Here when car in front of us applied brakes then it is slowing down due to frictional force on it
So here we can say that friction force on the car front of our car is given as

So the acceleration of car due to friction is given as



now it is given that


so here we have


so the car will accelerate due to brakes by a = - 8.52 m/s^2
V = IR
By completing the equation, i found that the total power equation is : 4.8,
Which means that it's not exceed the power rating.
So i believe the answer would be : The string will remain lit
hope this helps
The label means that the bulb will consume 75 joules of electrical energy in 1 sec
Answer:
θ = 13.16 °
Explanation:
Lets take mass of child = m
Initial velocity ,u= 1.1 m/s
Final velocity ,v=3.7 m/s
d= 22.5 m
The force due to gravity along the incline plane = m g sinθ
The friction force = (m g)/5
Now from work power energy
We know that
work done by all forces = change in kinetic energy
( m g sinθ - (m g)/5 ) d = 1/2 m v² - 1/2 m u²
(2 g sinθ - ( 2 g)/5 ) d = v² - u²
take g = 10 m/s²
(20 sinθ - ( 20)/5 ) 22.5 = 3.7² - 1.1²
20 sinθ - 4 =12.48/22.5
θ = 13.16 °
Answer:
kinetic energy will change by a factor of 1/2
Option C) 1/2 is the correct answer
Explanation:
Given the data in the question;
we know that;
Kinetic energy = 1/2.mv²
given that mass of the object is doubled; m1 = 2m
speed is halved; v1 = V/2
Now, New kinetic energy will be; 1/2.m1v1²
we substitute
Kinetic Energy = 1/2 × 2m × (v/2)²
Kinetic Energy = 1/2 × 2m × (v²/4)
Kinetic Energy = 1/2 × m × (v²/2)
Kinetic Energy = 1/2 [ 1/2mv² ]
Kinetic Energy = 1/2 [ KE ]
Therefore; kinetic energy will change by a factor of 1/2
Option C) 1/2 is the correct answer