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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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How much work will it take to lift a 2-kg pair of hiking boots 2 meters off the

AnnyKZ [126]

Answer:

Option C - 39.2 J

Explanation:

We are given that;

Mass; m = 2 kg.

Distance moved off the floor;d = 10 m.

Acceleration due to gravity;g = 9.8 m/s².

We want to find the work done.

Now, the Formula for work done is given by;

Work = Force × displacement.

In this case, it's force of gravity to lift up the boots, thus;

Formula for this force is;

Force = mass x acceleration due to gravity

Force = 2 × 9.8 = 19.2 N

∴ Work done = 19.6 × 2

Work done = 39.2 J.

Hence, the Work done to life the boot of 2 kg to a height of 2 m is 39.2 J.

7 0

3 years ago

Read 2 more answers

Consider an insulating sphere of radius 6 cm surrounded by a conducting sphere of inner radius 18 cm and outer radius 26 cm. Fur

dimulka [17.4K]

Answer:

$-1.7908787542\times 10^{-9}\ C$

$3.4260289211\times 10^{-9}\ C$

$1.7908787542\times 10^{-9}\ C$

$1.6351501669\times 10^{-9}\ C$

Explanation:

r = Radius

k = Coulomb constant = $8.99\times 10^{9}\ Nm^2/C^2$

Electric field is given by

$E=-\dfrac{kq}{r^2}\\\Rightarrow q=-\dfrac{Er^2}{k}\\\Rightarrow q=-\dfrac{1610\times 0.1^2}{8.99\times 10^9}\\\Rightarrow q=-1.7908787542\times 10^{-9}\ C$

The charge is $-1.7908787542\times 10^{-9}\ C$

$Q+q=\dfrac{Er^2}{k}\\\Rightarrow Q=\dfrac{Er^2}{k}-q\\\Rightarrow Q=\dfrac{120\times 0.35^2}{8.99\times 10^9}-(-1.7908787542\times 10^{-9})\\\Rightarrow Q=3.4260289211\times 10^{-9}\ C$