efficiency= [useful energy transferred ÷ total energy supply]×100%
So, [5500÷10000]×100%=0.55×100
=55%
Answer:
3.6 m
Explanation:
let x = horizontal distance between emily and allison should be for allison to catch the ball
Find horizontal speed of the ball
vx = 12 sin 30 = 12 x 0.5 = 6 m/s
To find time taken, we will use vertical values of the ball motion
Initial velocity in vertical direction
u = 12 cos 30 = 10.392 m/s
let a = g = 9.8m/s2
Use equation of motion
s = ut +1/2at^2
s = vertical distance = 8
8 = (10.392)t + (1/2)(9.8)t^2
8 = (10.392)t + (4.9)t^2
4.9t^2 + 10.392t - 8 = 0
Using formula of quadratic or calculator, we'll find
t = 0.6 and t = -2.72
We pick t=0.6s since it's not logical time in negative
Assuming no air resistance or external forces, the ball will move 6m/s horizontally. Hence using the formula of speed
speed vx = distance x / time
x = (vx)(t)
= 6 x 0.6
= 3.6 m
Answer:

Explanation:
Here we know that the glider is accelerated uniformly from rest to final speed of 25.7 m/s in total distance of d = 46.9 m
so we will have


d = 46.9
so for uniformly accelerated motion we have



now we will find the total work done given as change in kinetic energy



now power is given as



Answer:
-40 kJ
80 kJ
Explanation:
Work is equal to the area under the pressure vs volume graph.
W = ∫ᵥ₁ᵛ² P dV
2.27) Pressure and volume are linearly related. When we graph P vs V, the area under the line is a trapezoid. So the work is:
W = ½ (P₁ + P₂) (V₂ − V₁)
W = ½ (100 kPa + 300 kPa) (0.1 m³ − 0.3 m³)
W = -40 kJ
2.29) Pressure and volume are inversely proportional:
pV = k
The initial pressure and volume are 500 kPa and 0.1 m³. So the constant is:
(500) (0.1) = k
k = 50
The final pressure is 100 kPa. So the final volume is:
(100) V = 50
V = 0.5
The work is therefore:
W = ∫ᵥ₁ᵛ² P dV
W = ∫₀₁⁰⁵ (50/V) dV
W = 50 ln(V) |₀₁⁰⁵
W = 50 (ln 0.5 − ln 0.1)
W ≈ 80 kJ
Answer:
Obviously Lengthen...
or 
Explanation:
As we can observe from the equation, time period of a simple pendulum depends upon the length directly. When the gravitational acceleration increases the time period of the pendulum decreases and vice versa. So, by increasing the length, the time period can be adjusted...