The minimum potential difference must be supplied by the ignition circuit to start a car is -1800 V
<u>Explanation:</u>
Given data,
E= 3 ×10 ⁶ Δx=0.06/100
We have to find the minimum potential difference
E= -ΔV/Δx
ΔV=- E × Δx
ΔV =-3 ×10 ⁶ . 0.06/100
ΔV=-1800 V
The minimum potential difference must be supplied by the ignition circuit to start a car is -1800 V
Answer:
F₁ = F₂ = F₃ = 0 N
Explanation:
given,
Arrow 1 mass = 80 g speed = 10 m/s
Arrow 2 mass = 80 g speed = 9 m/s
Arrow 3 mass = 90 g speed = 9 m/s
Horizontal Force:- F₁ , F₂ and F₃
There is no air resistance.
If Air resistance is zero then the horizontal acceleration of the arrow also equal to zero.
We know,
According to newton's second law
F = m a
If Acceleration is equal to zero
Then Force is also equal to zero.
Hence, F₁ = F₂ = F₃ = 0 N
Answer:
d) 1.2 mT
Explanation:
Here we want to find the magnitude of the magnetic field at a distance of 2.5 mm from the axis of the coaxial cable.
First of all, we observe that:
- The internal cylindrical conductor of radius 2 mm can be treated as a conductive wire placed at the axis of the cable, since here we are analyzing the field outside the radius of the conductor. The current flowing in this conductor is
I = 15 A
- The external conductor, of radius between 3 mm and 3.5 mm, does not contribute to the field at r = 2.5 mm, since 2.5 mm is situated before the inner shell of the conductor (at 3 mm).
Therefore, the net magnetic field is just given by the internal conductor. The magnetic field produced by a wire is given by
where
is the vacuum permeability
I = 15 A is the current in the conductor
r = 2.5 mm = 0.0025 m is the distance from the axis at which we want to calculate the field
Substituting, we find:
The statement that could be made about the energy in this situation would be :
It being transferred from his arms muscles to the ball.
The muscle contraction from his arms created a force that could be used to lift the ball up.<span />
Answer:
Flow Rate = 80 m^3 /hours (Rounded to the nearest whole number)
Explanation:
Given
- Hf = head loss
- f = friction factor
- L = Length of the pipe = 360 m
- V = Flow velocity, m/s
- D = Pipe diameter = 0.12 m
- g = Gravitational acceleration, m/s^2
- Re = Reynolds's Number
- rho = Density =998 kg/m^3
- μ = Viscosity = 0.001 kg/m-s
- Z = Elevation Difference = 60 m
Calculations
Moody friction loss in the pipe = Hf = (f*L*V^2)/(2*D*g)
The energy equation for this system will be,
Hp = Z + Hf
The other three equations to solve the above equations are:
Re = (rho*V*D)/ μ
Flow Rate, Q = V*(pi/4)*D^2
Power = 15000 W = rho*g*Q*Hp
1/f^0.5 = 2*log ((Re*f^0.5)/2.51)
We can iterate the 5 equations to find f and solve them to find the values of:
Re = 235000
f = 0.015
V = 1.97 m/s
And use them to find the flow rate,
Q = V*(pi/4)*D^2
Q = (1.97)*(pi/4)*(0.12)^2 = 0.022 m^3/s = 80 m^3 /hours
