One car is sitting still the other is moving at a velocity of 4m/s. If two cars have a mass of 0.04 kg run into each other, what
1 answer:
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To solve this problem it is necessary to use the concepts related to the Hall Effect and Drift velocity, that is, at the speed that an electron reaches due to a magnetic field.
The drift velocity is given by the equation:
Where
I = current
n = Number of free electrons
A = Cross-Section Area
q = charge of proton
Our values are given by,
The hall voltage is given by
Where
B= Magnetic field
n = number of free electrons
d = distance
e = charge of electron
Then using the formula and replacing,
Answer:
Force, |F| = 2100 N
Explanation:
It is given that,
Water from a fire hose is directed horizontally against at a rate of 50.0 kg/s,
Initial speed, v = 42 m/s
The momentum is reduced to zero, final speed, v = 0
The relation between the force and the momentum is given by :
|F| = 2100 N
So, the magnitude of the force exerted on the wall is 2100 N. Hence, this is the required solution.
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