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saw5 [17]
3 years ago
14

Which describes what an outside observer at rest would observe about a spaceship that speeds up and approaches the speed of ligh

t?
Time moves faster on the spaceship, and the length of the spaceship decreases. Time moves faster on the spaceship, and the length of the spaceship increases. Time moves more slowly on the spaceship, and the length of the spaceship decreases.
Time moves more slowly on the spaceship, and the length of the spaceship increases.
Physics
2 answers:
Rzqust [24]3 years ago
7 0

Answer:

The correct answer choice is C.  

"Time moves more slowly on the spaceship, and the length of the spaceship decreases."

Explanation:

Answer on Q'let

Art [367]3 years ago
6 0
The outside observer, at rest relative to the spaceship, would see the spaceship
get shorter. and the clocks on the spaceship run slower than they should.

At the same time, the crew of the spaceship, looking back at the observer on
Earth, would see the observer on Earth get shorter, and the observer's clock
run slower than it should.

They would both be measuring what they see correctly.
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An egg is drown at a wall and breaks. A second egg is thrown at the same velocity and collides with a bed sheet. The egg does no
alexdok [17]

Answer:

The first egg is thrown at a wall and breaks because it has no time to decelerate and it breaks.

While, when a second egg is thrown at wall with same velocity and collides with a bed sheet, it does not break because it get more time to decelerate  or collide with the wall.  It reduces the amount of force exerted on the egg and it does not break.

5 0
3 years ago
If you throw an apple towards the sky why its going up in spite of gravitational pull?
AlexFokin [52]

Answer:

Force

Explanation:

The Force You Put Behind Said Apple Is Greater Than That Of Gravitational Pull

Example : The More Power You Put behind a basketball the higher the basketball will go

6 0
3 years ago
Read 2 more answers
Please help with 3 questions about acceleration
Goryan [66]

A great, helpful, useful definition of acceleration is

<em>A = (change in speed) / (time for the change)</em> .   <== you should memorize this

This simple tool will directly solve all 3 problems.

The REASON for assigning these problems for homework is NOT to find the answers.  It's to help YOU find out whether you know this definition, to let you go back and review it if you don't, and to give you a chance to practice using it if you do.  Noticed that if you get the answers from somebody else, you lose all of these benefits.

The only wrinkle anywhere here is in #3, because when you use this definition,      the unit of time has to be the same in both the numerator and the denominator.  

So for #3, you have to EITHER  change the km/hr to km/sec, OR change the 4sec to a fraction of an hour, before you plug anything into the definition.

5 0
3 years ago
Northrop aircraft developed and built a deceleration sled to test the effects of the extreme forces on humans and equipment. In
dolphi86 [110]
20 hours i think that what they do
4 0
3 years ago
An electron travels west to east with a kinetic energy of 10 keV. The Earth's magnetic field in Pittsburgh is 19,911.5 nT in the
ch4aika [34]

Explanation:

It is given that,

Kinetic energy of the electron, E_k=10\ keV=10^4\ eV=1.6\times 10^{-15}\ J

Let the east direction is +x direction, north direction is +y direction and vertical direction is +z direction.    

The magnetic field in north direction, B_y=19911.5\ nT

The magnetic field in west direction, B_x=-3257.1\ nT

The magnetic field in vertical direction, B_z=48381.8 \ nT

Magnetic field, B=(-3257.1i+19911.5j+48381.8k)\ nT

Firstly calculating the velocity of the electron using the kinetic energy formulas as :

E_k=\dfrac{1}{2}mv^2

v=\sqrt{\dfrac{2E_k}{m}}

v=\sqrt{\dfrac{2\times 1.6\times 10^{-15}}{9.1\times 10^{-31}}}

v=5.92\times 10^7 i\ m/s (as it is moving from west to east)

The force acting on the charged particle in the magnetic field is given by :

F=q(v\times B)

F=1.6\times 10^{-19}\times (5.92\times 10^7 i\times (-3257.1i+19911.5j+48381.8k)\times 10^{-9})

Since, i\times j=k\ \\j\times k=i\\k\times i=j

And, i\times i=j\times j=k\times k=0

F=1.6\times 10^{-19}\times [1178 k-2864.20j]

|F|=1.6\times 10^{-19}\times \sqrt{1178^2+2864.20^2}

F=4.95\times 10^{-16}\ N

(b) Let a is the acceleration of the electron. It can be calculated as :

a=\dfrac{F}{m}

a=\dfrac{4.95\times 10^{-16}}{9.1\times 10^{-31}}

a=5.43\times 10^{14}\ m/s^2

Hence, this is the required solution.

4 0
3 years ago
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