A 79-turn, 16.035-cm-diameter coil is at rest in a horizontal plane. A uniform magnetic field 43 degrees away from vertical increases from 0.997 T to 6.683 T in 56.691 s. Determine the emf induced in the coil.


Answer 1


The induced emf is 0.0888 V.


Given that,

Number of turns = 79

Diameter = 16.035 cm

Angle = 43

Change in magnetic field \Delta B=(6.683-0.997)= 5.686\ T

Time = 56.691 s

We need to calculate the induced emf

Using formula of induced emf

\epsilon=(NA\Delta B\cos\theta)/(\Delta T)

Where, N = number of turns

A = area

B = magnetic field

Put the value into the formula


\epsilon =0.0888\ V

Hence, The induced emf is 0.0888 V.

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A battery with an emf of 12.0 V shows a terminal voltage of 11.7 V when operating in a circuit with two lightbulbs, each rated at 4.0 W (at 12.0 V), which are connected in parallel.What is the battery's internal resistance?





The electromotive force (E) in the circuit is related to the terminal voltage(V), of the circuit and the internal resistance (r) of the battery as follows;

E = V + Ir                      --------------------(a)


I = current flowing through the circuit


V = I x Rₓ                    ---------------------(b)


Rₓ = effective or total resistance in the circuit.

First, let's calculate the effective resistance in the circuit:

The effective resistance (Rₓ) in the circuit is the one due to the resistances in the two lightbulbs.


R₁ = resistance in the first bulb

R₂ = resistance in the second bulb

Since the two bulbs are both rated at 4.0W ( at 12.0V), their resistance values (R₁ and R₂) are the same and will be given by the power formula;

P = (V^(2) )/(R)

=> R = (V^(2) )/(P)             -------------------(ii)


P = Power of the bulb

V = voltage across the bulb

R = resistance of the bulb

To get R₁, equation (ii) can be written as;

R₁ = (V^(2) )/(P)    --------------------------------(iii)


V = 12.0V

P = 4.0W

Substitute these values into equation (iii) as follows;

R₁ = (12.0^(2) )/(4)

R₁ = (144)/(4)

R₁ = 36Ω

Following the same approach, to get R₂, equation (ii) can be written as;

R₂ = (V^(2) )/(P)    --------------------------------(iv)


V = 12.0V

P = 4.0W

Substitute these values into equation (iv) as follows;

R₂ = (12.0^(2) )/(4)

R₂ = (144)/(4)

R₂ = 36Ω

Now, since the bulbs are connected in parallel, the effective resistance (Rₓ) is given by;

(1)/(R_(X) ) = (1)/(R_1) + (1)/(R_2)       -----------------(v)

Substitute the values of R₁ and R₂ into equation (v) as follows;

(1)/(R_X) = (1)/(36) + (1)/(36)

(1)/(R_X) = (2)/(36)

Rₓ = (36)/(2)

Rₓ = 18Ω

The effective resistance (Rₓ) is therefore, 18Ω

Now calculate the current I, flowing in the circuit:

Substitute the values of V = 11.7V and Rₓ = 18Ω into equation (b) as follows;

11.7 = I x 18

I = (11.7)/(18)

I = 0.65A

Now calculate the battery's internal resistance:

Substitute the values of E = 12.0, V = 11.7V and I = 0.65A  into equation (a) as follows;

12.0 = 11.7 + 0.65r

0.65r = 12.0 - 11.7

0.65r = 0.3

r = (0.3)/(0.65)

r = 0.46Ω

Therefore, the internal resistance of the battery is 0.46Ω


R_i_n_t=0.45 \Omega


Internal resistance is a concept that helps model the electrical consequences of the complex chemical reactions that occur within a battery. When a charge is applied to a battery, the internal resistance can be calculated using the following equation:

R_i_n_t=((V_N_L)/(V_F_L) -1)R_L


V_F_L=Load\hspace{3}voltage=11.7V\nV_N_L= O pen\hspace{3}circuit\hspace{3}voltage=12V\nR_L=Load\hspace{3}resistance

As you can see, we don't know the exactly value of the R_L. However we can calculated that value using the next simple operations:

The problem tell us that the power of each lightbulb is 4.0 W at 12.0 V, hence let's calculated the power at 11.7V using Cross-multiplication:

(12)/(11.7) =(4)/(P)

Solving for P :

P=(11.7*4)/(12) =3.9W

Now, the electric power is given by:





R_b=(V^2)/(P) =(11.7^2)/(3.9) =35.1\Omega

Now, because of the lightbulbs are connected in parallel the equivalent resistance is given by:

(1)/(R_L) =(1)/(R_b) +(1)/(R_b) =(2)/(R_b) \n\n R_L= (R_b)/(2) =(35.1)/(2)=17.55\Omega

Finally, now we have all the data, let's replace it into the internal resistance equation:

R_i_n_t=((12)/(11.7) -1)17.55=0.45\Omega

A curve of radius 35 m isbanked; therefore, no friction
is required at a speed of 7
m/s of a car. What is the?
banking angle



The banking angle is 23.98 degrees.


