Ask question

Ask question

All categories

3 years ago

6

For the circuit in the previous part, what happens to the maximum current if the frequency is doubled and the inductance is halv

ed

1 answer:

tamaranim1 [39]3 years ago

4 0

Answer:

Following are the responses to these question:

Explanation:

Since the $max^{m}$ is the current of ckt which depend on the reactance which inductor that also enables the ckt and inductor resistance $(X_L)$ for capacities

$\to X_{C}=\frac{1}{W L}$

for

$\to X_{L}=wL$

When $w \longrightarrow 2w$

$L\longrightarrow \frac{L}{2}$

then

$\to X_{L}=2 w \times \frac{L}{2}=wL$

therefore, $X_{L}$ remains at the same so, the maximum current remains the in same ckt.

You might be interested in

Does heating a cup of<br> Water allow it to dissolve more sugar ? Constant

Anestetic [448]

Answer:

Explanation:

Yes

5 0

3 years ago

Read 2 more answers

A mountain climber, in the process of crossing between two cliffs by a rope, pauses to rest. She weighs 514 N. As the drawing sh

n200080 [17]

Answer:

Part a)

$T_L = 155.4 N$

Part b)

$T_R = 379 N$

Explanation:

As we know that mountain climber is at rest so net force on it must be zero

So we will have force balance in X direction

$T_L cos65 = T_R cos80$

$T_L = 0.41 T_R$

now we will have force balance in Y direction

$mg = T_L sin65 + T_Rsin80$

$514 = 0.906T_L + 0.985T_R$

Part a)

so from above equations we have

$514 = 0.906T_L + 0.985(\frac{T_L}{0.41})$

$514 = 3.3 T_L$

$T_L = 155.4 N$

Part b)

Now for tension in right string we will have

$T_R = \frac{T_L}{0.41}$

$T_R = 379 N$

3 0

3 years ago

uniform disk with mass 40.0 kg and radius 0.200 m is pivoted at its center about a horizontal, frictionless axle that is station

Alex787 [66]

Answer:

The magnitude of the tangential velocity is $v= 0.868 m/s$

The magnitude of the resultant acceleration at that point is $a = 4.057 m/s^2$

Explanation:

From the question we are told that

The mass of the uniform disk is $m_d = 40.0kg$

The radius of the uniform disk is $R_d = 0.200m$

The force applied on the disk is $F_d = 30.0N$

Generally the angular speed i mathematically represented as

$w = \sqrt{2 \alpha \theta}$

Where $\theta$ is the angular displacement given from the question as

$\theta = 0.2000 rev = 0.2000 rev * \frac{2 \pi \ rad }{1 rev}$

$=1.257\ rad$

$\alpha$ is the angular acceleration which is mathematically represented as

$\alpha = \frac{torque }{moment \ of \ inertia} = \frac{F_d * R_d}{I}$

The moment of inertial is mathematically represented as

$I = \frac{1}{2} m_dR^2_d$

Substituting values

$I = 0.5 * 40 * 0.200^2$

$= 0.8kg \cdot m^2$

Considering the equation for angular acceleration

$\alpha = \frac{torque }{moment \ of \ inertia} = \frac{F_d * R_d}{I}$

Substituting values

$\alph$ $\alpha = \frac{(30.0)(0.200)}{0.8}$

$= 7.5 rad/s^2$

Considering the equation for angular velocity

$w = \sqrt{2 \alpha \theta}$

Substituting values

$w =\sqrt{2 * (7.5) * 1.257}$

$= 4.34 \ rad/s$

The tangential velocity of a given point on the rim is mathematically represented as

$v = R_d w$

Substituting values

$= (0.200)(4.34)$

$v= 0.868 m/s$

The radial acceleration at hat point is mathematically represented as

$\alpha_r = \frac{v^2}{R}$

$= \frac{0.868^2}{0.200^2}$

$= 3.7699 \ m/s^2$

The tangential acceleration at that point is mathematically represented as

$\alpha _t = R \alpha$

Substituting values

$\alpha _t = (0.200) (7.5)$

$= 1.5 m/s^2$

The magnitude of resultant acceleration at that point is

$a = \sqrt{\alpha_r ^2+ \alpha_t^2 }$

Substituting values

$a = \sqrt{(3.7699)^2 + (1.5)^2}$

$a = 4.057 m/s^2$

7 0

3 years ago

A car has the velocity of 2.35 after 4.67 it's velocity is 9.89 what is the average acceleration

kiruha [24]

Average acceleration = (change in speed) / (time for the change) .

Change in speed = (ending speed) - (beginning speed)

=  (9.89 miles/hour) - (2.35 yards/second) = 26,839.2 ft/hr

Acceleration = (26,839.2 ft/hr) / (4.67 days) = 2,873.58 inch/hour²

6 0

3 years ago

What wavelength would a ripple in water have if the frequency is 1.8 Hz and a

Yuki888 [10]

Explanation:

825m/s / 1.8Hz = 458.33m

6 0

3 years ago

Read 2 more answers