If objects travelling at near the speed of light gain mass because their

If objects travelling at near the speed of light gain mass because their energy of their motion is being converted into mass, the does a photon gain mass?

2 months ago

Solution 1

Guest Guest #9847890
2 months ago
A photon travels only at the speed of light, and has no mass at any lower speed.

If that makes sense to you, then there is a problem.  But that's the way it is.

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A people-moving conveyor-belt moves a 600-newton person a distance of 100 meters through the airport.
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b. The same 600-newton person lifts his 100-newton carry-on bag upward a distance of 1 meter. They 
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Solution 1
In a), no work is done to move horizontally 'across" gravity. In b), he does (m) (g) (h) = (100)(9.8)(1) = 980 joules of work to lift the bag, and then no more work is done by the belt to move both of them horizontally.
What is the displacement after 6 seconds when a car is moving at 12m/s and coasts up a hill with a uniform aacceleration of -1.6 m/s^2
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The kinematic formula using displacement, time, initial velocity, and acceleration:
x = v t+.5 a t^2
x= (12m/s)(6s) + .5(-1.6m/s^2)(6s)^2
x= 72m - 28.8m
x = 43.2 m
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If a dune Buggy is traveling for 20 seconds at a speed of 8.5m/s how do I calculate the unknown quantities
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Right now I can think of about 50 unknown quantities, and the one you need to find is the most unknown of all of them, because you haven't even sait WHAT quantity you need to find.
What are two things that the amount of gravitational force between two object depends on
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1).  the product of the two masses being gravitationally attracted to each other

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The equation behind this question is:

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How far will a free falling object have fallen from a position of rest when it's instantaneous speed is 10m/s?
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Use kinematic equation:

vf² = vo² + 2g(xf - xo)

vf = final velocity = 10 m/s
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g = 9.81 m/s²
xf - xo = distance traveled = d

*rearrange equation to solve for d

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d = (100 m²/s²)/(2*9.81 m/s²)

d = 5.097 m

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Answer: B forces Hope this helps!!

A fugitive tries to hop a freight train traveling at a constant speed Of 4.5 m/s. Just as a empty box car passes him, the fugitive starts from rest and accelerates at a= 3.6 m/s squared to his maximum speed of 8.0 m/s. How long does it take him to catch up to the empty box car? What is the distance traveled to reach the box car?
Solution 1
We use the kinematics equation:
Vf = Vi + a*t
8 = 0 + 3.6 * t
t=2.222s to reach 8.0 m/s

At that time the train has moved
4.5 m/s * 2.222s = 9.999 m

He travelled (another kinematics equation)
Vf^2 = Vi^2 + (2*a*d)
(8.0)^2 = (0)^2 + (2 * 3.6 * d)
d=8.888 m

The train is 9.999m, the fugitive is 8.888m,
He still needs to travel
9.999-8.888= 1.111m

He needs to cover the rest of the distance in a smaller amount of time, however hes at his maximum velocity, so...
8m/s(man) - 4.5m/s(train) = 3.5 m/s more

(1.111m) / (3.5m/s) = .317seconds more to reach the train

So if it takes 2.222 seconds to approach the train at 8.888m, it should take
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Last but not least is to figure out the total distance traveled in that time frame:

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How do you find displacement when all you have is time with no velocity, and then in order to find velocity you have to have displacement but you don't have that either?
Solution 1
If you really have nothing else but time, then you can't. There must be some other shred of information. Search around. Look under rocks.