Answer:

**Answer:**

The induced emf is 0.0888 V.

**Explanation:**

**Given that, **

Number of turns = 79

Diameter = 16.035 cm

Angle = 43

Change in magnetic field

Time = 56.691 s

**We need to calculate the induced emf**

**Using formula of induced emf**

Where, N = number of turns

A = area

B = magnetic field

Put the value into the formula

**Hence, The induced emf is 0.0888 V.**

A concrete highway is built of slabs 18.0 m long (at 25 °C). How wide should the expansion cracks be (at 25 °C) between the slabs to prevent buckling if the annual extreme temperatures are −32 °C and 52 °C?(the coefficient of linear expansion of concrete is 1.20 × 10 − 5 °C-1) g

Does lighting striking the earth considered the speed of light?

The index of refraction for red light in water is 1.331, and that for blue light is 1.340. If a ray of white light enters the water at an angle of incidence of 83.00o , what are the underwater angles of refraction for the blue and red components of the light

Resonances of the ear canal lead to increased sensitivity of hearing, as we’ve seen. Dogs have a much longer ear canal—5.2 cm—than humans. What are the two lowest frequencies at which dogs have an increase in sensitivity? The speed of sound in the warm air of the ear is 350 m/s.A. 1700 Hz, 3400 HzB. 1700 Hz, 5100 HzC. 3400 Hz, 6800 HzD. 3400 Hz, 10,200 Hz

The electric field at a point 2.8 cm from a small object points toward the object with a strength of 180,000 N/C. What is the object's charge q? ( k = 1/4πε 0 = 8.99 × 10 9 N ∙ m 2/C 2)

Does lighting striking the earth considered the speed of light?

The index of refraction for red light in water is 1.331, and that for blue light is 1.340. If a ray of white light enters the water at an angle of incidence of 83.00o , what are the underwater angles of refraction for the blue and red components of the light

Resonances of the ear canal lead to increased sensitivity of hearing, as we’ve seen. Dogs have a much longer ear canal—5.2 cm—than humans. What are the two lowest frequencies at which dogs have an increase in sensitivity? The speed of sound in the warm air of the ear is 350 m/s.A. 1700 Hz, 3400 HzB. 1700 Hz, 5100 HzC. 3400 Hz, 6800 HzD. 3400 Hz, 10,200 Hz

The electric field at a point 2.8 cm from a small object points toward the object with a strength of 180,000 N/C. What is the object's charge q? ( k = 1/4πε 0 = 8.99 × 10 9 N ∙ m 2/C 2)

0.46Ω

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)

Where;

I = current flowing through the circuit

But;

V = I x Rₓ ---------------------(b)

Where;

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.

Let;

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 =

=> R = -------------------(ii)

Where;

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 = --------------------------------(iii)

Where;

V = 12.0V

P = 4.0W

Substitute these values into equation (iii) as follows;

R₁ =

R₁ =

R₁ = 36Ω

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

R₂ = --------------------------------(iv)

Where;

V = 12.0V

P = 4.0W

Substitute these values into equation (iv) as follows;

R₂ =

R₂ =

R₂ = 36Ω

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

= + -----------------(v)

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

= +

=

Rₓ =

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 =

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 =

r = 0.46Ω

Therefore, the internal resistance of the battery is **0.46Ω**

**Answer:**

**Explanation:**

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:

Where:

As you can see, we don't know the exactly value of the . 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:

Solving for :

Now, the electric power is given by:

Where:

So:

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

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

is required at a speed of 7

m/s of a car. What is the?

banking angle

**Answer:**

The banking angle is 23.98 degrees.

**Explanation:**

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

So, the banking angle is 23.98 degrees.

**Answer:**

v₀ₓ = 14.34 m / s

**Explanation:**

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 = 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.

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.

Answer:

1, 2, 4, 5 are correct

Explanation:

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.

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)

then 80 km southwest,

* s* = (-80/√2 km)

so that its total displacement is

* r* +

* r* +

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**.

**Answer:**

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

**Explanation:**

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

where

'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

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

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