We have,

Radius of a curve is 35 m

Speed of a car is 7 m/s

It is required to find the banking angle. At equilibrium, net force is equal to the centripetal force between vehicle and the road such that the banking angle is given by :


g is acceleration due to gravity

\tan\theta=(7^2)/(35* (22)/(7))\n\n\theta=\tan^(-1)\left(0.445\right)\n\n\theta=23.98^(\circ)

So, the banking angle is 23.98 degrees.

You are standing on the edge of a ravine that is 17.5 m wide. You notice a cave on the opposite wall whose ceiling is 7.3 m below your feet. The cave is 4.2 m deep, and has a vertical back wall. You decide to kick a rock across the ravine into the cave. What initial horizontal velocity must you give the rock so that the rock barely misses the overhang? The acceleration of gravity is 9.8 m/s 2 .



v₀ₓ = 14.34 m / s


We can solve this problem using the projectile launch equations.

Let's look for the time it takes to descend to the height of the cave

    y = v_(oy) t - ½ g t²

As it rises horizontally the initial vertical speed is zero  

    y = 0 - ½ gt²

    t = √2 y / g

    t = √2 7.3 / 9-8

    t = 1.22 s

This is the same time to cross the ravine

    x = v₀ₓ t

    v₀ₓ = x / t

    v₀ₓ = 17.5 / 1.22

    v₀ₓ = 14.34 m / s

This is the minimum speed.

Which statementsabout a neutral atom are correct? Check all that apply.1. A neutral atom is composed of bothpositively and negatively charged particles.
2. Positively charged protons are located in the tiny, massivenucleus.
3. The positively chargedparticles in the nucleus are positrons.
4. The negatively chargedelectrons are spread out in a "cloud" around thenucleus.
5. The electrons areattracted to the positively charged nucleus.
6. The radius of the electroncloud is twice as large as the radius of the nucleus.



1, 2, 4, 5 are correct


1) This is true because In a neutral atom, the number of positive charges (protons) is equal to the number of negative charges (electrons).

2) This is true because the mass of the atom which is made up of the protons and neutrons, is located in the tiny nucleus.

3) This is not true because the positively charged particles in the nucleus are called protons.

4) This is true because electrons move around the nucleus in diffuse areas known as orbitals.

5) This is true because opposite charges attract each other. And electron is a negative charge.

6) This is not true because the radius of the electron cloud is normally 10,000 times larger than the radius of the nucleus.

Please helpa car is driven 200 km west and then 80 km southwest. what is the displacement of the car from the point of orgin (magnitude and direction)? draw a diagram. ​


Let's take east and west to be positive and negative, respectively, and north and south to be positive and negative, respectively. Then in terms of vectors (using ijk notation), the car first moves 200 km west,

r = (-200 km) j

then 80 km southwest,

s = (-80/√2 km) i + (-80/√2 km) j

so that its total displacement is

r + s = (-80/√2 km) i + ((-200 - 80/√2) km) j

r + s ≈ (-56.6 km) i + (-256.6 km) j

This vector has magnitude

√((-56.6 km)² + (-256.6 km)²) ≈ 262.7 km

and direction θ such that

tan(θ) = (-256.6 km) / (-56.6 km)  ==>  θ ≈ -102.4º

relative to east, or about 12.4º west of south.

A basketball player jumps 76cm to get a rebound. How much time does he spend in the top 15cm of the jump (ascent and descent)?



The time for final 15 cm of the jump equals 0.1423 seconds.


The initial velocity required by the basketball player to be able to jump 76 cm can be found using the third equation of kinematics as



'v' is the final velocity of the player

'u' is the initial velocity of the player

'a' is acceleration due to gravity

's' is the height the player jumps

Since the final velocity at the maximum height should be 0 thus applying the values in the above equation we get

0^2=u^2-2* 9.81* 0.76\n\n\therefore u=√(2* 9.81* 0.76)=3.86m/s

Now the veocity of the palyer after he cover'sthe initial 61 cm of his journey can be similarly found as

v^(2)=3.86^2-2* 9.81* 0.66\n\n\therefore v=√(3.86^2-2* 9.81* 0.66)=1.3966m/s

Thus the time for the final 15 cm of the jump can be found by the first equation of kinematics as


where symbols have the usual meaning

Applying the given values we get