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Thepotemich [5.8K]
3 years ago
8

Distance measurements based on the speed of light; used for objects in space

Physics
1 answer:
Vika [28.1K]3 years ago
4 0

Answer : Yes, distance measurements based on the speed of light used for objects in space.

Explanation :  A light year is measurement  of distance  that light travel in a one year.

In a one year light travels 9460000000000 kilometer.

We know that, speed of light is 3\times10^{8}\ m/s

and  time is 31536000 seconds in 1 year

so, distance = speed of light X time

Now, the light year is 9.4608\times10^{15} meter

Example : The nearest star to earth is about 4.3 light year away.

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A beta particle. Hoped I help. Sorry if it wrong.

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What happens to a liquid beyond its certain temperature
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2 years ago
An electric vehicle starts from rest and accelerates at a rate of 2.4 m/s2 in a straight line until it reaches a speed of 27 m/s
DENIUS [597]
A) use v=u+at for both

First section, v=27, u=0, a=2.4. You should get 11seconds.
Second section, v=0, u=27, a=-1.3. You should get 21seconds.
This means that the total time is 22seconds.

b) You can either use s=ut+0.5at^2 or v^2=u^2+2as. Personally, I would use the second one as you are not relying on your previous answer.

First section, v=27, u=0, a=2.4. You should get 152m.
Second section, v=0, u=27, a=-1.3. You should get 280m.
This makes your overall displacement 432m.
6 0
3 years ago
A jet, sitting on the runway, takes off and accelerates at 8.0 m/s for 16s. How far did the jet travel down the runway?
nydimaria [60]

Answer:

2.4 m/s". 1

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A jet with mass m = 8 x 10* kg jet accelerates down the runway for takeoff at 2.4 m/s". 1

8 0
3 years ago
A 500 kg dragster accelerates from rest to a final speed of 100 m/s in 400 m (about a quarter of a mile) and encounters an avera
Fantom [35]
In order to accelerate the dragster at a speed v_f = 100 m/s, its engine must do a work equal to the increase in kinetic energy of the dragster. Since it starts from rest, the initial kinetic energy is zero, so the work done by the engine to accelerate the dragster to 100 m/s is
W= K_f - K_i = K_f =  \frac{1}{2}mv_f^2=2.5 \cdot 10^6 J

however, we must take into account also the fact that there is a frictional force doing work against the dragster, and the work done by the frictional force is:
W_f = F_f d = -(1200 N)(400 m)= -4.8 \cdot 10^5 J
and the sign is negative because the frictional force acts against the direction of motion of the dragster.

This means that the total work done by the dragster engine is equal to the work done to accelerate the dragster plus the energy lost because of the frictional force, which is -W_f:
W_t = W + (-W_f)=2.5 \cdot 10^6 J+4.8 \cdot 10^5 J=2.98 \cdot 10^6 J

So, the power delivered by the engine is the total work divided by the time, t=7.30 s:
P= \frac{W}{t}= \frac{2.98 \cdot 10^6 J}{7.30 s}=4.08 \cdot 10^6 W

And since 1 horsepower is equal to 746 W, we can rewrite the power as
P=4.08 \cdot 10^6 W \cdot  \frac{1 hp}{746 W} =547 hp



3 0
3 years ago
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