Answer:
Power, P = 30 W
Explanation:
We have,
Voltage drop of a circuit is 60 V
Resistance of the resistor is 120 ohms
Current across the circuit is 0.5 A
It is required to find the power conducted by the resistor. Power conducted by a resistor is given by :
V and I are voltage and current
So, the power conducted by the resistor is 30 watts.
Answer:
b = 242 m
Explanation:
A = 24200 m²
a = 100 m
b = ?
A seguinte fórmula é aplicada
A = a*b
⇒ b = A / a
⇒ b = (24200 m²) / (100 m)
⇒ b = 242 m
22. a - (vf^2 - vi^2)/(2d)
a = (0 - 23^2)/(170)
a = -3.1 m/s^2
23. Find the time (t) to reach 33 m/s at 3 m/s^2
33-0/t = 3
33 = 3t
t = 11 sec to reach 33 m/s^2
Find the av velocuty: 33+0/2 = 16.5 m/s
Dist = 16.5 * 11 = 181.5 meters to each 33m/s speed. Runway has to be at least this long.
24. The sprinter starts from rest. The average acceleration is found from:
(Vf)^2 = (Vi)^2 = 2as ---> a = (Vf)^2 - (Vi)^2/2s = (11.5m/s)^2-0/2(15.0m) = 4.408m/s^2 estimated: 4.41m/s^2
The elapsed time is found by solving
Vf = Vi + at ----> t = vf-vi/a = 11.5m/s-0/4.408m/s^2 = 2.61s
25. Acceleration of car = v-u/t = 0ms^-1-21.0ms^-1/6.00s = -3.50ms^-2
S = v^2 - u^2/2a = (0ms^-1)^2-(21.0ms^-1)^2/2*-3.50ms^-2 = 63.0m
26. Assuming a constant deceleration of 7.00 m/s^2
final velocity, v = 0m/s
acceleration, a = -7.00m/s^2
displacement, s - 92m
Using v^2 = u^2 - 2as
0^2 - u^2 + 2 (-7.00) (92)
initial velocity, u = sqrt (1288) = 35.9 m/s
This is the speed pf the car just bore braking.
I hope this helps!!
Answer:
Push and pull are both forces, but the difference is in their direction at which it is applied. If the force applied in the direction of motion of the particle then we call it push. If that force applied in the direction OPPOSITE to the motion of the particle then it it termed as pull
