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:
b. 9.5°C
Explanation:
= Mass of ice = 50 g
= Initial temperature of water and Aluminum = 30°C
= Latent heat of fusion = 
= Mass of water = 200 g
= Specific heat of water = 4186 J/kg⋅°C
= Mass of Aluminum = 80 g
= Specific heat of Aluminum = 900 J/kg⋅°C
The equation of the system's heat exchange is given by

The final equilibrium temperature is 9.50022°C
Answer:
x = 1,185 m
, t = 4/3 s
, F = - 4 N
Explanation:
For this exercise we use Newton's second law
F = m a = m dv /dt
β - α t = m dv / dt
dv = (β – α t) dt
We integrate
v = β t - ½ α t²
We evaluate between the lower limits v = v₀ for t = 0 and the upper limit v = v for t = t
v-v₀ = β t - ½ α t²
the farthest point of the body is when v = v₀ = 0
0 = β t - ½ α t²
t = 2 β / α
t = 2 4/6
t = 4/3 s
Let's find the distance at this time
v = dx / dt
dx / dt = v₀ + β t - ½ α t2
dx = (v₀ + β t - ½ α t2) dt
We integrate
x = v₀ t + ½ β t - ½ 1/3 α t³
x = v₀ 4/3 + ½ 4 (4/3)² - 1/6 6 (4/3)³
The body comes out of rest
x = 3.5556 - 2.37
x = 1,185 m
The value of force is
F = β - α t
F = 4 - 6 4/3
F = - 4 N
Answer:

Explanation:
First of all, we need to find the volume of the room, which is given by

Now we can find the mass of the air by using

where
is the density of the air
is the volume of the room
Substituting,
