Answer:
Explanation:
a) Force of friction = μ R where μ is coefficient of kinetic friction and R is reaction force
R = mg where m is mass of the block
Force of friction F = μ x mg
= .173 x 12.2 x 9.8
= 20.68 N
b ) Only force of friction is acting on the body so
deceleration = force / mass = 20.68 / 12.2 = 1.7 m /s²
acceleration = - 1.7 m /s²
c )
v² = u² - 2 a s
v = 0 , u = 3.9 m /s
a = 1.7 m /s
0 = 3.9² - 2 x 1.7 x s
s = 4.47 m
When a force causes a body to move, work is done on the object by the force. Work is the measure of the energy transfer when a force 'F' moves an object through a distance 'd'. So we say that energy is transferred from one energy store to another when work is done, and therefore, energy transferred = work done.
Answer:
1.4m/s
Explanation:
Average velocity is the total distance covered divided by the total time taken.
Average velocity = 
Total time taken = 5s + 6s = 11s
The first distance covered = velocity x time = 1.4 x 5 = 7m
second distance covered = velocity x time = 1.4 x 6 = 8.4m
So;
Average velocity = 
= 1.4m/s
Answer:
0° C
Explanation:
Given that
Mass of ice, m = 50g
Mass of water, m(w) = 50g
Temperature of ice, T(i) = 0° C
Temperature of water, T(w) = 80° C
Also, it is known that
Specific heat of water, c = 1 cal/g/°C
Latent heat of ice, L(w) = 89 cal/g
Let us assume T to be the final temperature of mixture.
This makes the energy balance equation:
Heat gained by ice to change itself into water + heat gained by melted ice(water) to raise its temperature at T° C = heat lost by water to reach at T° C
m(i).L(i) + m(i).c(w)[T - 0] = m(w).c(w)[80 - T], on substituting, we have
50 * 80 + 50 * 1(T - 0) = 50 * 1(80 - T)
4000 + 50T = 4000 - 50T
0 = 100 T
T = 0° C
Thus, the final temperature is 0° C
Answer:
240 V
Explanation:
number of turns in primary coil, Np = 10
Number of loops in secondary coil, Ns = 20
Voltage in primary coil, Vp = 120 V
Let the voltage in secondary coil is Vs.
So, Vs / Vp = Ns / Np
Vs / 120 = 20 / 10
Vs / 120 = 2
Vs = 240 V
Thus, the voltage in secondary coil is 240 Volt.
