Ask question

Ask question

All categories

3 years ago

14

Rotor windings in wound rotor motors are connected to ? on the shaft of the motor. Brushes allow those windings to be connected

to external resistors to control the speed of the motor.

1 answer:

Mashutka [201]3 years ago

6 0

Answer:

Slip Rings

Explanation:

The wound rotor motor has a three-phase winding with each one connected to seperate slip rings. These slip rings contain brushes which form a secondary circuit where resistance can be inserted and this will allow for the rotor current to run more in phase with the stator current which will result in increased torque that is created

You might be interested in

If you found yourself on the see-through side of this one-way mirror what is the best way you could prevent someone on the other

Taya2010 [7]

Answer:

If there is any sheets or padded material in this room you can cover the window, you could turn off all the lights if there is a light switch in the room, you could try to bring a bright flashlight in and shine it into the other room(try to annoy the person watching you so they leave), act really boring and hopefully make the other person lose interest.

Explanation:

(hint) If you actually get in a situation like this place your fingernail against the mirror or glass you think could possibly be a one-way mirror. If there's a gap between your nail and the mirror, it's most likely a genuine mirror :)

7 0

3 years ago

Alpha radiation is ______.

KonstantinChe [14]

B because alpha radiation is just the amount of alpha particles, c could also be plausible but i think gamma rays are more deadly

3 0

3 years ago

explain why when water is poured on a dry class slab it spreads uniformly but it forms spherical droplets on a waxed glass slab

Mars2501 [29]

Answer and Explanation:

The texture of the slabs affects the type of bonding the water molecules has with each other.

The dry glass slab allows the water to form cohesive bonds with each other, making the water stick to itself and not the slab, which makes it flow uniformly, or regularly.

The Waxed glass slab has a different texture that forms an adhesive-type bonding with water, making the water stick to the slab at certain parts/forming the spherical droplets.

<u><em>#teamtrees #PAW (Plant And Water)</em></u>

8 0

3 years ago

A thin spherical spherical shell of radius R which carried a uniform surface charge density σ. Write an expression for the volum

ozzi

Answer:

Explanation:

From the given information:

We know that the thin spherical shell is on a uniform surface which implies that both the inside and outside the charge of the sphere are equal, Then

The volume charge distribution relates to the radial direction at r = R

∴

$\rho (r) \ \alpha \ \delta (r -R)$

$\rho (r) = k \ \delta (r -R) \ \ at \ \ (r = R)$

$\rho (r) = 0\ \ since \ r< R \ \ or \ \ r>R---- (1)$

To find the constant k, we examine the total charge Q which is:

$Q = \int \rho (r) \ dV = \int \sigma \times dA$

$Q = \int \rho (r) \ dV = \sigma \times4 \pi R^2$

∴

$\int ^{2 \pi}_{0} \int ^{\pi}_{0} \int ^{R}_{0} \rho (r) r^2sin \theta \ dr \ d\theta \ d\phi = \sigma \times 4 \pi R^2$

$\int^{2 \pi}_{0} d \phi* \int ^{\pi}_{0} \ sin \theta d \theta * \int ^{R}_{0} k \delta (r -R) * r^2dr = \sigma \times 4 \pi R^2$

$(2 \pi)(2) * \int ^{R}_{0} k \delta (r -R) * r^2dr = \sigma \times 4 \pi R^2$

Thus;

$k * 4 \pi \int ^{R}_{0} \delta (r -R) * r^2dr = \sigma \times R^2$

$k * \int ^{R}_{0} \delta (r -R) r^2dr = \sigma \times R^2$

$k * R^2= \sigma \times R^2$

$k = R^2 --- (2)$

Hence, from equation (1), if k = $\sigma$

$\mathbf{\rho (r) = \delta* \delta (r -R) \ \ at \ \ (r=R)}$

$\mathbf{\rho (r) =0 \ \ at \ \ rR}$

To verify the units:

$\mathbf{\rho (r) =\sigma \ * \ \delta (r-R)}$

↓ ↓ ↓

c/m³ c/m³ × 1/m

Thus, the units are verified.

The integrated charge Q

$Q = \int \rho (r) \ dV \\ \\ Q = \int ^{2 \ \pi}_{0} \int ^{\pi}_{0} \int ^R_0 \rho (r) \ \ r^2 \ \ sin \theta \ dr \ d\theta \ d \phi \\ \\ Q = \int ^{2 \pi}_{0} \ d \phi \int ^{\pi}_{0} \ sin \theta \int ^R_{0} \rho (r) r^2 \ dr$

$Q = (2 \pi) (2) \int ^R_0 \sigma * \delta (r-R) r^2 \ dr$

$Q = 4 \pi \sigma \int ^R_0 * \delta (r-R) r^2 \ dr$

$Q = 4 \pi \sigma *R^2$ since $( \int ^{xo}_{0} (x -x_o) f(x) \ dx = f(x_o) )$

$\mathbf{Q = 4 \pi R^2 \sigma }$

6 0

3 years ago

A simple pendulum with a length of 2.23 m and a mass of 6.74 kg is given an initial speed of 2.06 m/s at its equi- librium posit

solniwko [45]

Answer:

a) T = 2.997 s

b) K = 14.3 J

c) φ = 0.444 rad

Explanation:

a) Determine its period

The pendulum simple’s period is:

T = 2π $\sqrt{\frac{l}{g} }$

Where l: Pendulum’s length

g = 9.8 m/s2

T = 2π $\sqrt{\frac{2.23}{9.8} }$

T = 2.997 s

b) Total energy

Initially his total energy is kinetic

K = $\frac{mv^{2} }{2}$

K = $\frac{(6.74)(2.06)^{2} }{2}$

K = 14.3 J

c) Maximum angular displacement

φ = $cos^{-1}(1-\frac{E}{mgl} )$

φ = $cos^{-1}(1-\frac{14.3}{(6.74)(9.8)(2.23)} )$

φ = 0.444 rad

4 0

3 years ago