The correct answer to the question is : C) The horizontal momentum and the vertical momentum are both conserved.
EXPLANATION :
Before coming into any conclusion, first we have to understand the law of conservation of momentum.
As per the law of conservation of momentum, the total linear as well as angular momentum of an isolated system is always conserved . The law of conservation of energy is a universal fact.
Hence, during any type of collision, the total momentum is always conserved.
Hence, the total horizontal momentum as well as total vertical momentum are always conserved during both elastic as well as inelastic collision.
Answer:
at T = 0ºC the change of state is from the solid state to the gaseous state
Explanation:
In this exercise we are asked about the changes of state, from the data we will assume that the material is water.
Water can exist in three solid states, liquid and gas, in a graph of pressure ℗ against temperature (T) there is a point called triple at T = 0.01ºC, below this point the curve has two states at high pressure solid and low pressure gas.
As a result of the previous ones at T = 0ºC the change of state is from the solid state to the gaseous state
a) The kinetic energy (KE) of an object is expressed as the product of half of the mass (m) of the object and the square of its velocity (v²):
It is given:
v = 8.5 m/s
m = 91 kg
So:
b) We can calculate height by using the formula for potential energy (PE):
PE = m*g*h
In this case, h is eight, and PE is the same as KE:
PE = KE = 3,287.4 J
m = 91 kg
g = 9.81 m/s² - gravitational acceleration
h = ? - height
Now, let's replace those:
3,287.4= 91 * 9.81 * h
⇒ h = 3,287.4/(91*9.81) = 3,287.4/892.7 = 3.7 m
Answer:
Explanation:
Potential energy on the surface of the earth
= - GMm/ R
Potential at height h
= - GMm/ (R+h)
Potential difference
= GMm/ R - GMm/ (R+h)
= GMm ( 1/R - 1/ R+h )
= GMmh / R (R +h)
This will be the energy needed to launch an object from the surface of Earth to a height h above the surface.
Extra energy is needed to get the same object into orbit at height h
= Kinetic energy of the orbiting object at height h
= 1/2 x potential energy at height h
= 1/2 x GMm / ( R + h)
