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2 years ago

If E1 = 13.0 V and E2 = 5.0 V , calculate the current I2 flowing in emf source E2.

Physics

1 answer:

swat322 years ago

3 0

The solution would be like this for this specific problem:

KCL at Junction a. 
+ I1 + I2 + I3 = 0 (1) 

KVL
+ 13 V - 0.2 R I1 - 0.025 R I1 - 5 V + 0.02 R I2 = 0 (2) 
8 + 0.02 I2 = 0.225 I1 
I1 = 35.6 + 0.0889 I2 (2A)

KVL (bottom loop - CCW direction) 
- 0.02 R I2 + 5 V + 0.5 R I3 = 0 (3) 
0.5 I3 = -5 + 0.02 I2 
I3 = -10 + 0.04 I2 (3A)

Replace 2A and 3A into 1. 

+ I1 + I2 + I3 = 0 
( 35.6 + 0.0889 I2 ) + I2 + ( -10 + 0.04 I2 ) = 0 
1.129 I2 = -25.6 
I2 = -22.6A 

Solve 2A and 3A for other currents. 

I1 = 35.6 + 0.0889 I2 = 35.6 + 0.0889 * -22.6 = 33.5A 
I3 = -10 + 0.04 I2 = -10 + 0.04 * -22.6 = -10.9A

So the answer is letter D.

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What is the rate of heat transfer by radiation, with an unclothed person standing in a dark room whose ambient temperature is 22

SIZIF [17.4K]

Answer:

$5.45\times 10^{-4}$ W

Explanation:

$T_{r}$ = Temperature of the room = 22.0 °C = 22 + 273 = 295 K

$T_{s}$ = Temperature of the skin = 33.0 °C = 33 + 273 = 306 K

$A$ = Surface area = 1.50 m²

$\epsilon$ = emissivity = 0.97

$\sigma$ = Stefan's constant = 5.67 x 10⁻⁸ Wm⁻² K⁻⁴

Rate of heat transfer is given as

$R = \epsilon \sigma A (T_{s}^{2} - T_{r}^{2})$

$R = (0.97)(5.67\times 10^{-8}) (1.50) ((306)^{2} - (295)^{2})$

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2 years ago

1. For each of the following scenarios, describe the force providing the centripetal force for the motion: a. a car making a tur

GenaCL600 [577]

Complete Question

For each of the following scenarios, describe the force providing the centripetal force for the motion:

a. a car making a turn

b. a child swinging around a pole

c. a person sitting on a bench facing the center of a carousel

d. a rock swinging on a string

e. the Earth orbiting the Sun.

Answer:

Considering a

The force providing the centripetal force is the frictional force on the tires \

i.e $\mu mg = \frac{mv^2}{r}$

where $\mu$ is the coefficient of static friction

Considering b

The force providing the centripetal force is the force experienced by the boys hand on the pole

Considering c

The force providing the centripetal force is the normal from the bench due to the boys weight

Considering d

The force providing the centripetal force is the tension on the string

Considering e

The force providing the centripetal force is the force of gravity between the earth and the sun

Explanation:

6 0

2 years ago

What is (Fnet3)x, the x-component of the net force exerted by these two charges on a third charge q3 = 55.0 nC placed between q1

notka56 [123]

Complete Question

Part of the question is shown on the first uploaded image

The rest of the question

What is (Fnet3)x, the x-component of the net force exerted by these two charges on a third charge q3 = 55.0 nC placed between q1 and q2 at x3 = -1.220 m ? Your answer may be positive or negative, depending on the direction of the force. Express your answer numerically in newtons to three significant figures.

Answer:

The net force exerted on the third charge is $F_{net}= 3.22*10^{-5} \ J$

Explanation:

From the question we are told that

The third charge is $q_3 = 55 nC = 55 *10^{-9} C$

The position of the third charge is $x = -1.220 \ m$

The first charge is $q_1 = -16 nC = -16 *10^{-9} \ C$

The position of the first charge is $x_1 = -1.650m$

The second charge is $q_2 = 32 nC = 32 *10^{-9} C$

The position of the second charge is $x_2 = 0 \ m$

The distance between the first and the third charge is

$d_{1-3} = -1.650 -(-1.220)$

$d_{1-3} = -0.43 \ m$

The force exerted on the third charge by the first is

$F_{1-3} = \frac{k q_1 q_3}{d_{1-3}^2}$

Where k is the coulomb's constant with a value $9*10^{9} \ kg\cdot m^3\cdot s^{-4}\cdot A^2.$

substituting values

$F_{1-3} = \frac{9*10^{9}* 16 *10^{-9} * (55*10^{-9})}{(-0.43)^2}$

$F_{1-3} = 4.28 *10^{-5} \ N$

The distance between the second and the third charge is

$d_{2-3} = 0- (-1.22)$

$d_{2-3} =1.220 \ m$

The force exerted on the third charge by the first is mathematically evaluated as

$F_{2-3} = \frac{k q_2 q_3}{d_{2-3}^2}$

substituting values

$F_{2-3} = \frac{9*10^{9} * (32*10^{-9}) *(55*10^{-9})}{(1.220)^2}$

$F_{2-3} = 1.06*10^{-5} N$

The net force is

$F_{net} = F_{1-3} -F_{2-3}$

substituting values

$F_{net} = 4.28 *10^{-5} - 1.06*10^{-5}$

$F_{net}= 3.22*10^{-5} \ J$

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2 years ago

A spring with a spring constant of 0.70 N/m is stretched 1.5 m. What was the force?

Talja [164]

Answer:

1.05 N

Explanation:

K = 0.7 N/m

e = 1.5 m

F = ?

from Hooke's law of elasticity

F = Ke

= 0.7×1.5

= 1.05 N

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2 years ago

Energy conservation with conservative forces: Two identical balls are thrown directly upward, ball A at speed v and ball B at sp

MatroZZZ [7]

Answer:

E) True. Ball B will go four times as high as ball A because it had four times the initial kinetic energ

Explanation:

To answer the final statements, let's pose the solution of the exercise

Energy is conserved

Initial

Em₀ = K

Em₀ = ½ m v²

Final

Emf = U = mg h

Em₀ = emf

½ m v² = mgh

h = v² / 2g

For ball A

h_A = v² / 2g

For ball B

h_B = (2v)² / 2g

h_B = 4 (v² / 2g) = 4 h_A

Let's review the claims

A) False. The neck acceleration is zero, it has the value of the acceleration of gravity

B) False. Ball B goes higher

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1 year ago