Ask question

Ask question

All categories

3 years ago

8

A coil of conducting wire carries a current of i(t) = 14.0 sin(1.15 ✕ 103t), where i is in amperes and t is in seconds. A second

coil is placed in close proximity to the first, and the mutual inductance of the coils is 130 µH. What is the peak emf (in V) in the second coil?

1 answer:

Advocard [28]3 years ago

8 0

Answer:

The peak emf in second coil is 1.876 V

Explanation:

Given :

Inductance $L = 130 \times 10^{-6}$ H

The current $I(t) = 14 \sin(1.15\times 10^{3} t)$

We compare above equation with standard equation,

$I(t) = I_{o} \sin (\omega t + \phi)$

From above equation we have,

$\omega = 10^{3}$ and $\phi = 1.15$

Find the inductive resistance,

$X_{L} = \omega L$

$X_{L} = 10^{3} \times 130 \times 10^{-6}$

$X_{L} = 0.134$

The peak emf in second coil is,

$V = I_{o} X_{L}$

$V = 14 \times 0.134$

$V = 1.876$ V

Therefore, the peak emf in second coil is 1.876 V

You might be interested in

Whats the answer???????????

Leviafan [203]

Answer: Less than 4 ohms

Explanation:

We have three resistors with the following resistance:

$R_{1}=4\Omega$

$R_{2}=6\Omega$

$R_{3}=8\Omega$

Now, when the resistors are connected in parallel, the total resistance $R$ is calculated as follows:

$\frac{1}{R}=\frac{1}{R_{1}}+\frac{1}{R_{2}}+\frac{1}{R_{3}}$

Isolating $R$ :

$R=\frac{R_{1}R_{2}R_{3}}{R_{3}(R_{1}+R_{2})+R_{1}R_{2}}$

Rewriting with th known values:

$R=\frac{(4\Omega)(6\Omega)(8\Omega)}{8\Omega(4\Omega+6\Omega)+(4\Omega)(6\Omega)}$

Finally:

$R=1.84 \Omega$

Hence, the correct option is less than 4 ohms.

4 0

3 years ago

A charge 3q is at the origin, and a charge −2q is on the positive x axis at x=a. part a where on the x-axis would you place a th

Mila [183]

Answer:

$x = a + \frac{a\sqrt2}{\sqrt3 - \sqrt2}$

Explanation:

Position of charge 3q is x = 0

position of charge -2q is x = a

so here we know that

when two charges are of opposite nature then the electric field will be zero on the line joining the two charges at the position near to smaller magnitude charge

So here the electric field will be zero if the field due to 3q is counterbalanced by field due to -2q

so here we can say

$\frac{k(3q)}{(r+a)^2} + \frac{k(-2q)}{r^2} = 0$

$\frac{3}{(r + a)^2} = \frac{2}{r^2}$

$\frac{r+a}{r} = \sqrt{\frac{3}{2}}$

$1 + \frac{a}{r} =\sqrt{\frac{3}{2}}$

$\frac{a}{r} = \frac{\sqrt3 - \sqrt2}{\sqrt2}$

so we will have

$r = \frac{a\sqrt2}{\sqrt3 - \sqrt2}$

so the x coordinate of this position is given as

$x = a + \frac{a\sqrt2}{\sqrt3 - \sqrt2}$

6 0

3 years ago

When energy is changed from one form to another, some of the energy is lost to the surroundings as waste thermal energy. The mor

Taya2010 [7]

The answer is B Solar cell just took the test

7 0

4 years ago

Read 2 more answers

True.or false A railroad track runs southwest to northeast.

telo118 [61]

Answer:

ns for high-speed rail in the United States date back to the High Speed Ground Transportation Act of 1965. Various state and federal proposals have followed. Despite being one of the world's first countries to get high-speed trains (the Metroliner service in 1969), it failed to spread. Definitions of what constitutes high-speed rail vary, including a range of speeds over 110 mph (180 km/h) and dedicated rail lines. Inter-city railwith top speeds between 90 and 125 mph (140 and 200 km/h) is sometimes referred to in the United States as higher-speed rail.[1]

Amtrak's Acela Express (reaching 150 mph, 240 km/h), Silver Star, Northeast Regional, Keystone Service, Vermonter and certain MARC Penn Line express trains (all five reaching 125 mph, 201 km/h) are the only high-speed services in the country.

As of 2020, the California High-Speed Rail Authority is working on the California High-Speed Rail project and construction is under way on sections traversing the Central Valley. The Central Valley section is planned to open in 2029 and Phase I is planned for completion in 2031.[2]

Contents

1 Definitions in American context

2 History

2.1 Faster inter-city trains: 1920–1941

2.2 Post-war period: 1945–1960

2.3 First attempts: 1960–1992

2.4 Renewed interest: 1993–2008

2.5 Plans for 2008–2013

3 Current state and regional efforts

3.1 The Northeast

3.1.1 Northeast Corridor: Next Generation High-Speed Rail

3.1.1.1 Proposed routes

3.1.2 Northeast Maglev proposal

3.1.3 New Jersey–New York City upgrades

3.1.4 New York

3.1.5 Pennsylvania

3.2 Western States

3.2.1 California

3.2.2 Pacific Northwest

3.2.3 Arizona

3.3 Mid-Atlantic and the South

3.3.1 Florida

3.3.2 Southeast

3.3.3 Texas

3.4 Midwest

3.4.1 Illinois and the Midwest

3.5 The Southwest

4 Federal high-speed rail initiatives

4.1 American Recovery and Reinvestment Act of 2009

4.1.1 Strategic plan

4.2 2009 federal grant funding

4.3 2010 allocation

4.3.1 Cancellation of funds for Wisconsin, Ohio, and Florida

4.4 2011 and 2012 proposals and rejections of funding

5 See also

6 Notes

7 Further reading

8 External links

Explanation:

3 0

3 years ago

A NASA explorer spacecraft with a mass of 1,000 kg takes off in a positive direction from a stationary asteroid. If the velocity

Jet001 [13]

Before the launch, the momentum of the (spacecraft + asteroid) was zero. So after the launch, the momentum of the (spacecraft + asteroid) has to be zero.

Momentum = (mass) x (velocity)

Momentum after the launch:

Spacecraft: (1,000 kg) x (250 m/s) = 250,000 kg-m/s

Asteroid: (mass) x (-25 m/s)

Their sum: 250,000 - 25(mass) .

Their sum must be zero, so 250,000 kg-m/s = (25 m/s) x (mass)

Divide each side by 25 : 10,000 kg-m/s = (1 m/s) x (mass)

Divide each side by (1 m/s) : 10,000 kg = mass

3 0

3 years ago

Read 2 more answers