Answer:
0.36 A.
Explanation:
We'll begin by calculating the equivalent resistance between 35 Ω and 20 Ω resistor. This is illustrated below:
Resistor 1 (R₁) = 35 Ω
Resistor 2 (R₂) = 20 Ω
Equivalent Resistance (Rₑq) =?
Since, the two resistors are in parallel connections, their equivalence can be obtained as follow:
Rₑq = (R₁ × R₂) / (R₁ + R₂)
Rₑq = (35 × 20) / (35 + 20)
Rₑq = 700 / 55
Rₑq = 12.73 Ω
Next, we shall determine the total resistance in the circuit. This can be obtained as follow:
Equivalent resistance between 35 Ω and 20 Ω (Rₑq) = 12.73 Ω
Resistor 3 (R₃) = 15 Ω
Total resistance (R) in the circuit =?
R = Rₑq + R₃ (they are in series connection)
R = 12.73 + 15
R = 27.73 Ω
Finally, we shall determine the current. This can be obtained as follow:
Total resistance (R) = 27.73 Ω
Voltage (V) = 10 V
Current (I) =?
V = IR
10 = I × 27.73
Divide both side by 27.73
I = 10 / 27.73
I = 0.36 A
Therefore, the current is 0.36 A.
Drag Force = bv^2 = ma; a = g = 9.81 m/s^2
b = mg/v^2 = (0.0023×9.81)/(9.4^2)
b = 0.000255
Answer:
the correct solution is 13 s
Explanation:
This is a kinematic problem, let's use accelerated rectilinear motion relationships.
For the first car it has an accelerometer of 2.0 m/s²
x = v₀₁ t + ½ a₁ t²
The second car leaves the same point, but 4.0 seconds later
x = v₀₂ (t-4) + ½ a₂ (t-4)²
With this form we use the same time for both cars.
The initial speeds are zero for both vehicles leave the rest, at the point where they are located has the same position
x = ½ a₁ t²
x = ½ a₂ (t-4)²
Let's solve
a₁ t² = a₂ (t-4)²
a₁/a₂ t² = t² -2 4 t + 16
t² (1- 2.0 / 4.0) - 8 t +16
t² 0.5 - 8 t +16 = 0
t² -16 t + 32 = 0
Let's solve the second degree equation
t = [16 ±√( 16² - 4 32)] / 2
t = ½ (16 ± 11,3)
Solutions
t1 = 13.66 s
t2 = 2.34 s
These are the mathematical solutions for the meeting point, but car 2 leaves after 4 seconds, so the only solution is 13.66 s
the correct solution is 13 s, if you have to select one the nearest 12s
Answer:
115, 80, 15m
Explanation
t1 = 14s
t2 = 18s
change in time = 4s (18-14)
r(final) = r(initial) + (average velocity) x (change in time)
multiply the average velocity with the change in time
= (4, 0, -3) x 4 = 16, 0, -12
now we'll add this value to the initial position of the car
(99, 80, 27)m + (16, 0, -12)m = (115, 80, 15)m
The statement that describes how work and power are similar is D. you must know time and energy to calculate both.
I am not completely sure though, so I hope this helps. :)
