Does anyone know the answer to all of these questions

What is the voltage across each resistor?

kogti [31]

Answer:

see solution below

Explanation:

The given resistors are connected in series.

Equivalent resistance in series = 30 + 55 + 15

Equivalent resistance in series Rt = 100 ohms

Since the potential difference in the circuit = 36V

Get the current in the circuit first

I = V/Rt

I = 36/100

I = 0.36A

Get the voltage across 30ohms resistor;

V30 = 0.36 * 30

V30 = 10.8volts

Hence the voltage across the 30ohms resistor is 10.8volts

Get the voltage across 55ohms resistor;

V55 = 0.36 * 55

V55 = 19.8volts

Hence the voltage across the 55ohms resistor is 19.8volts

Get the voltage across 15ohms resistor;

V15 = 0.36 * 15

V15 = 5.4volts

Hence the voltage across the 15ohms resistor is 5.4volts

4 0

3 years ago

A point charge q1q1 is held stationary at the origin. A second charge q2q2 is placed at point aa, and the electric potential ene

Yuri [45]

Explanation:

The given data is as follows.

$U_{a} = 5.4 \times 10^{-8} J$

$W_{/text{a to b}} = -1.9 \times 10^{-8} J$

Electric potential energy ( $U_{b}$ ) = ?

Formula to calculate electric potential energy is as follows.

$U_{b} = U_{a} - W_{/text{a to b}}$

= $5.4 \times 10^{-8} J - (-1.9 \times 10^{-8} J)$

= $7.3 \times 10^{-8} J$

Thus, we can conclude that the electric potential energy of the pair of charges when the second charge is at point b is $7.3 \times 10^{-8} J$ .

6 0

3 years ago

Identify characteristics of energy from the Sun. Check all that apply.

xxMikexx [17]

Answer:

First one, third one, and fourth one

6 0

3 years ago

Read 2 more answers

PLEASE HELP The graph shows the amplitude of a passing wave over time in seconds (s). What is the approximate frequency of the w

d1i1m1o1n [39]

Answer:

$F = 0.3\ Hz$

Explanation:

Given

See attachment for the graph

Required

Determine the frequency

Frequency (F) is calculated as:

$F = \frac{1}{T}$

Where

T = Time to complete a period

From the attachment, the wave complete a cycle or period in 3 seconds..

So:

$F = \frac{1}{3s}$

$F = 0.333\ Hz$

$F = 0.3\ Hz$ --- Approximated

7 0

3 years ago

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A +71 nC charge is positioned 1.9 m from a +42 nC charge. What is the magnitude of the electric field at the midpoint of these c

viva [34]

Answer:

The net Electric field at the mid point is 289.19 N/C

Given:

Q = + 71 nC = $71\times 10^{- 9} C$

Q' = + 42 nC = $42\times 10^{- 9} C$

Separation distance, d = 1.9 m

Solution:

To find the magnitude of electric field at the mid point,

Electric field at the mid-point due to charge Q is given by:

$\vec{E} = \frac{Q}{4\pi\epsilon_{o}(\frac{d}{2})^{2}}$

$\vec{E} = \frac{71\times 10^{- 9}}{4\pi\8.85\times 10^{- 12}(\frac{1.9}{2})^{2}}$

$\vec{E} = 708.03 N/C$

Now,

Electric field at the mid-point due to charge Q' is given by:

$\vec{E'} = \frac{Q'}{4\pi\epsilon_{o}(\frac{d}{2})^{2}}$

$\vec{E'} = \frac{42\times 10^{- 9}}{4\pi\8.85\times 10^{- 12}(\frac{1.9}{2})^{2}}$

$\vec{E'} = 418.84 N/C$

Now,

The net Electric field is given by:

$\vec{E_{net}} = \vec{E} - \vec{E'}$

$\vec{E_{net}} = 708.03 - 418.84 = 289.19 N/C$

5 0

3 years ago