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mars1129 [50]
3 years ago
6

There is an upper limit on the speed of a particle. This means there is also an upper limit on its

Physics
1 answer:
Dafna11 [192]3 years ago
5 0

Answer:

As speed of a particle has a limit so does its momentum has a limit.

Explanation:

As we know that in this Universe no information can travel faster than the speed of light.

The speed of light is the fastest possible speed and hence the quantities dependent on it also have certain limits.

The momentum of a body at normal non-relativistic speed is given as:

p=m.v

The momentum of a body at relativistic speed (a speed of the order of speed of light) is given as:

p=m.v.\gamma

here

\gamma =\frac{1}{\sqrt{1-\frac{v^2}{c^2} }}  Lorentz factor

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If the back of the truck is 1.3 m above the ground and the ramp is inclined at 22 ∘ , how much time do the workers have to get t
solong [7]
Refer to the diagram shown below.

Assume that
(a) The piano rolls down on frictionless wheels,
(b) Wind resistance is negligible.

The distance along the ramp is
d = (1.3 m)/sin(22°) = 3.4703 m

The component of the piano's weight along the ramp is
mg sin(22°)
If the acceleration down the ramp is a, then
ma = mg sin(22°)
a = g sin(22°) = (9.8 m/s²) sin(22°) = 3.671 m/s²

The time, t, to travel down the ramp from rest is given by
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3 0
3 years ago
A cat dozes on a stationary merry-go-round, at a radius of 4.4 m from the center of the ride. The operator turns on the ride and
monitta

Answer:

The coefficient of static friction is 0.29

Explanation:

Given that,

Radius of the merry-go-round, r = 4.4 m

The operator turns on the ride and brings it up to its proper turning rate of one complete rotation every 7.7 s.

We need to find the least coefficient of static friction between the cat and the merry-go-round that will allow the cat to stay in place, without sliding. For this the centripetal force is balanced by the frictional force.

\mu mg=\dfrac{mv^2}{r}

v is the speed of cat, v=\dfrac{2\pi r}{t}

\mu=\dfrac{4\pi^2r}{gt^2}\\\\\mu=\dfrac{4\pi^2\times 4.4}{9.8\times (7.7)^2}\\\\\mu=0.29

So, the least coefficient of static friction between the cat and the merry-go-round is 0.29.

4 0
3 years ago
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