Pressure is the amount of force per unit area. In formula it
is,
P = F ÷ A
P = 7000 N ÷ 0.4 m2
P = 17,500 N/m2
The amount of pressure the truck exerts on the piston is
17,500 N/m2
Answer:
The detailed calculations are shown below;
Explanation:
a)The maximum acceleration of the particle:
It is seen that the maximum change in velocity is at the time between 8s to 10s.
Maximum acceleration: 
= 
= 10 m/
b) The deceleration of the particle
The velocity of particle is decreased after 10s so,
deceleration = - 
= - 6.67 m/
c)The total distance traveled by the particle = Area under the curve
=
* 4*20 + 4*20 +
* 2*20+ 2*20+
* 40*16
= 290 m
d)The average velocity of the particle = 
= 
= 18.12 m/s
Answer:
(a) 43.2 kC
(b) 0.012V kWh
(c) 0.108V cents
Explanation:
<u>Given:</u>
- i = current flow = 3 A
- t = time interval for which the current flow =

- V = terminal voltage of the battery
- R = rate of energy = 9 cents/kWh
<u>Assume:</u>
- Q = charge transported as a result of charging
- E = energy expended
- C = cost of charging
Part (a):
We know that the charge flow rate is the electric current flow through a wire.

Hence, 43.2 kC of charge is transported as a result of charging.
Part (b):
We know the electrical energy dissipated due to current flow across a voltage drop for a time interval is given by:

Hence, 0.012V kWh is expended in charging the battery.
Part (c):
We know that the energy cost is equal to the product of energy expended and the rate of energy.

Hence, 0.108V cents is the charging cost of the battery.
Answer:
The correct answer is "6.96 rad/s".
Explanation:
The given values are:
Length,
L = 0.6 m
Mass,
m₁ = 0.5 kg
m₂ = 0.2 kg
Initial velocity,
V = 8 m/s
Now,
The final angular velocity will be:
⇒ 
By substituting the values, we get
⇒ 
⇒ 
⇒ 