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almond37 [142]
2 years ago
9

When used as an energy source in a nuclear power plant, uranium is burned in a similar way as one would burn wood or coal for en

ergy.
a. True
b. False
Physics
1 answer:
zavuch27 [327]2 years ago
3 0
False because nuclear fission is used on uranium in nuclear power station and this reaction is continuous since a neutron hits a uranium atom and splits it. Then 2-3 neutrons are released which continues the cycle or hitting other uranium atoms. If this reaction isn't contained, it could get out of control and become a nuclear bomb. So it can't be burned like wood nor coal.
Hope this is helpful. <span />
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The moon Phobos orbits Mars
podryga [215]

Explanation:

For a circular orbit v= \sqrt{\frac{G.m}{r} } with G = 6.6742 × 10^{-11}

Given m = 6.42 x 10^23 kg and  r=9.38 x 10^6 m

=> v = 2137.3 m/s

I hope this is the correct way to solve

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2 years ago
A 42 Watt light bulb is connected to a 12 volt source of potential
abruzzese [7]

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The natural medium emanating from the Sun and other very hot sources (now recognised as electromagnetic radiation with a wavelength of 400-750 nm), within which vision is possible.

Explanation:

just the way it is

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3 years ago
A rock climber stands on top of a 50 m -high cliff overhanging a pool of water. He throws two stones vertically downward 1.0 s a
bagirrra123 [75]
<h2><em>Answer: b) What was the initial speed of the second stone?</em></h2>

Explanation:

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3 years ago
A Michelson interferometer uses red light with a wavelength of 656.45 nm from a hydrogen discharge lamp. Part A How many bright-
kati45 [8]

Answer:

51793 bright-dark-bright fringe shifts are observed when the mirror M2 moves through 1.7cm

Explanation:

The number of maxima appearing when the mirror M moves through distance \Delta L is given as follows,

\Delta m = \frac{\Delta L}{\frac{\lambda}{2}}

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\Delta L= 0.017m

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3 0
2 years ago
The driver of a car going 86.0 km/h suddenly sees the lights of a barrier 44.0 m ahead. It takes the driver 0.75 s to apply the
Lynna [10]

Part a

Answer: NO

We need to calculate the distance traveled once the brakes are applied. Then we would compare the distance traveled and distance of the barrier.

Using the second equation of motion:

s=ut+0.5at^2

where s is the distance traveled, u is the initial velocity, t is the time taken and a is the acceleration.

It is given that, u=86.0 km/h=23.9 m/s, t=0.75 s, a=-10.0 m/s^2

\Rightarrow s=23.9 m/s \times 0.75s+0.5\times (-10.0 m/s^2)\times (0.75 s)^2=15.11 m

Since there is sufficient distance between position where car would stop and the barrier, the car would not hit it.

Part b

Answer: 29.6 m/s

The maximum distance that car can travel is s=44.0 m

The acceleration is same, a=-10.0 m/s^2

The final velocity, v=0

Using the third equation of motion, we can find the maximum initial velocity for car to not hit the barrier:

v^2-u^2=2as

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Hence, the maximum speed at which car can travel and not hit the barrier is 29.6 m/s.





7 0
3 years ago
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