Answer:
a) When its length is 23 cm, the elastic potential energy of the spring is
0.18 J
b) When the stretched length doubles, the potential energy increases by a factor of four to 0.72 J
Explanation:
Hi there!
a) The elastic potential energy (EPE) is calculated using the following equation:
EPE = 1/2 · k · x²
Where:
k = spring constant.
x = stretched lenght.
Let´s calculate the elastic potential energy of the spring when it is stretched 3 cm (0.03 m).
First, let´s convert the spring constant units into N/m:
4 N/cm · 100 cm/m = 400 N/m
EPE = 1/2 · 400 N/m · (0.03 m)²
EPE = 0.18 J
When its length is 23 cm, the elastic potential energy of the spring is 0.18 J
b) Now let´s calculate the elastic potential energy when the spring is stretched 0.06 m:
EPE = 1/2 · 400 N/m · (0.06 m)²
EPE = 0.72 J
When the stretched length doubles, the potential energy increases by a factor of four to 0.72 J
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 moon region
Explanation:
This is because there is little to no gravity on the moon. That is where the astronaut would feel the lightest.
Because weight W = M g, the ratio of weights equals the ratio of masses.
(M_m g)/ (M_w g) = [ (p^2 Man )/ (2 K_man)] / [ (p^2 Woman )/ (2 K_woman)
but p's are equal, so
K_m/K_m = (M_w g)/(M_m g) = W_woman / W_man = 450/680 = 0.662
Answer:
0.02
Explanation:
coefficient of kinetic friction = μ
force of friction = Ff
Normal Force = FN, but
FN = -W
Ff = -μFN
so μ = Ff/FN
= 4N/200N
= 0.02.
