct to say that a radio wave can be considered a low-frequency light wave.  Radio and light are both electromagnetic waves.  The only difference is that radio waves have much much much longer wavelengths, and much much much lower frequencies, than light waves have.  But they're both the same physical phenomenon.

However, a radio wave CAN'T also be considered to be a sound wave.  These two things are as different as two waves can be.

-- Radio is an electromagnetic wave. Sound is a mechanical wave.

-- Radio waves travel more than 800 thousand times faster than sound waves do.

-- Radio waves are transverse waves. Sound waves are longitudinal waves.

-- Radio waves can travel through empty space. Sound waves need material stuff to travel through.

-- Radio waves can be detected by radio, TV, and microwave receivers. Sound waves can't.

-- Sound waves can be detected by our ears. Radio waves can't.

-- Sound waves can be generated by talking, or by hitting a frying pan with a spoon. Radio waves can't.

-- Radio waves can be generated by an alternating current flowing through an isolated wire. Sound waves can't.

4 0
3 years ago
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