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3 years ago

Equivalent resistance between A and B.A) 2.4 ohmsB)18 ohmsC) 6 ohmsD) 36 ohms

Physics

2 answers:

IgorLugansk [536]3 years ago

7 0

<h3>Answer :</h3>

First of all see the attachment for better understanding $.$ :D

Eq. resistance between C and D :

➝ 1/Rp = 1/6 + 1/6

➝ 1/Rp = 2/6

➝ Rp = 3Ω

Eq. resistance of upper part :

(All three are in series)

➝ Rs = 3 + 3 + 3

➝ Rs = 9Ω

Eq. resistance of lower part :

(Both are in series)

➝ Rs' = 3 + 3

➝ Rs' = 6Ω

Eq. resistance between A and B :

(Rs and Rs' are in parallel)

➝ 1/Req = 1/Rs + 1/Rs'

➝ 1/Req = 1/9 + 1/6

➝ 1/Req = (2 + 3)/18

➝ Req = 18/5

labwork [276]3 years ago

3 0

Answer:

Explanation:

Eq. resistance between C and D :

➝ Rp = 3Ω

Eq. resistance of upper part :

➝ Rs = 9Ω

Eq. resistance of lower part :

➝ Rs' = 6Ω

Eq. resistance between A and B :

➝ Req = 18/5

➝ Req = 3.6Ω

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6000 kg train moving 5 m/sec to east collides with 5000 kg train moving 3 m/sec to west. What is their velocity

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The final velocity is 1.37 m/s east

Explanation:

We can solve this problem by using the law of conservation of momentum: in fact, in absence of external forces, the total momentum of the two trains must be conserved before and after the collision.

So we can write:

$p_i = p_f\\m_1 u_1 + m_2 u_2 = (m_1+m_2)v$

where:

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$u_1 = 5 m/s$ is the initial velocity of the first train (we take east as positive direction)

$m_2 = 5000 kg$ is the mass of the second train

$u_2 = -3 m/s$ is the initial velocity of the second train

$v$ is the final combined velocity of the two trains

Re-arranging the equation and substituting the values, we find:

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3 years ago

PRACTICE ANOTHER A cube of wood having an edge dimension of 19.7 cm and a density of 647 kg/m3 floats on water. (a) What is the

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CHECK COMPLETE QUESTION

The cube of wood having an edge dimension of 19.7cm and a density of 647kg/m3 floats on water

(a) What is the distance from the horizontal top surface of the cube to the water level answer in cm

(b) What mass of lead should be placed on the cube so that the top of the cube will be just level with the water surface answer in kg

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a)6.29cm

b)2.78 kg

Explanation:

Given:

Let us calculate the volume first, we were given dimension as 19.7cm=0.197m

Volume is (0.197 meters)³ = 0.00764m³

Then we can calculate the mass as;

Given mass is 647 kg/m³ x 0.00774m³ = 4.947kg

The weight = mass × acceleration due to gravity

weight = 4.947 x 9.8 N/kg = 48.44N

By Floating we can say the the buoyancy force has to equal the weight (48.44 N) which has

which is equal to the weight of volume of the displaced water. Or the mass, the calculation is the same.

We know that density of fresh water at 20ºC is 998 kg/m³

Then we can calculate the volume of displaced water as

4.947 kg / 998 kg/m³ = 0.00496 m³

We know that the displaced water has a shape of a rectangular solid with 0.197 meters on the two horizontal dimensions, and h as the height to the surface then

V = 0.197²h = 0.00496

0.00496= 0.197²h

h = 0.1278 meters or 12.78 cm

Then the the distance exposed, would be 19.7–12.78 = 6.29 cm

b) ifthe cube is fully submerged, the volume of the displaced water is 0.00774m³

mass of displaced water is 0.00774m³ x 998 kg/m³ = 7.724 kg

Added mass is the mass of the displaced water – mass of block

= 7.724–4.947 = 2.78 kg

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Equivalent resistance between A and B.A) 2.4 ohmsB)18 ohmsC) 6 ohmsD) 36 ohms ​

Equivalent resistance between A and B.A) 2.4 ohmsB)18 ohmsC) 6 ohmsD) 36 ohms