In the physics of motion a force acting over a distance is?

Two coils are wound around the same cylindrical form. When the current in the first coil is decreasing at a rate of -0.245 A/s ,

Airida [17]

Answer:

Complete question:

c.If the current in the second coil increases at a rate of 0.365 A/s , what is the magnitude of the induced emf in the first coil?

a. $M= 6.53\times10^{-3} H$

b.flux through each turn = Ф = $4.08\times10^{-4} Wb$

c.magnitude of the induced emf in the first coil = e= $2.38\times10^{-3} V$

Explanation:

a. rate of current changing = $\frac{di}{dt}=[tex]M=\frac{1.60\times10^{-3} V}{0.240\frac{A}{s} }}$ [/tex]

Induced emf in the coil =e= $1.60\times10^{-3} V$

For mutual inductance in which change in flux in one coil induces emf in the second coil given by the farmula based on farady law

$e=-M\frac{di}{dt}$

$M=\frac{e}{\frac{di}{dt} }$

$M=\frac{1.60\times10^{-3} }{-0.245}$

$M= 6.53\times10^{-3} H$

b.

Flux through each turn=?

Current in the first coil =1.25 A

Number of turns = 20

using MI = NФ

flux through each turn = Ф = $\frac{6.53\times10^{-3}\times1.25}{20}$

flux through each turn = Ф = $4.08\times10^{-4} Wb$

c.

second coil increase at a rate = 0.365 A/s

magnitude of the induced emf in the first coil =?

using $e=-M\frac{di_{2} }{dt}$

$e= 6.53\times10^{-3} \times 0.365$

magnitude of the induced emf in the first coil = e= $2.38\times10^{-3} V$

4 0

3 years ago

Starting fom rest, a car accelerates at a constant rate, reaching 88 km/h in 12 s. (a) What is its acceleration? (b) How far doe

JulsSmile [24]

Answer:

(A) $a=2.0.37m/sec^2$

(B) s = 146.664 m

Explanation:

We have given car starts from the rest so initial velocity u = 0 m /sec

Final velocity v = 88 km/hr

We know that 1 km = 1000 m

And 1 hour = 3600 sec

So $88km/hr=88\times \frac{1000}{3600}=24.444m/sec$

Time is given t = 12 sec

(A) From first equation of motion v = u+at

So $24.444=0+a\times 12$

$a=2.0.37m/sec^2$

So acceleration of the car will be $a=2.0.37m/sec^2$

(b) From third equation of motion $v^2=u^2+2as$

So $24.444^2=0^2+2\times 2.037\times s$

s = 146.664 m

Distance traveled by the car in this interval will be 146.664 m

6 0

3 years ago

Calculate the ratio of the kinetic energy of an electron to that of a proton if their wavelengths are equal. Assume that the spe

goblinko [34]

Answer:

the ratio of the kinetic energy of an electron to that of a proton if their wavelengths are equal is 1835.16 .

Explanation:

We know, wavelength is expressed in terms of Kinetic Energy by :

$\lambda=\dfrac{h}{\sqrt{2mE}}$

Therefore , $E=\dfrac{h^2}{2 \lambda^2 m}$

It is given that both electron and proton have same wavelength.

Therefore,

$E_e=\dfrac{h^2}{2 \lambda^2 m_e}$ .... equation 1.

$E_p=\dfrac{h^2}{2 \lambda^2 m_p}$ .... equation 2.

Now, dividing equation 1 by 2 .

We get ,

$\dfrac{E_e}{E_p}=\dfrac{\dfrac{h^2}{2 \lambda^2 m_e}}{\dfrac{h^2}{2 \lambda^2 m_p}}\\\\\\\dfrac{E_e}{E_p}=\dfrac{m_p}{m_e}$

Putting value of mass of electron = $9.1\times 10^{-31}\ kg$ and mass of proton = $1.67\times 10^{-27}\ kg.$

We get :

$\dfrac{E_e}{E_p}=\dfrac{1.67\times 10^{-27}\ kg}{9.1\times 10^{-31}\ kg}=1835.16$

Hence , this is the required solution.

4 0

4 years ago

Read 2 more answers

Two balloons (m = 0.012 kg) are separated by a distance of d = 15 m. They are released from rest and observed to have an instant

Evgesh-ka [11]

Answer:

1.492*10^14 electrons

Explanation:

Since we know the mass of each balloon and the acceleration, let’s use the following equation to determine the total force of attraction for each balloon.

F = m * a = 0.012 * 1.9 = 0.0228 N

Gravitational forces are negligible

Charge force = 9 * 10^9 * q * q ÷ 225

= 9 * 10^9 * q^2 ÷ 225 = 0.0228

q^2 = 5.13 ÷ 9 * 10^9

q = 2.387 *10^-5

This is approximately 2.387 *10^-5 coulomb of charge. The charge of one electron is 1.6 * 10^-19 C

To determine the number of electrons, divide the charge by this number.

N =2.387 *10^-5 ÷ 1.6 * 10^-19 = 1.492*10^14 electrons

3 0

3 years ago

What is the electrical force between q2 and q3? recall that k = 8.99 × 109 n•meters squared over coulombs squared.. 1.0 × 1011 n

max2010maxim [7]

The magnitude of the electrical force between q2 and q3 is given as a ratio between the product of their charges and the square of the distance of separation.

<h3>What is the magnitude of electrical forces between two charges?</h3>

The magnitude of the electrical force between two charges refers to the attractive or repulsive forces that exists between two charges separated by a given distance in an electric field.

The magnitude of the electrical force, F between the two charges q2 and q3 is given be my the formula below

$F = \frac{K \times q_2 \times q_3}{d^{2}}$

Therefore, the magnitude of the electrical force between q2 and q3 is given as a ratio between the product of their charges and the square of the distance of separation.

Learn more about electrical force at: brainly.com/question/17692887

#SPJ4

6 0

2 years ago

In the physics of motion a force acting over a distance is?​

In the physics of motion a force acting over a distance is?