The power of batteries in a circuit. The key functions of a battery and bulb in a circuit are explained. A battery is a source of energy which provides a push - a voltage - of energy to get the current flowing in a circuit. A bulb uses the electrical energy provided by the battery, but does not use current.
HOPE IT HELPS YOU OUT!
Conservation of energy explains that energy can only be transferred between different forms of energy
Solution :
Given :
M = 0.35 kg

Total mechanical energy = constant
or 
But
and 
Therefore, potential energy at the top = kinetic energy at the bottom


(h = 35 cm = 0.35 m)
= 2.62 m/s
It is the velocity of M just before collision of 'm' at the bottom.
We know that in elastic collision velocity after collision is given by :

here, 
∴ 

= 0.33 m/s
Therefore, velocity after the collision of mass M = 0.33 m/s
Mechanical digestion is chewing, and chemical digestion is the saliva in your mouth breaking down food.
Answer:
A u = 0.36c B u = 0.961c
Explanation:
In special relativity the transformation of velocities is carried out using the Lorentz equations, if the movement in the x direction remains
u ’= (u-v) / (1- uv / c²)
Where u’ is the speed with respect to the mobile system, in this case the initial nucleus of uranium, u the speed with respect to the fixed system (the observer in the laboratory) and v the speed of the mobile system with respect to the laboratory
The data give is u ’= 0.43c and the initial core velocity v = 0.94c
Let's clear the speed with respect to the observer (u)
u’ (1- u v / c²) = u -v
u + u ’uv / c² = v - u’
u (1 + u ’v / c²) = v - u’
u = (v-u ’) / (1+ u’ v / c²)
Let's calculate
u = (0.94 c - 0.43c) / (1+ 0.43c 0.94 c / c²)
u = 0.51c / (1 + 0.4042)
u = 0.36c
We repeat the calculation for the other piece
In this case u ’= - 0.35c
We calculate
u = (0.94c + 0.35c) / (1 - 0.35c 0.94c / c²)
u = 1.29c / (1- 0.329)
u = 0.961c