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Shkiper50 [21]
3 years ago
6

Ask Your Teacher Two long, straight wires are parallel and 11 cm apart. One carries a current of 2.9 A, the other a current of 5

.3 A. (a) If the two currents flow in opposite directions, what is the magnitude (in N/m) and direction of the force per unit length of one wire on the other? magnitude N/m direction (b) What is the magnitude (in N/m) and direction of the force per unit length if the currents flow in the same direction? magnitude N/m direction
Physics
1 answer:
dsp733 years ago
7 0

Answer:

The magnitude of force per unit length of one wire on the other is 2.7945\times 10^{-5}\ N and the direction is away from one another

The magnitude of force per unit length of one wire on the other is 2.7945\times 10^{-5}\ N and the direction is towards each other.

Explanation:

\mu_0 = Vacuum permeability = 4\pi\times 10^{-7}\ N/A^2

i_1 = Current in first wire = 2.9 A

i_2 = Current in second wire = 5.3 A

r = Gap between the wires = 11 cm

Force per unit length

F_{12}=F_{21}=\frac{\mu_0 i_1i_2}{2\pi r}\\ =\frac{4\pi\times 10^{-7}\times 2.9\times 5.3}{2\pi 0.11}\\ =2.7945\times 10^{-5}\ N

The magnitude of force per unit length of one wire on the other is 2.7945\times 10^{-5}\ N and the direction is away from one another

F_{12}=F_{21}=\frac{\mu_0 i_1i_2}{2\pi r}\\ =\frac{4\pi\times 10^{-7}\times 2.9\times 5.3}{2\pi 0.11}\\ =2.7945\times 10^{-5}\ N

The magnitude of force per unit length of one wire on the other is 2.7945\times 10^{-5}\ N and the direction is towards each other.

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26. Keenan found the mass of a book to be 4.56*10^ -2 kg . What is the mass of the book in milligrams?​
vagabundo [1.1K]

Taking into account the rule of three for the change of units, the mass of the book is 45600 miligrams.

First of all, the rule of three is a mathematical tool that helps you quickly solve proportionality problems.

Having three known values ​​and one unknown, a proportional relationship is established between all of them in order to find the fourth term of the proportion.

If the relationship between the magnitudes is direct (when one magnitude increases, so does the other; or when one magnitude decreases, so does the other), the rule of three is applied as follows, where a, b and c are known values ​​and x is the unknown to calculate:

a → b

c → x

So: x=\frac{cxb}{a}

Being 1 kg equivalent to 1000000 milligrams, In this case the rule of three is applied as follows: if 1 kg equals 1000000 milligrams, 4.56×10⁻² kg equals how many milligrams?

1 kg → 1000000 milligrams

4.56×10⁻² kg  → x

So:

x=\frac{4.56x10^{-2} kg x1000000 miligrams }{1 kg}

<u><em>x=45600 miligrams</em></u>

In summary, the mass of the book is 45600 miligrams.

Learn more:

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3 years ago
An athlete at the gym holds a 3.0 kg steel ball in his hand. His arm is 60 cm long and has a mass of 3.8 kg, with the center of
Serggg [28]

Answer:

(a) τ = 26.58 Nm

(b) τ = 18.79 Nm

Explanation:

(a)

First we find the torque due to the ball in hand:

τ₁ = F₁d₁

where,

τ₁ = Torque due to ball in hand = ?

F₁ = Force due to ball in hand = m₁g = (3 kg)(9.8 m/s²) = 29.4 N

d₁ = perpendicular distance between ball and shoulder = 60 cm = 0.6 m

τ₁ = (29.4 N)(0.6 m)

τ₁ = 17.64 Nm

Now, we calculate the torque due to the his arm:

τ₁ = F₁d₁

where,

τ₂ = Torque due to arm = ?

F₂ = Force due to arm = m₂g = (3.8 kg)(9.8 m/s²) = 37.24 N

d₂ = perpendicular distance between center of mass and shoulder = 40% of 60 cm = (0.4)(60 cm) = 24 cm = 0.24 m

τ₂ = (37.24 N)(0.24 m)

τ₂ = 8.94 Nm

Since, both torques have same direction. Therefore, total torque will be:

τ = τ₁ + τ₂

τ = 17.64 Nm + 8.94 Nm

<u>τ = 26.58 Nm</u>

<u></u>

(b)

Now, the arm is at 45° below horizontal line.

First we find the torque due to the ball in hand:

τ₁ = F₁d₁

where,

τ₁ = Torque due to ball in hand = ?

F₁ = Force due to ball in hand = m₁g = (3 kg)(9.8 m/s²) = 29.4 N

42.42 cm = 0.4242 m

τ₁ = (29.4 N)(0.4242 m)

τ₁ = 12.47 Nm

Now, we calculate the torque due to the his arm:

τ₁ = F₁d₁

where,

τ₂ = Torque due to arm = ?

F₂ = Force due to arm = m₂g = (3.8 kg)(9.8 m/s²) = 37.24 N

d₂ = perpendicular distance between center of mass and shoulder = 40% of (60 cm)(Cos 45°) = (0.4)(42.42 cm) = 16.96 cm = 0.1696 m

τ₂ = (37.24 N)(0.1696 m)

τ₂ = 6.32 Nm

Since, both torques have same direction. Therefore, total torque will be:

τ = τ₁ + τ₂

τ = 12.47 Nm + 6.32 Nm

<u>τ = 18.79 Nm</u>

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The work done by the normal force n when the box slides down a frictionless incline and gaining speed is zero.

<h3>What is normal force?</h3>

The force of contact is called the normal force. When the two surfaces are in contact with each other, then the normal force acts.

This force is applied by the solid bodies on each other in order to prevent the passing through each other.

A box slides down a frictionless incline, gaining speed. For this box, the value of work done by normal force has to be found out. Let's analyze the given condition.

  • The body is gaining the speed, which means there is a change in kinetic energy.
  • The change in kinetic energy is equal to the work done.
  • The friction force is the product of coefficient of the friction and normal force.
  • The friction force for the given case is zero. Thus, the normal force must be equal to the zero.

Thus, the work done by the normal force n when the box slides down a frictionless incline and gaining speed is zero.

Learn more about the normal force here;

brainly.com/question/10941832

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