Mark the correct statement: a) Charges flow through a circuit b) voltage is applied across the circuit c) voltage is the ratio o
1 answer:
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Answer:
Part a)
Part b)
Direction = upwards
Explanation:
When ball is dropped from height h = 4.0 m
then the speed of the ball just before it will strike the ground is given as
Now ball will rebound to height h = 2.00 m
so the velocity of ball just after it will rebound is given as
Part a)
Average acceleration is given as
Part B)
As we know that ball rebounds upwards after collision while before collision it is moving downwards
So the direction of the acceleration is vertically upwards
The emf induced in the second coil is given by:
V = -M(di/dt)
V = emf, M = mutual indutance, di/dt = change of current in the first coil over time
The current in the first coil is given by:
i = i₀
i₀ = 5.0A, a = 2.0×10³s⁻¹
i = 5.0e^(-2.0×10³t)
Calculate di/dt by differentiating i with respect to t.
di/dt = -1.0×10⁴e^(-2.0×10³t)
Calculate a general formula for V. Givens:
M = 32×10⁻³H, di/dt = -1.0×10⁴e^(-2.0×10³t)
Plug in and solve for V:
V = -32×10⁻³(-1.0×10⁴e^(-2.0×10³t))
V = 320e^(-2.0×10³t)
We want to find the induced emf right after the current starts to decay. Plug in t = 0s:
V = 320e^(-2.0×10³(0))
V = 320e^0
V = 320 volts
We want to find the induced emf at t = 1.0×10⁻³s:
V = 320e^(-2.0×10³(1.0×10⁻³))
V = 43 volts
<u>Answer:</u> The Young's modulus for the wire is
<u>Explanation:</u>
Young's Modulus is defined as the ratio of stress acting on a substance to the amount of strain produced.
The equation representing Young's Modulus is:
where,
Y = Young's Modulus
F = force exerted by the weight =
m = mass of the ball = 10 kg
g = acceleration due to gravity =
l = length of wire = 2.6 m
A = area of cross section =
r = radius of the wire = (Conversion factor: 1 m = 1000 mm)
= change in length = 1.99 mm =
Putting values in above equation, we get:
Hence, the Young's modulus for the wire is