Answer:
a) 2.5m/s
b) 0.91m/s
c) 0m/s
Explanation:
Average velocity can be said to be the ratio of the displacement with respect to time.
Average speed on the other hand is the ratio of distance in relation to time
Thus, to get the average velocity for the first half of the swim
V(average) = displacement of first trip/time taken on the trip
V(average) = 50/20
V(average) = 2.5m/s
Average velocity for the second half of the swim will be calculated in like manner, thus,
V(average) = 50/55
V(average) = 0.91m/s
Average velocity for the round trip will then be
V(average) = 0/75, [50+25]
V(average) = 0m/s
A.) We use the famous equation proposed by Albert Einstein written below:
E = Δmc²
where
E is the energy of the photon
Δm is the mass defect, or the difference of the mass before and after the reaction
c is the speed of light equal to 3×10⁸ m/s
Substituting the value:
E = (1.01m - m)*(3×10⁸ m/s) = 0.01mc² = 3×10⁶ Joules
b) The actual energy may be even greater than 3×10⁶ Joules because some of the energy may have been dissipated. Not all of the energy will be absorbed by the photon. Some energy would be dissipated to the surroundings.
Flow rate = 220*0.355 l/m = 78.1 l/min = 1.3 l/s = 0.0013 m^3/s
Point 2:
A2= 8 cm^2 = 0.0008 m^2
V2 = Flow rate/A2 = 0.0013/0.0008 = 1.625 m/s
P1 = 152 kPa = 152000 Pa
Point 1:
A1 = 2 cm^2 = 0.0002 m^2
V1 = Flow rate/A1 = 0.0013/0.0002 = 6.5 m/s
P1 = ?
Height = 1.35 m
Applying Bernoulli principle;
P2+1/2*V2^2/density = P1+1/2*V1^2/density +density*gravitational acceleration*height
=>152000+0.5*1.625^2*1000=P1+0.5*6.5^2*1000+1000*9.81*1.35
=> 153320.31 = P1 + 34368.5
=> P1 = 1533210.31-34368.5 = 118951.81 Pa = 118.95 kPa
Answer: deceleration of 
Explanation:
Given
Car is traveling at a speed of u=20 m/s
The diameter of the car is d=70 cm
It slows down to rest in 300 m
If the car rolls without slipping, then it must be experiencing pure rolling i.e. 
Using the equation of motion

Insert 

Write acceleration as 

So, the car must be experiencing the deceleration of
.
Time taken by the water balloon to reach the bottom will be given as

here we know that


now by the above formula



now in the same time interval we can say the distance moved by it will be


so it will fall at a distance 15.7 m from its initial position