That ratio is called"efficiency". It doesn't need to be a percent.
It can just as well be a fraction or a decimal number.
Answer:
The answer is D
Explanation:
I'm too lz to explain everything.
sorry.
Answer:
Explanation:
Given that:
the initial angular velocity 
angular acceleration
= 4.44 rad/s²
Using the formula:

Making t the subject of the formula:

where;

∴

t = 0.345 s
b)
Using the formula:

here;
= angular displacement
∴



Recall that:
2π rad = 1 revolution
Then;
0.264 rad = (x) revolution

x = 0.042 revolutions
c)
Here; force = 270 N
radius = 1.20 m
The torque = F * r

However;
From the moment of inertia;

given that;
I = 84.4 kg.m²

For re-tardation; 
Using the equation



t = 0.398s
The required time it takes= 0.398s
Answer:
9.43 m/s
Explanation:
First of all, we calculate the final kinetic energy of the car.
According to the work-energy theorem, the work done on the car is equal to its change in kinetic energy:

where
W = -36.733 J is the work done on the car (negative because the car is slowing down, so the work is done in the direction opposite to the motion of the car)
is the final kinetic energy
is the initial kinetic energy
Solving,

Now we can find the final speed of the car by using the formula for kinetic energy

where
m = 661 kg is the mass of the car
v is its final speed
Solving for v, we find

Answer:
the radii of curvature is 30 cm.
Explanation:
given,
object is place at = 45 cm
image appears at = 90 cm
focal length = ?
refractive index = 1.5
radii of curvature = ?


f = 30 cm
using lens formula





R = 30 cm
hence, the radii of curvature is 30 cm.