Why aren’t homes wired in series

In each of two coils the rate of change of the magnetic flux in a single loop is the same. The emf induced in coil 1, which has

irinina [24]

Answer:

Coil 2 have 235 loops

Explanation:

Given

The number of loops in coil 1 is n ₁= 159

The emf induced in coil 1 is ε ₁ = 2.78 V

The emf induced in coil 2 is ε ₂ = 4.11 V

Let

n ₂ is the number of loops in coil 2.

Given, the emf in a single loop in two coils are same. That is,

ϕ ₁/n ₁= ϕ ₂ n ₂⟹ 2.78/159 = 4.11/ n ₂

n₂= $\frac{159 * 4.11}{2.78}$

n₂=235

Therefore, the coil 2 has n ₂= 235 loops.

4 0

3 years ago

Read 2 more answers

A flywheel with a diameter of 1.63 m is rotating at an angular speed of 79.9 rev/min. (a) What is the angular speed of the flywh

Studentka2010 [4]

Answer:

(a) 8.362 rad/sec

(b) 6.815 m/sec

(c) 9.446 $rad/sec^2$

(d) 396.22 revolution

Explanation:

We have given that diameter d = 1.63 m

So radius $r=\frac{d}{2}=\frac{1.63}{2}=0.815m$

Angular speed N = 79.9 rev/min

(a) We know that angular speed in radian per sec

$\omega =\frac{2\pi N}{60}=\frac{2\times 3.14\times 79.9}{60}=8.362rad/sec$

(b) We know that linear speed is given by

$v=r\omega =0.815\times 8.362=6.815m/sec$

(c) We have given final angular velocity $\omega _f=675rev/min$

And $\omega _i=79.9rev/min$

Time t = 63 sec

Angular acceleration is given by $\alpha =\frac{\omega _f-\omega _i}{t}=\frac{675-79.9}{63}=9.446rad/sec^2$

(d) Change in angle is given by

$\Theta =\frac{1}{2}(\omega _i+\omega _f)t=\frac{1}{2}(675+79.9)\times 1.05=396.22rev$

7 0

3 years ago

Vector A → has magnitude 8.78 m at 37.0 ∘ from the + x axis. Vector B → has magnitude 8.26 m at 135.0 ∘ from the + x axis. Vecto

kodGreya [7K]

Answer:

R = (- 3.72î + 8.29j)

Magnitude of R = 9.09 m

Explanation:

Let î and j represent unit vectors along the x and y axis respectively.

Vector A --> magnitude 8.78 m, direction 37.0° from the +x-axis

Let the x and y components of this vector be Aₓ and Aᵧ

A = (Aₓî + Aᵧj) m

The components given magnitude and direction from the +x-axis are calculated as

Aₓ = A cos θ and Aᵧ = A sin θ

Aₓ = (8.78 cos 37°) = 7.01 m

Aᵧ = (8.78 sin 37°) = 5.28 m

A = (7.01î + 5.28j) m

Vector B has magnitude 8.26 m and direction 135° from the +x-axis

B = (Bₓî + Bᵧj) m

Bₓ = (8.26 cos 135°) = - 5.84 m

Bᵧ = (8.26 sin 135°) = 5.84 m

B = (-5.84î + 5.84j) m

Vector C has magnitude 5.65 m and direction 210° from the +x-axis

C = (Cₓî + Cᵧj) m

Cₓ = (5.65 cos 210°) = - 4.89 m

Cᵧ = (5.65 sin 210°) = - 2.83 m

C = (- 4.89î - 2.83j) m

The resultant force is a vector sum of all the forces. Let the resultant force be R

R = (Rₓî + Rᵧj) m

R = A + B + C = (7.01î + 5.28j) + (-5.84î + 5.84j) + (- 4.89î - 2.83j)

Summing the î and j components seperately,

R = (- 3.72î + 8.29j) m

To get its magnitude,

Magnitude of R = √(Rₓ² + Rᵧ²) = √((-3.72)² + (8.29)²) = 9.09 m

8 0

4 years ago

A very long uniform line of charge has charge per unit length λ1 = 4.68 μC/m and lies along the x-axis. A second long uniform li

Kitty [74]

Answer:

$E_{net} = 6.44 \times 10^5 N/C$

Explanation:

As we know that electric field due to infinite line charge distribution at some distance from it is given as

$E = \frac{2k \lambda}{r}$

now we need to find the electric field at mid point of two wires

So here we need to add the field due to two wires as they are oppositely charged

Now we will have

$E_{net} = \frac{2k\lambda_1}{r} + \frac{2k\lambda_2}{r}$

now plug in all data

$\lambda_1 = 4.68 \muC/m$

$\lambda_2 = 2.48 \mu C/m$

$r = 0.200 m$

now we have

$E_{net} = \frac{2k}{r}(4.68 + 2.48)$

$E_{net} = \frac{2(9\times 10^9)}{0.200}(7.16 \times 10^{-6})$

$E_{net} = 6.44 \times 10^5 N/C$

8 0

3 years ago

A skater is using very low friction rollerblades. A friend throws a Frisbee at her, on the straight line along which she is coas

kupik [55]

Answer:

a) perfectly inelastic, b) collision is inelastic, c) elastic

Explanation:

In this exercise, it is asked to identify what type of shock occurs between the skater and the frisbee, for this we must define a system formed by the skater and the fribee, so that the forces during the crash have been internal and the amount of movement is preserved

Initial instant. Before the skater touches the frisbee

p₀ = M v₁ + m v₂

where M and m are the masses of the skater and frisbee, respectively

for the final moment they give us several possibilities, in all case the moment is conserved

p₀ = $p_{f}$

case a)

Final instant. grabs the frisbee and holds it

p_{f} = (M + m) v '

p₀ = p_{f}

We can see that this shock is perfectly inelastic, it holds the fressbee

case b)

final instant.

This case is similar to the previous one, but the final speed of fresbee is zero, therefore this collision is inelastic and the kinetic energy is not conserved.

case c)

final instant. Grab the fressbee and resend it

$p_{f} = M v_{1f} + m v_{2f}$

this is an elastic Shock since the equivalent of a rebound of the fressbee, the kinetic energy is conserved.

5 0

3 years ago