Ask question

Ask question

All categories

3 years ago

9

If the magnetic field strength is found to be zero between the two wires at a distance of 3.0 cm from the first wire, what is th

e magnitude of the current in the second wire?

1 answer:

Taya2010 [7]3 years ago

4 0

Answer: The magnitude of the current in the second wire 2.67A

Explanation:

Here is the complete question:

Two straight parallel wires are separated by 7.0 cm. There is a 2.0-A current flowing in the first wire. If the magnetic field strength is found to be zero between the two wires at a distance of 3.0 cm from the first wire, what is the magnitude of the current in the second wire?

Explanation: Please see the attachments below

You might be interested in

A person kicks a ball, giving it an initial velocity of 20.0 m/s up a wooden ramp. When the ball reaches the top, it becomes air

Bess [88]

Answer:

Explanation:

given,

initial velocity of the ball = 20 m/s

angle of ramp = 22°

ball travel at a distance = 5 m

a) for friction less

$\dfrac{1}{2}mv^2 = \dfrac{1}{2}mu^2 - mgh$

$v^2 = u^2 - 2gh$

$v = \sqrt{u^2- 2 g h cos 22^0}$

$v = \sqrt{20^2- 2\times 9.8 \times 5 cos 22^0 }$

v = 17.58 m/s

b) considering the friction

$\dfrac{1}{2}mv^2 = \dfrac{1}{2}mu^2 - mgh-\mu_kmgl$

$v^2 = u^2 - 2gh-2\mu_kmgl$

$v = \sqrt{u^2- 2 g h cos 22^0-2\mu_kgl}$

$v = \sqrt{20^2- 2\times 9.8 \times 5 cos 22^0-2\times 0.15\times 9.8 \times 5 }$

v = 17.16 m/s

7 0

3 years ago

Billy picks up a 40 lb. dumbbell (mass = 18.14 kg). The center of his hand, where the dumbbell is held, is 56 cm (0.56 m) from t

umka21 [38]

Answer:

<h2>Force due to biceps is given as</h2><h2> $F = 1991.05 N$ </h2>

Explanation:

For balancing the force we know that

Torque due to weight hold on his hand = torque due to force applied by biceps

So here we will have

$mg \times L = F \times d$

so we have

$18.14 \times 9.8 \times 0.56 = F \times (0.05)$

$F = 1991.05 N$

8 0

3 years ago

An 67-kg jogger is heading due east at a speed of 2.3 m/s. A 70-kg jogger is heading 61 ° north of east at a speed of 1.3 m/s. F

ololo11 [35]

Answer

given,

mass of jogger = 67 kg

speed in east direction = 2.3 m/s

mass of jogger 2 = 70 Kg

speed = 1.3 m/s in 61 ° north of east.

jogger one

$P_1 = m_1 v_1 \hat{i}$

$P_1 = 67 \times 2.3\hat{i}$

$P_1 = 154.1 \hat{i}$

$P_2 = m_2 v_2 \hat{i} +m_2 v_2 \hat{j}$

$P_2 = 70\times v cos \theta \hat{i} +70\times v sin \theta \hat{j}$

$P_2 = 70\times 1.3 cos 61^0 \hat{i} +70\times 1.3 sin 61^0\hat{j}$

$P_2 = 44.12\hat{i} +79.59\hat{j}$

now

P = P₁ + P₂

$P = 198.22 \hat{i} +79.59 \hat{j}$

magnitude

$P = \sqrt{198.22^2 + 79.59^2}$

$P =213.60 kg.m/s$

$\theta = tan^{-1}\dfrac{79.59}{198.22}$

$\theta = 21.87$

the angle is $\theta = 21.87$ north of east

7 0

3 years ago

Sics and Chemistry A (2nd Six Weeks) - SC3200SFA/01

Andreyy89

Force = Mass * Acceleration therefore the red ball with the higher mass will have more force and greater acceleration

3 0

3 years ago

Which of the following statements best describes the second law of

Serga [27]

Answer:

B

Explanation:

Heat flows from hot to cold to lower the temperature of hot areas and increase temperature of cold areas. The end result is that the 2 areas have the same temperature, thus increasing entropy.

7 0

3 years ago