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ziro4ka [17]
3 years ago
9

A circuit is constructed with six resistors and two batteries as shown. the battery voltages are v1 = 18 v and v2 = 12 v. the po

sitive terminals are indicated with a + sign, the values for the resistors are: r1 = r5 = 58 Ï, r2 = r6 = 124 Ï r3 = 47 Ï, and r4 = 125 Ï. the positive directions for the currents i1, i2 and i3 are indicated by the directions of the arrows. 1) what is v4, the magnitude of the voltage across the resistor r4?
Physics
1 answer:
VladimirAG [237]3 years ago
5 0

Answer:

V4=9.197v

Explanation:

Given:

V1= 18v ,V2= 12v ,r1=r5=58ohms ,r2=r6=124ohms , r3=47ohms ,r4= 125ohms

V4= I4R4 = V2/(R4 + R5)×R4

V4= 12×125 /(125 + 58)

V4=1500/183 =9.197v

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Refer to the first diagram. What is the weight of the person hanging on the end of the seesaw in Newtons?
irina1246 [14]

Due to equilibrium of moments:

1) The weight of the person hanging on the left is 250 N

2) The 400 N person is 3 m from the fulcrum

3) The weight of the board is 200 N

Explanation:

1)

To solve the problem, we use the principle of equilibrium of moments.

In fact, for the seesaw to be in equilibrium, the total clockwise moment must be equal to the total anticlockwise moment.

The moment of a force is defined as:

M=Fd

where

F is the magnitude of the force

d is the perpendicular distance of the force from the fulcrum

In the first diagram:

- The clockwise moment is due to the person on the right is

M_c = W_2 d_2

where W_2 = 500 N is the weight of the person and d_2 = 2 m is its distance from the fulcrum

- The anticlockwise moment due to the person hanging on the left is

M_a = W_1 d_1

where W_1 is his weight and d_1 = 4 m is the distance from the fulcrum

Since the seesaw is in equilibrium,

M_c = M_a

So we can find the weight of the person on the left:

W_1 d_1 = W_2 d_2\\W_1 = \frac{W_2 d_2}{d_1}=\frac{(500)(2)}{4}=250 N

2)

Again, for the seesaw to be in equilibrium, the total clockwise moment must be equal to the total anticlockwise moment.

- The clockwise moment due to the person on the right is

M_c = W_2 d_2

where W_2 = 400 N is the weight of the person and d_2 is its distance from the fulcrum

- The anticlockwise moment due to the person on the left is

M_a = W_1 d_1

where W_1 = 300 N is his weight and d_1 = 4 m is the distance from the fulcrum.

Since the seesaw is in equilibrium,

M_c = M_a

So we can find the distance of the person on the right:

W_1 d_1 = W_2 d_2\\d_2 = \frac{W_1 d_1}{W_2}=\frac{(300)(4)}{400}=3 m

3)

As before, for the seesaw to be in equilibrium, the total clockwise moment must be equal to the total anticlockwise moment.

- The clockwise moment around the fulcrum this time is due to the weight of the seesaw:

M_c = W_2 d_2

where W_2 is the weight of the seesaw and d_2 = 3 m is the distance of its centre of mass from the fulcrum

- The anticlockwise moment due to the person on the left is

M_a = W_1 d_1

where W_1 = 600 N is his weight and d_1 = 1 m is the distance from the fulcrum

Since the seesaw is in equilibrium,

M_c = M_a

So we can find the weight of the seesaw:

W_1 d_1 = W_2 d_2\\W_2 =\frac{W_1 d_1}{d_2}= \frac{(600)(1)}{3}=200 N

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ivolga24 [154]

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Explanation:

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