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3 years ago

Help i forgot to complete this worksheet

Physics

1 answer:

Naily [24]3 years ago

6 0

just search up the answer/ definition to all of them, rephrase into own words, then do the same for examples.

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Two generators use the same magnetic field and operate at the same frequency. Each has a single-turn circular coil. One generato

Volgvan

Answer:

The coil radius of other generator is 5.15 cm

Explanation:

Consider the equation for induced emf in a generator coil:

EMF = NBAω Sin(ωt)

where,

N = No. of turns in coil

B = magnetic field

A = Cross-sectional area of coil = π r²

ω = angular velocity

t = time

It is given that for both the coils magnetic field, no. of turn and frequency is same. Since, the frequency is same, therefore, the angular velocity, will also be same. As, ω = 2πft.

Therefore, EMF for both coils or generators will be:

EMF₁ = NBπr₁²ω Sin(ωt)

EMF₂ = NBπr₂²ω Sin(ωt)

dividing both the equations:

EMF₁/EMF₂ = (r₁/r₂)²

r₂ = r₁ √(EMF₂/EMF₁)

where,

EMF₁ = 1.8 V

EMF₂ = 3.9 V

r₁ = 3.5 cm

r₂ = ?

Therefore,

r₂ = (3.5 cm)√(3.9 V/1.8 V)

r₂ = 5.15 cm

3 0

3 years ago

In a standard tensile test, a steel rod of 7 8-in. diameter is subjected to a tension force of 17 kips. Knowing that ν = 0.30 an

natali 33 [55]

Answer:

(a) Elongation of the rod==5.61×10⁻⁹m

(b) Change in diameter=1.640×10⁻⁸m

Explanation:

Given data

Diameter d=78 in=1.9812 m

Cross Area is:

$A=(\pi /4)d^{2} \\A=(\pi /4)(1.9812m)^{2}\\A=3.08m^{2}$

Applied Load P=17 KN=17×10³N

E=29 × 106 psi=1.99947961×10¹¹Pa

Stress and Strain in x direction

Stress in x direction

σ=P/A

$=\frac{17*10^{3}N }{3.08m^{2} }\\ =5517.25Pa$

σ=5517.25 Pa

Strain in x direction

ε=σ/E

$=\frac{5517.25}{1.99947961*10^{11} } \\=2.76*10^{-8}$

ε=2.76×10⁻⁸

Part (a)

Elongation of the rod=Lε

=(0.2032)(2.76×10⁻⁸)

Elongation of the rod==5.61×10⁻⁹m

Part(b) Change in diameter

Strain in y direction

ε₁= -vε

ε₁= -(0.30)(2.76×10⁻⁸)

ε₁=-8.28×10⁻⁹

Change in diameter=d×ε₁

Change in diameter=(1.9812m)×(-8.28×10⁻⁹)

Change in diameter=1.640×10⁻⁸m

4 0

3 years ago

What are the applications of pascal's principle

Murrr4er [49]

Explanation:

The applications are, hydraulic lift- to transmit equal pressure throughout a fluid.
Hydraulic jack- used in the braking system of cars.
use of a straw- to suck fluids, which goes because of air pressure.