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3 years ago

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A starship passes Earth at 80% of the speed of light and sends a drone ship forward at half the speed of light rela- tive to its

elf. Show that the drone travels at 93% of the speed of light relative to Earth.

1 answer:

alisha [4.7K]3 years ago

8 0

Answer:

Check Explanation

Explanation:

Let the speed of the drone relative to earth be u = ?

Let the speed of the drone relative to the starship be u' = 0.5c

Let the speed of the starship relative to earth be v = 0.8c

In the theory of relativity,

The 3 speeds are related thus

u = (u' + v)/[1 + (u'v/c²)]

u = (0.5c + 0.8c)[1 + ((0.8c × 0.5c)/c²)]

u = 1.3c/[1 + (0.4c²/c²)]

u = 1.3c/1.4 = 0.9286 c = 0.93 c = 93% c

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Which refers to the amount of heat required to change the temperature of 1 gram of a substance by 1°C and is related to the chem

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The CERN particle accelerator is circular with a circumference of 7.0 km.

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Answer:

$a_c=2.0196\times 10^{13}\ m/s^2$

$F=3.37273\times 10^{-14}\ N$

Explanation:

m = Mass of proton = $1.67\times 10^{-27}\ kg$

v = Speed of proton = 0.5c = $0.5\times 3\times 10^8=1.5\times 10^8\ m/s$

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Centripetal acceleration is $2.0196\times 10^{13}\ m/s^2$

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Force on protons is $3.37273\times 10^{-14}\ N$

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3 years ago

If the coefficient of kinetic friction between a 22 kg kg crate and the floor is 0.27, what horizontal force is required to move

alina1380 [7]

Answer:

58.27 N

Explanation:

the data we have is:

mass: $m=22kg$

coefficient of friction: $\mu =0.27$

and we also know the acceleration of gravity is $g=9.81m/s^2$

We need to do an analysis of horizontal and vertical forces acting on the object:

-------

Vertically the forces acting on the object:

Normal force $N$ (acting up from the object)

weight: $w=mg$ (acting down from)

so the sum of forces in the vertical axis "y" are:

$F_{y}=N-w\\F_{y}=N-mg$

from Newton's second Law we know that $F=ma$ , so:

$ma_{y}=N-mg$

and since the object is not accelerating in the vertical direction (the movement is only horizontal) $a_{y}=0$ , and:

$0=N-mg\\N=mg$

-----------

now let's analyze the horizontal forces

frictional force: $f= \mu N$ and since $N=mg$ --> $f=\mu mg$

force to move the object: $F$

and the two forces just mentioned must be opposite, thus the sum of forces in the "x" axis is:

$F=ma_{x}=F-f\\ma_{x}=F-\mu mg$

and we are told that the crate moves at a steady speed, thus there is no acceleration: $a_{x}=0$

and we get:

$0=F-\mu mg\\F=\mu mg$

substituting known values:

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A train is moving along a horizontal track. A pendulum suspended from the roof makes an angle of 4° with the vertical. If g=10m/

nataly862011 [7]

Answer:

Train accaleration = 0.70 m/s^2

Explanation:

We have a pendulum (presumably simple in nature) in an accelerating train. As the train accelerates, the pendulum is going move in the opposite direction due to inertia. The force which causes this movement has the same accaleration as that of the train. This is the basis for the problem.

Start by setting up a free body diagram of all the forces in play: The gravitational force on the pendulum (mg), the force caused by the pendulum's inertial resistance to the train(F_i), and the resulting force of tension caused by the other two forces (F_r).

Next, set up your sum of forces equations/relationships. Note that the sum of vertical forces (y-direction) balance out and equal 0. While the horizontal forces add up to the total mass of the pendulum times it's accaleration; which, again, equals the train's accaleration.

After doing this, I would isolate the resulting force in the sum of vertical forces, substitute it into the horizontal force equation, and solve for the acceleration. The problem should reduce to show that the acceleration is proportional to the gravity times the tangent of the angle it makes.

I've attached my work, comment with any questions.

Side note: If you take this end result and solve for the angle, you'll see that no matter how fast the train accelerates, the pendulum will never reach a full 90°!

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