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Ket [755]
3 years ago
15

If the photon scatters in the backward direction, what is the magnitude of the linear momentum of the electron just after the co

llision with the photon
Physics
1 answer:
Strike441 [17]3 years ago
8 0
Momentum is a vector quantity, and is always conserved. Whenever a collision occurs between two objects, the objects behave under the principle of conservation of momentum. Therefore, if an object moves in the direction opposite to its original direction after a collision, then this indicates that the momentum of the colliding object was greater than the object under consideration. 
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Using Ohm’s Law, explain how resistance changes in relation to current, assuming that voltage remains constant.
Cerrena [4.2K]
V=IR so voltage is directly proportional to current. So for a given resistance increasing the voltage will result in a high current as well. This is because resistance is proportional to the voltage over the current. Ex: I=V/R
Hope this helped. THANKYOU for asking. <span />
7 0
3 years ago
4. How much work is done when you lift a 25 N box up to a 2 m shelf?
Semmy [17]

Answer:

<h2>50 J</h2>

Explanation:

The work done by an object can be found by using the formula

workdone = force × distance

From the question

force = 25 N

distance = 2 m

We have

workdone = 25 × 2 = 50

We have the final answer as

<h3>50 J</h3>

Hope this helps you

6 0
3 years ago
Which astronomer spent 20 years plotting the positions of the planets
klio [65]
That was Tycho Brahe, and I thought it was actually more years than that.
5 0
3 years ago
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Explain how the isotopes of an element are alike and how are they different
leva [86]
All isotopes of the same element have the same number of protons
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3 0
3 years ago
A 450g mass on a spring is oscillating at 1.2Hz. The totalenergy of the oscillation is 0.51J. What is the amplitude.
Volgvan

Answer:

A=0.199

Explanation:

We are given that  

Mass of spring=m=450 g==\frac{450}{1000}=0.45 kg

Where 1 kg=1000 g

Frequency of oscillation=\nu=1.2Hz

Total energy of the oscillation=0.51 J

We have to find the amplitude of oscillations.

Energy of oscillator=E=\frac{1}{2}m\omega^2A^2

Where \omega=2\pi\nu=Angular frequency

A=Amplitude

\pi=\frac{22}{7}

Using the formula

0.51=\frac{1}{2}\times 0.45(2\times \frac{22}{7}\times 1.2)^2A^2

A^2=\frac{2\times 0.51}{0.45\times (2\times \frac{22}{7}\times 1.2)^2}=0.0398

A=\sqrt{0.0398}=0.199

Hence, the amplitude of oscillation=A=0.199

4 0
3 years ago
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