Answer:
1%
Explanation:
Percent error can be found by dividing the absolute error (difference between measure and actual value) by the actual value, then multiplying by 100.

The measured value is 2.02 meters and the actual value is 2.00 meters.


First, evaluate the fraction. Subtract 2.00 from 2.02

Next, divide 0.02 by 2.00

Finally, multiply 0.01 and 100.

The percent error is 1%.
Answer:
|v| = 8.7 cm/s
Explanation:
given:
mass m = 4 kg
spring constant k = 1 N/cm = 100 N/m
at time t = 0:
amplitude A = 0.02m
unknown: velocity v at position y = 0.01 m

1. Finding Ф from the initial conditions:

2. Finding time t at position y = 1 cm:

3. Find velocity v at time t from equation 2:

Answer:
The velocity of the man is 0.144 m/s
Explanation:
This is a case of conservation of momentum.
The momentum of the moving ball before it was caught must equal the momentum of the man and the ball after he catches the ball.
Mass of ball = 0.65 kg
Mass of the man = 54 kg
Velocity of the ball = 12.1 m/s
Before collision, momentum of the ball = mass x velocity
= 0.65 x 12.1 = 7.865 kg-m/s
After collision the momentum of the man and ball system is
(0.65 + 54)Vf = 54.65Vf
Where Vf is their final common velocity.
Equating the initial and final momentum,
7.865 = 54.65Vf
Vf = 7.865/54.65 = 0.144 m/s
Answer:
40 MJ (D)
Explanation:
Quantity of heat (Qh) = 100 MJ
temperature of steam (Th) = 450°c = 450 + 273 = 723 K
emperature of water (TI) = 20 °c = 20 + 273 = 293 k
efficiency = (Qh-Qi)/Qh = (Th-Ti)/Th

- Qi= 0.5947 x 
- (0.5947 x
) = Qi
Qi = 40.5 MJ equivalent to 40 MJ (D)
Answer:
It cancels recoil.
Explanation:
For each action there is an equal an opposite reaction.
The principle of conservation of momentum tell us that if a single spore were ejected the fern would suffer a recoil from it. This recoil would take energy and speed from the spore. But if they are ejected in pairs the recoil is canceled and all the energy is transferred to the spores resulting in higher speeds.