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zaharov [31]
1 year ago
12

an electron is moving with a speed of 0.85c in a direction opposite to that of photon. calculate the relative velocity of the el

ectron and photon. ​
Physics
1 answer:
ololo11 [35]1 year ago
5 0

Answer:

1.85c

Explanation:

a photon moves at c, the electron is moving at 0.85c, and since they are moving in opposing directions, the relative speed would be 1.85c

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If the Sun and volcanoes were controlling the climate, the climate would
Harman [31]

Answer:

<em>Undergo global warming at a faster rate than what we are seeing currently</em>

Explanation:

Climate can be described as the average weather of a place. The climate of a particular place can be described after looking at the temperature of the place for a year or more.

If factors, such as the Sun and volcanoes controlled climates then there would be an increase in the temperature and more global warming. Volcanoes can be described as heat erupting from mountains which will, of course, lead to global warming.

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3 years ago
Define Newton’s three laws of motion and how they apply to everyday situations.
son4ous [18]
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2 years ago
Which of these steps would most likely be part of a lab procedure?
Snezhnost [94]

Answer:

These are a part of lab procedures:

1. Write a hypothesis to answer a question.

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3. Record the time to complete a chemical reaction.

These are NOT a part of lab procedures:

1. Create a question on the cause of a chemical reaction.

8 0
2 years ago
Determine the gravitational field 300km above the surface of the earth. How does this compare to the field on the earth's surfac
Serjik [45]
The strength of the gravitational field is given by:
g= \frac{GM}{r^2}
where
G is the gravitational constant
M is the Earth's mass
r is the distance measured from the centre of the planet.

In our problem, we are located at 300 km above the surface. Since the Earth radius is R=6370 km, the distance from the Earth's center is:
r=R+h=6370 km+300 km=6670 km= 6.67 \cdot 10^{6} m

And now we can use the previous equation to calculate the field strength at that altitude:
g= \frac{GM}{r^2}= \frac{(6.67 \cdot 10^{-11} m^3 kg^{-1} s^{-2})(5.97 \cdot 10^{24} kg)}{(6.67 \cdot 10^6 m)^2}  = 8.95 m/s^2

And we can see this value is a bit less than the gravitational strength at the surface, which is g_s = 9.81 m/s^2.
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Answer:

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Explanation:

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