A star with a mass like the Sun which will soon die is observed to be surrounded by a large amount of dust and gas -- all materi
1 answer:
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Answer:
0.0613°C
Explanation:
the given parameters are m=15gm=15×10⁻³ V₁=865m/s V₂=534m/s
the bullet moves with different kinetic energies before and after the penetration, therefore
Kinetic energy before - kinetic energy after = 1/2 × m × ( V₁² - V₂²)
= × 15×10⁻³ × (865² - 534²)
= 3.47 × 10⁻³J
this loss in energy is transferred to the water, therefore
change in temperature =
where c = heat capacity of water = 4.19 x 10^3
m = mass of water = 13.5 kg
= {3.47 × 10⁻³} / {13.5 x 4.19 x 10^3 }
=0.0613°C
Answer:
Final velocity v = 8.944 m/sec
Explanation:
We have given distance S = 40 meters
Time t = 10 sec
As it starts from rest so initial velocity u = 0
From second equation of motion
Now from first equation of motion , here v is final velocity, u is initial velocity, a is acceleration and t is time
So