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DENIUS [597]
3 years ago
11

A constant voltage of 3.00 V has been observed over a certain time interval across a 3.00 H inductor. The current through the in

ductor, measured as 3.00 A at the beginning of the time interval, was observed to increase at a constant rate to a value of 8.00 A at the end of the time interval. How long was this time interval?
Physics
1 answer:
frozen [14]3 years ago
5 0

Answer:

Explanation:

If L be inductance and i be current , flux associated with it

Φ = L i

emf induced = dΦ / dt = L di / dt

Putting the given values

emf induced = 3 V , L = 3 H

di = 8 - 3 = 5 A

dt = ?

3 = 3 x 5 / dt

dt or time duration = 3 x 5 / 3

= 5 sec .

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What type of configuration is used in the HEVs available in the US today?
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3 years ago
Read 2 more answers
A block of 250-mm length and 54 × 40-mm cross section is to support a centric compressive load P. The material to be used is a b
musickatia [10]

Answer:

P = 17.28*10⁶ N

Explanation:

Given

L = 250 mm = 0.25 m

a = 0.54 m

b = 0.40 m

E = 95 GPa = 95*10⁹ Pa

σmax = 80 MPa = 80*10⁶ Pa

ΔL = 0.12%*L = 0.0012*0.25 m = 3*10⁻⁴ m

We get A as follows:

A = a*b = (0.54 m)*(0.40 m) = 0.216 m²

then, we apply the formula

ΔL = P*L/(A*E)  ⇒ P = ΔL*A*E/L

⇒  P = (3*10⁻⁴ m)*(0.216 m²)*(95*10⁹ Pa)/(0.25 m)

⇒  P = 24624000  N = 24.624*10⁶ N

Now we can use the equation

σ = P/A

⇒  σ = (24624000  N)/(0.216 m²) = 114000000 Pa = 114 MPa > 80 MPa

So σ > σmax  we use σmax

⇒  P = σmax*A = (80*10⁶ Pa)*(0.216 m²) = 17280000 N = 17.28*10⁶ N

7 0
3 years ago
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