Answer:
Explanation:
Initial speed, v = 10 x 10^3 m/s
Mass of the earth, M = 6 x 10^24 kg
Radius of the earth, R = 6.4 x 10^6 m
Maximum from the surface of earth, h = ?
Let m = Mass of the projectile
Solution:
Potential energy at maximum height = ( Potential + Kinetic energy ) at the surface
=
=
(1 cal/g °C) x (4000 g) x (45 - 25)°C = 80000 cal = 80 kcal. So the answer is 80 kcal .
Answer:
speed of the bullet before it hit the block is 200 m/s
Explanation:
given data
mass of block m1 = 1.2 kg
mass of bullet m2 = 50 gram = 0.05 kg
combine speed V= 8.0 m/s
to find out
speed of the bullet before it hit the block
solution
we will apply here conservation of momentum that is
m1 × v1 + m2 × v2 = M × V .............1
here m1 is mass of block and m2 is mass of bullet and v1 is initial speed of block i.e 0 and v2 is initial speed of bullet and M is combine mass of block and bullet and V is combine speed of block and bullet
put all value in equation 1
m1 × v1 + m2 × v2 = M × V
1.2 × 0 + 0.05 × v2 = ( 1.2 + 0.05 ) × 8
solve it we get
v2 = 200 m/s
so speed of the bullet before it hit the block is 200 m/s
Answer:
A - Crest, B - amplitude, C - wavelength, D - trough
Explanation:
That would be
0 degrees Celsius aka the melting point of water.... If you look at the diagram I attached you notice that at 0 degrees Celsius it is flat, this is because much heat is needed at this point for water to rise to 1 degree... It is the same for the boiling point (100)<span />
