Respuesta :
Answer:
a) [tex]I=646.9298\ A[/tex]
b) [tex]I=1942.018\ A[/tex]
Explanation:
Given:
- efficiency of the motor, [tex]\eta=0.95[/tex]
- voltage of the battery, [tex]V=12\ V[/tex]
- mass of the car, [tex]m=708\ kg[/tex]
a)
initial velocity, [tex]u=0\ m.s^{-1}[/tex]
final velocity, [tex]v=25\ m.s^{-1}[/tex]
time taken for the acceleration, [tex]t=1\ min=60\ s[/tex]
Now we know by the Newton's second law of motion:
[tex]F=m.a[/tex]
[tex]F=708\times \frac{(25-0)}{60}[/tex]
[tex]F=295\ N[/tex]
Now the power will be :
[tex]P=F.v[/tex]
[tex]P=295\times 25[/tex]
[tex]P=7375\ W[/tex]
According to the question:
0.95 times of the electrical power should yield this mechanical power.
[tex]P=V.I\times 0.95[/tex]
[tex]7375=12\times I\times 0.95[/tex]
[tex]I=\frac{7375}{12\times 0.95}[/tex]
[tex]I=646.9298\ A[/tex]
b)
height climbed by the car, [tex]h=200\ m[/tex]
velocity of climb, [tex]v=25\ m.s^{-1}[/tex]
time taken to climb the height, [tex]t=2\ min=120\ s[/tex]
force exerted to overcome air and frictional resistances, [tex]f=423\ N[/tex]
Now the Power required to climb the hill:
[tex]P=\frac{m.g.h}{t} +f\times v[/tex]
[tex]P=\frac{708\times 9.8\times 200}{120}+423\times 25[/tex]
[tex]P=22139\ W[/tex]
Now according to the electrical efficiency:
[tex]P=V.I\times 0.95[/tex]
[tex]22139=12\times I\times 0.95[/tex]
[tex]I=1942.018\ A[/tex]
Answer:
(a). The current is 323.4 A.
(b). The current is 1942 A.
Explanation:
Given that,
Efficiency = 95.0 %
Voltage = 12.0 V
Mass of electric car= 708 Kg
Height = 200 m
We need to calculate the change in kinetic energy
Using formula of kinetic energy
[tex]\Delta K=K_{f}-K_{i}[/tex]
Put the value into the formula
[tex]\Delta K=\dfrac{1}{2}mv^2-0[/tex]
[tex]\Delta K=\dfrac{1}{2}\times708\times(25)^2-0[/tex]
[tex]\Delta K=221250\ J[/tex]
We need to calculate the output power
Using formula of power
[tex]P_{o}=\dfrac{\Delta K}{t}[/tex]
[tex]P_{o}=\dfrac{221250}{60}[/tex]
[tex]P_{o}=3687.5\ W[/tex]
We need to calculate the current
Using formula of electric power
[tex]P_{in}=iV[/tex]
[tex]P_{o}=0.95P_{in}[/tex]
[tex]P_{0}=0.95\times iV[/tex]
Put the value into the formula
[tex]3687.5=0.95\times i\times12.0[/tex]
[tex]i=\dfrac{3687.5}{0.95\times12.0}[/tex]
[tex]i=323.4\ A[/tex]
The current is 323.4 A.
(b). We need to calculate the distance
[tex]d=vt[/tex]
Put the value into the formula
[tex]d=25\times2\times60[/tex]
[tex]d=3000\ m[/tex]
We need to calculate the force
Using formula of force
[tex]F=mg\sin\theta[/tex]
Put the value into the formula
[tex]F=708\times9.8\times\dfrac{200}{3000}[/tex]
[tex]F=462.56\ N[/tex]
We need to calculate the power
Using formula of power
[tex]P=F\times v[/tex]
Put the value into the formula
[tex]P=(462.56+423)\times25[/tex]
[tex]P=22139\ W[/tex]
We need to calculate the current
Using formula of current
[tex]I=\dfrac{P}{V}[/tex]
Put the value into the formula
[tex]I=\dfrac{22139\times100}{95\times12}[/tex]
[tex]I=1942.0\ A[/tex]
The current is 1942 A.
Hence, (a). The current is 323.4 A.
(b). The current is 1942 A.
