Answer:
Power output = 96.506 watts
Explanation:
Drag coefficient (Cd) = 0.9
V = 7.3 m/s
Air density (ρ) = 1.225 kg/m^(3)
Area (A) = 0.45 m^2
Let's find the drag force ;
Fd=(1/2)(Cd)(ρ)(A)(v^(2))
So Fd = (1/2)(0.9)(1.225)(0.45)(7.3^(2)) = 13.22N
Drag power = Drag Force x Drag velocity.
Thus drag power, = 13.22 x 7.3 = 96.506 watts
Despite current has a magnitude and a direction, like vectors, it is a scalar because it doesn't obey laws of vector addition. For instance, if we consider a junction of

in a circuit, and two currents entering this junction, we know that the resultant current is just the algebraic sum of the two currents, not the vector sum, so it is not a vector quantity.
Answer:
8000J
Explanation:
The kinetic energy of the car lost during breaking are converted to thermal energy and are gained by the brakes.
Kinetic energy loss by car = thermal energy gained by brakes.
∆K.E = ∆T.E ....1
The Kinetic energy loss by car can be expressed as;
∆K.E = K.E1 - K.E2
Initial K.E = K.E1 = 10000J
Final K.E = K.E2 = 2000J
∆K.E= 10000J - 2000J = 8000J
From equation 1,
∆K.E = ∆T.E
∆T.E = 8,000J
thermal energy gain by brakes = 8,000J
Answer: A
Explanation:
Molecules speed up as heat is added
For example when water is heated as the water gets hotter the molecules speed up causing the water to boil and change phases into a gas (this is called evaporation)
In an ice cube the water molecules are frozen (barely moving compressed tight together) as the ice cube heats up the molecules start speeding up and moving further apart as the ice cube turns into liquid form. So as heat is added molecules speed up, move faster and spread further apart
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