Answer: 66.29%
Explanation:
Efficiency = work output / work input x 100 %
Work output = 822J
Work input = 1240 L
E = 822/1240 x 100
E = 0.6629 x 100
E = 66.29%
The mass slows to a rest from an initial speed of 10.0 m/s in a matter of 1.50 s. If we assume constant acceleration, then the mass has acceleration a such that
0 = 10.0 m/s + a (1.50 s) ⇒ a ≈ -(10.0 m/s)/(1.50 s)
The net force acting on the mass as it slows down is
∑ F[horizontal] = -F[friction] = ma
where m = 15.0 kg, and we take the direction in which friction is acting to be negative. Then
-F[friction] = -(15.0 kg) (10.0 m/s)/(1.50 s) ⇒ F[friction] = 100 N
<h2>
The child swing through the swing's equilibrium position 6 times during the course of 3 periods.</h2>
Explanation:
One period means time taken to complete one revolution.
In case of swings in one period time it travels the same position through two times.
Here we need to find how many times does the child swing through the swing's equilibrium position during the course of 3 period(s) of motion.
For 1 period = 2 times
For 3 periods = 3 x For 1 period
For 3 periods = 3 x 2 times
For 3 periods = 6 times
The child swing through the swing's equilibrium position 6 times during the course of 3 periods.
Answer:
27.2 cm
Explanation:
Note: For the student to see an object clearly at 25 cm, from the eye the image must be formed 309 cm on the same side of the lens, at his near point.
From mirror Formula,
1/f = 1/v+1/u
Where f = focal length of the glasses, v = image distance, u = object distance.
f = vu/(v+u) ................. Equation 1
Given: u = 25 cm, v = -309 cm ( virtual image)
Substitute into equation 1
f = [25×(-309)]/[25+(-309)]
f = -7725/-284
f = 27.2 cm.
Hence the focal length of the glass = 27.2 cm.