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babunello [35]
3 years ago
9

When the net force on an object is zero, what do we know about the motion of that object

Physics
1 answer:
aev [14]3 years ago
6 0

Answer:

The objects speed or motion does not change if the net force is 0.

Explanation: There basically doesn't have anything acting on it so it stays put.

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How many degrees are in each quadrant <br> A: 90°<br> B: 30°<br> C: 180°<br> D: 360°
Airida [17]

In one quadrant there are 90 degrees.

8 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
3 years ago
A battery supplies E volts to a circuit containing a resistor R. How could the current in the resistor be increased? Select all
anastassius [24]

Answer:

I=V/R

so larger bsttery or small resistor can increase the current value.

5 0
3 years ago
An 3.7 lb hammer head, traveling at 5.8 ft/s strikes a nail and is brought to a stop in 0.00068 s. The acceleration of gravity i
CaHeK987 [17]

Answer:

31677.2 lb

Explanation:

mass of hammer (m) = 3.7 lb

initial velocity (u) = 5.8 ft/s

final velocity (v) = 0

time (t) = 0.00068 s

acceleration due to gravity (g) 32 ft/s^{2}

force = m x ( a + g )

where

  • m is the mass = 3.7 lb
  • g is the acceleration due to gravity = 32 ft/s^{2}
  • a is the acceleration of the hammer

       from v = u + at

       a = (v-u)/ t

       a = (0-5.8)/0.00068 = -8529.4 ( the negative sign showa the its decelerating)

we can substitute all required values into force= m x (a+g)

force = 3.7 x (8529.4 + 32) = 31677.2 lb

       

4 0
2 years ago
Which has a greater kinetic energy, a truck, or a train.
Oliga [24]

Answer:

a train

Explanation:

the train is longer the longer something is the more power it will have

6 0
2 years ago
Read 2 more answers
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