A 0.600 kg toy car is powered by one 9V battery (9.00 V) connected directly to a small DC motor. The car has an effective energy
1 answer:
Answer:
The average current I supplied to the cars motor is 0.225 Amperes.
Explanation:
The mass of the toy car is given as, m = 0.6 kg
The toy car operates at a steady speed of, v = 3.43 m/s.
It achieves the steady speed after a time, t = 5.69 seconds.
The total kinetic energy of the car is given by,
E =
= 0.5 × 0.6 ×
= 3.53 Joules
Since the efficiency of the motor is 30.6% = 0.306
The electrical energy equivalent is supplied to motor = .
The electrical energy is given by P × t = I × V × t = 11.54 Joules
P is the electric power where P = I × V where V is the voltage of the cell and is the average current, I.
Therefore we can write I × 9 × 5.69 = 11.54 Joules.
So the average current I = = 0.225 Amperes.
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Answer:
y = 17 m
Explanation:
For this projectile launch exercise, let's write the equation of position
x = v₀ₓ t
y = t - ½ g t²
let's substitute
45 = v₀ cos θ t
10 = v₀ sin θ t - ½ 9.8 t²
the maximum height the ball can reach where the vertical velocity is zero
v_{y} = v_{oy} - gt
0 = v₀ sin θ - gt
0 = v₀ sin θ - 9.8 t
Let's write our system of equations
45 = v₀ cos θ t
10 = v₀ sin θ t - ½ 9.8 t²
0 = v₀ sin θ - 9.8 t
We have a system of three equations with three unknowns for which it can be solved.
Let's use the last two
v₀ sin θ = 9.8 t
we substitute
10 = (9.8 t) t - ½ 9.8 t2
10 = ½ 9.8 t2
10 = 4.9 t2
t = √ (10 / 4.9)
t = 1,429 s
Now let's use the first equation and the last one
45 = v₀ cos θ t
0 = v₀ sin θ - 9.8 t
9.8 t = v₀ sin θ
45 / t = v₀ cos θ
we divide
9.8t / (45 / t) = tan θ
tan θ = 9.8 t² / 45
θ = tan⁻¹ ( 9.8 t² / 45 )
θ = tan⁻¹ (0.4447)
θ = 24º
Now we can calculate the maximum height
v_y² = - 2 g y
vy = 0
y = v_{oy}^2 / 2g
y = (20 sin 24)²/2 9.8
y = 3,376 m
the other angle that gives the same result is
θ‘= 90 - θ
θ' = 90 -24
θ'= 66'
for this angle the maximum height is
y = v_{oy}^2 / 2g
y = (20 sin 66)²/2 9.8
y = 17 m
thisis the correct