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fgiga [73]
3 years ago
7

Which has more momentum, a heavy truck moving at 30 miles per hour or a light truck moving at 30 miles per hour?

Physics
2 answers:
denis-greek [22]3 years ago
8 0

Answer: The heavy truck

Explanation: Momentum is defined as the amount of movement.

It can be written as P = m*v

Where P is the momentum, m is the mass of the object and v is the velocity.

So you can see that if the mass increases, also does the momentum, so if both trucks move at the same velocity, then the truck with the bigger mas will have a bigger momentum.

Then the correct answer is the heavy truck

valentina_108 [34]3 years ago
7 0
A heavy truck moving a 30 mph. It has more mass.
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Do you think it is possible to control the magnetic properties of a magnet? Can a magnet be turned on and off?
Sunny_sXe [5.5K]

Answer:

Yes it is possible to control to some extent.

Explanation:

In general there are two types of magnets : permanent and temporary (electromagnets).

Electromagnets can be controlled since it basically depends on electricity. By switching on and off the electric supply the magnets also can be switched on and off respectively. We can also control the intensity of magnetic power.

On the other hand permanent magnet cannot be switched on and off but the magnetic properties can be altered event to an extent when it loses all its magnetic properties. It can be caused by high temperature, physical impact and also exposure to other magnetic fields. For every element there is a point of temperature called curie temperature above which the permanent magnet loses its magnetic properties. This can be brought back again by induced magnetism. The only issue is that induced magnetism work in most cases but not in all.

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4 years ago
What is the difference between a physical change and a chemical change
Contact [7]
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6 0
3 years ago
Read 2 more answers
can somebody help me answer this question? A car went from 110 m/s to 80 m/s in 20 seconds. What was the acceleration of the car
gogolik [260]

Answer:

-1.5m/s²

Explanation:

Acceleration can be thought of as [Change in Velocity]/[Change in time]. To find these changes, you simply subtract the initial quantity from the final quantity.

So for this question you have:

  • V_i = 110m/s
  • V_f = 80m/s
  • t_i = 0s
  • t_f = 20s

which means that the acceleration = (80-110)/(20-0)[m/s²] = (-30/20)m/s² = -1.5m/s²

4 0
3 years ago
A photon of wavelength 2.78 pm scatters at an angle of 147° from an initially stationary, unbound electron. What is the de Brogl
Elena-2011 [213]

Answer:

2.07 pm

Explanation:

The problem given here is the very well known Compton effect which is expressed as

\lambda^{'}-\lambda=\frac{h}{m_e c}(1-cos\theta)

here, \lambda is the initial photon wavelength, \lambda^{'} is the scattered photon wavelength, h is he Planck's constant, m_e is the free electron mass, c is the velocity of light, \theta is the angle of scattering.

Given that, the scattering angle is, \theta=147^{\circ}

Putting the respective values, we get

\lambda^{'}-\lambda=\frac{6.626\times 10^{-34} }{9.11\times 10^{-31}\times 3\times 10^{8} } (1-cos147^\circ ) m\\\lambda^{'}-\lambda=2.42\times 10^{-12} (1-cos147^\circ ) m.\\\lambda^{'}-\lambda=2.42(1-cos147^\circ ) p.m.\\\lambda^{'}-\lambda=4.45 p.m.

Here, the photon's incident wavelength is \lamda=2.78pm

Therefore,

\lambda^{'}=2.78+4.45=7.23 pm

From the conservation of momentum,

\vec{P_\lambda}=\vec{P_{\lambda^{'}}}+\vec{P_e}

where,\vec{P_\lambda} is the initial photon momentum, \vec{P_{\lambda^{'}}} is the final photon momentum and \vec{P_e} is the scattered electron momentum.

Expanding the vector sum, we get

P^2_{e}=P^2_{\lambda}+P^2_{\lambda^{'}}-2P_\lambda P_{\lambda^{'}}cos\theta

Now expressing the momentum in terms of De-Broglie wavelength

P=h/\lambda,

and putting it in the above equation we get,

\lambda_{e}=\frac{\lambda \lambda^{'}}{\sqrt{\lambda^{2}+\lambda^{2}_{'}-2\lambda \lambda^{'} cos\theta}}

Therefore,

\lambda_{e}=\frac{2.78\times 7.23}{\sqrt{2.78^{2}+7.23^{2}-2\times 2.78\times 7.23\times cos147^\circ }} pm\\\lambda_{e}=\frac{20.0994}{9.68} = 2.07 pm

This is the de Broglie wavelength of the electron after scattering.

6 0
4 years ago
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