Ask question

Ask question

All categories

3 years ago

9

Which of the following are true for acceleration?

1 answer:

Fittoniya [83]3 years ago

6 0

The SI unit for acceleration is m/s2 ( D)

You might be interested in

Answer attachment below

polet [3.4K]

Answer:

second option is the answer.

6 0

3 years ago

An engineer in India (standard household voltage = 220 volts) is designing a transformer for use on her

Gnom [1K]

He should a step-up transformer with k=220/120=1.83 so output coil must have 240*1.83=440 turns

5 0

3 years ago

krek1111 [17]

Explanation:

the table and the wooden block

6 0

3 years ago

Read 2 more answers

The current in a series circuit is 19.3 A. When an additional 7.40-Ω resistor is inserted in series, the current drops to 13.4 A

Bogdan [553]

Answer:

16.8ohms

Explanation:

According to ohm's law which states that the current passing through a metallic conductor at constant temperature is directly proportional to the potential difference across its ends.

Mathematically, V = IRt where;

V is the voltage across the circuit

I is the current

R is the effective resistance

For a series connected circuit, same current but different voltage flows through the resistors.

If the initial current in a circuit is 19.3A,

V = 19.3R... (1)

When additional resistance of 7.4-Ω is added and current drops to 13.4A, our voltage in the circuit becomes;

V = 13.4(7.4+R)... (2)

Note that the initial resistance is added to the additional resistance because they are connected in series.

Equating the two value of the voltages i.e equation 1 and 2 to get the resistance in the original circuit we will have;

19.3R = 13.4(7.4+R)

19.3R = 99.16+13.4R

19.3R-13.4R = 99.16

5.9R = 99.16

R= 99.16/5.9

R = 16.8ohms

The resistance in the original circuit will be 16.8ohms

5 0

3 years ago

In an LC circuit at one time the charge stored by the capacitor is 10 mC and the current is 3.0 A. If the frequency of the circu

Ronch [10]

Answer:

$i_2=3.61\ A$

Explanation:

<u>LC Circuit</u>

It's a special circuit made of three basic elements: The AC source, a capacitor, and an inductor. The charge, current, and voltage are oscillating when there is an interaction between the electric and magnetic fields of the elements. The following variables will be used for the formulas:

$q, q_1, q_2$ = charge of the capacitor in any time $t, t_1, t_2$

$q_o$ = initial charge of the capacitor

$\omega$ =angular frequency of the circuit

$i, i_1, i_2$ = current through the circuit in any time $t, t_1, t_2$

The charge in an LC circuit is given by

$q(t) = q_0 \, cos (\omega t )$

The current is the derivative of the charge

$\displaystyle i(t) = \frac{dq(t)}{dt} = - \omega q_0 \, sin(\omega t).$

We are given

$q_1=10\ mc=0.01\ c, i_1=3\ A,\ q_2=6\ mc=0.006\ c\ ,\ f=\frac{1000}{4\pi}$

It means that

$q(t_1) = q_0 \, cos (\omega t_1 )=q_1\ .......[eq 1]$

$i(t_1) = - \omega q_0 \, sin(\omega t_1)=i_1.........[eq 2]$

From eq 1:

$\displaystyle cos (\omega t_1 )=\frac{q_1}{q_0}$

From eq 2:

$\displaystyle sin(\omega t_1)=-\frac{i_1}{\omega q_0}$

Squaring and adding the last two equations, and knowing that

$sin^2x+cos^2x=1$

$\displaystyle \left ( \frac{q_1}{q_0} \right )^2+\left ( \frac{i_1}{\omega q_0} \right )^2=1$

Operating

$\displaystyle \omega^2q_1^2+i_1^2=\omega^2q_o^2$

Solving for $q_o$

$\displaystyle q_o=\frac{\sqrt{\omega^2q_1^2+i_1^2}}{\omega}$

Now we know the value of $q_0$ , we repeat the procedure of eq 1 and eq 2, but now at the second time $t_2$ , and solve for $i_2$

$\displaystyle \omega^2q_2^2+i_2^2=\omega^2q_o^2$

Solving for $i_2$

$\displaystyle i_2=w\sqrt{q_o^2-q_2^2}$

Now we replace the given values. We'll assume that the placeholder is a pi for the frequency, i.e.

$\displaystyle f=\frac{1}{4\pi}\ KHz$

$w=2\pi f=500\ rad/s$

$\displaystyle q_o=\frac{\sqrt{(500)^2(0.01)^2+3^2}}{500}$

$q_0=0.01166\ c$

Finally

$\displaystyle i_2=500\sqrt{0.01166^2-.006^2}$

$i_2=5\ A$

3 0

3 years ago