Respuesta :
Answer:
I = 0.160 A
Explanation:
For this exercise we must calculate the power supplied to the motor
power is the work done between the time interval
P = W / t
W = F x
P = [tex]\frac{F \ x}{t}[/tex]
the force is
F = ma
let's use kinematics
v = v₀ + a t
as part of rest v₀ = 0
a = v / t
a = 1.81 / 7.15
a = 0.253 m / s
therefore the applied force is
F = m a
F = 0.836 0.253
F = 0.212 N
to calculate the work, let's find the distance traveled
x = v₀ t + ½ a t²
x = 0+ ½ a t²
x = ½ 0.253 7.15²
x = 6.457 m
therefore The power developed in this movement by the motor
P = 0.212 6.457/7.15
P = 0.19145 W
Let's use a direct proportion rule. If the for a supplied power of Ps = 1W to use 0.471W, what power is united when the effective is 0.19145 W
P_s = 0.19145 W (0.471W/1W)
P_s = 0.4786 W
P_s = 0.479 W
how electrical power is
P = I V
we assume that the batteries are not discharged during the experiment
I = P / V
electrical power equals supplied power
P = Ps
calulate
I = 0.479 / 3.00
I = 0.1597 A
I = 0.160 A
The ratio of power delivered to the electric potential is equal to the average current flow.
The required magnitude of the average current supplied to the car's motor is 0.064 A.
What is the average current?
The amount of electric charges passing through a specific point in a certain period of time is known as an average current.
Given data -
The mass of the toy car is, m = 0.836 kg.
The combined electric potential of two cells is, V' = 3.00 V.
The energy conversion efficiency is, [tex]\eta =47.1 \%[/tex].
The time interval is, t = 7.15 s.
The final speed of car is, v = 1.81 m/s.
The power produced by the two cell is,
P = W / t
P = (F × d)/t
Here,
F is the force exerted on the toy car.
d is the distance traveled by car in the given time interval.
The magnitude of the force is given as,
F = ma
here, a is the magnitude of acceleration.
Now, using the first kinematic equation of motion as,
v = v₀ + a t
As part of rest v₀ = 0
a = v / t
a = 1.81 / 7.15
a = 0.253 m / s²
Therefore the applied force is
F = m a
F = 0.836 × 0.253
F = 0.212 N
To calculate the work, let's find the distance traveled from the second kinematic equation of motion,
d = v₀ t + ½ at²
d = 0+ ½ a t²
d = ½ 0.253 7.15²
d = 6.457 m
Finally, the power developed in this movement by the motor is,
P = (0.212 × 6.457)/7.15
P = 0.19145 W
Now, the expression for the power developed is also given as,
P = V × I
Here,
I is the magnitude of the average current.
Solving as,
I = P / V
I = 0.19145/3.00
I = 0.064 A
Thus, we can conclude that the required magnitude of the average current supplied to the car's motor is 0.064 A.
Learn more about the average current here:
https://brainly.com/question/10597501
