Answer:
Speed will be equal to 1.40 m/sec
Explanation:
Mass of the rubber ball m = 5.24 kg = 0.00524 kg
Spring is compressed by 5.01 cm
So x = 5.01 cm = 0.0501 m
Spring constant k = 8.08 N/m
Frictional force f = 0.031 N
Distance moved by ball d = 15.8 cm = 0.158 m
Energy gained by spring
Energy lost due to friction
So remained energy to move the ball = 0.0101 - 0.0048 = 0.0052 J
This energy will be kinetic energy
v = 1.40 m/sec
The equilibrium condition allows finding the correct answer for the force that is resisting the weight of the apple is:
3. Normal force
Newton's second law gives the relationship between <em>force, mass</em> and acceleration of bodies, in the special case that the acceleration is is called the equilibrium condition.
∑ F = 0
Where F is the external force.
The free body diagram is a diagram of the forces on bodies without the details of the shape of the body, in the attached we can see a scheme of the forces.
Let's write the equilibrium condition for the apple
N - W = 0
N = W
.
We can see that the only forces acting on the apple are its weights and reaction from the table called Normal.
Let's analyze the different answers:
1. False. The apple is not moving therefore the resistance is zero
2. False. The apple is not moving so friction it with the table is
3. True. The free-body diagram shows that the normal and the weight are equal
4. False. There is nothing to pull the apple so there is no tension
5. False. Gravity is the weight of the apple that is applied to the table, not from the table to the apple.
In conclusion using the equilibrium condition we can find the correct result for the force that is resisting the weight of the apple is:
3. Normal force
Learn more here: brainly.com/question/2872207
Answer:
a) ΔV₁ = 21.9 V, b) U₀ = 99.2 10⁻¹² J, c) U_f = 249.9 10⁻¹² J, d) W = 150 10⁻¹² J
Explanation:
Let's find the capacitance of the capacitor
C =
C = 8.85 10⁻¹² (8.00 10⁻⁴) /2.70 10⁻³
C = 2.62 10⁻¹² F
for the initial data let's look for the accumulated charge on the plates
C =
Q₀ = C ΔV
Q₀ = 2.62 10⁻¹² 8.70
Q₀ = 22.8 10⁻¹² C
a) we look for the capacity for the new distance
C₁ = 8.85 10⁻¹² (8.00 10⁻⁴) /6⁴.80 10⁻³
C₁ = 1.04 10⁻¹² F
C₁ = Q₀ / ΔV₁
ΔV₁ = Q₀ / C₁
ΔV₁ = 22.8 10⁻¹² /1.04 10⁻¹²
ΔV₁ = 21.9 V
b) initial stored energy
U₀ =
U₀ = (22.8 10⁻¹²)²/(2 2.62 10⁻¹²)
U₀ = 99.2 10⁻¹² J
c) final stored energy
U_f = (22.8 10⁻¹²) ² /(2 1.04 10⁻⁻¹²)
U_f = 249.9 10⁻¹² J
d) the work of separating the plates
as energy is conserved work must be equal to energy change
W = U_f - U₀
W = (249.2 - 99.2) 10⁻¹²
W = 150 10⁻¹² J
note that as the energy increases the work must be supplied to the system