Incomplete question as there is so much information is missing.The complete question is here
A car sits on an entrance ramp to a freeway, waiting for a break in the traffic. Then the driver accelerates with constant acceleration along the ramp and onto the freeway. The car starts from rest, moves in a straight line, and has a speed of 24 m/s (54 mi/h) when it reaches the end of the 120-m-long ramp. The traffic on the freeway is moving at a constant speed of 24 m/s. What distance does the traffic travel while the car is moving the length of the ramp?
Answer:
Distance traveled=240 m
Explanation:
Given data
Initial velocity of car v₀=0 m/s
Final velocity of car vf=24 m/s
Distance traveled by car S=120 m
To find
Distance does the traffic travel
Solution
To find the distance first we need to find time, for time first we need acceleration
So
As we find acceleration.Now we need to find time
So
Now for distance
So
The practice of playing top hits several times a day on radio stations is called rotation.
Answer:
1660 V
Explanation:
Resistance should be determined and then voltage drop across the power line can be determined.
R = ρ L /A
Here ρ = Resistivity of aluminum = 
L = length = 32 km = 32,000 m
Area of cross section = A = π r² = π (0.027/2)² = 0.00057255 m²
Resistance = R =
( (32,000)/(0.00057255) = 1.5090 Ohms.
Voltage drop = V = I R = (1100)(1.5090) = 1659.9 V.
(If resistivity value is different, then the resistance will be different and hence final answer for voltage will also vary ).
Answer:
The potential enwrgy associated with charge decreases.
The ele ric field does negative work on the charge.
Explanation:
Answer:
1.8 m/s²
36 N
34.8 N
Explanation:
For the monkey :
m₁ = mass of monkey = 4.50 kg
T₁ = Tension force in the rope on monkey's side
a = acceleration
From the force diagram, force equation for the motion of monkey is given as
m₁ g - T₁ = m₁ a
(4.50 x 9.8) - T₁ = 4.5 a
T₁ = 44.1 - 4.5 a eq-1
For the bunch of bananas :
m₂ = mass of bunch of bananas = 3 kg
T₂ = tension force in the rope on the side of banana
From the force diagram, force equation for the motion of bananas is given as
T₂ - m₂ g = m₂ a
T₂ - (3 x 9.8) = 3 a
T₂ = 29.4 + 3 a eq-2
m = mass of the pulley = 1.50 kg
r = radius of the pulley = 0.090 m
α = angular acceleration of pulley = a/r
Torque equation for the pulley is given as
(T₁ - T₂ )r = I α
(T₁ - T₂ )r = I (a/r)
T₁ - T₂ = (0.5 m r²) (a/r²)
T₁ - T₂ = (0.5) ma
using eq-1 and eq-2
44.1 - 4.5 a - (29.4 + 3 a) = (0.5) ma
44.1 - 4.5 a - (29.4 + 3 a) = (0.5) (1.50) a
a = 1.8 m/s²
Using eq-1
T₁ = 44.1 - 4.5 a
T₁ = 44.1 - 4.5 (1.8)
T₁ = 36 N
using eq-2
T₂ = 29.4 + 3 a
T₂ = 29.4 + 3 (1.8)
T₂ = 34.8 N
