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snow_tiger [21]
3 years ago
7

Two concentric current loops lie in the same plane. The outer loop has twice the diameter of the inner loop. The inner loop carr

ies a 1.0 A current in a clockwise direction. What current, and which direction, should the outer loop carry such that the magnetic field in the center of the loops is zero?

Physics
1 answer:
Flauer [41]3 years ago
7 0

Answer: The outer loop should carry a current of 2.0 A.

The current should flow in the counter-clockwise direction

Explanation: Please see the attachments below

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Mariana [72]

Answer:

135g

Explanation:

6 0
3 years ago
A fisherman is fishing from a bridge and is using a "44.0-N test line." In other words, the line will sustain a maximum force of
vitfil [10]

Answer:

a) 4.485 kg b) 3.94 kg

Explanation:

since the maximum tension the line can stand is 44 N and for question a the speed is constant (acceleration must be zero since the velocity or speed is not changing), F(tension) = mass * acceleration due to gravity (g) .

44 = m * 9.81m/s^2

m = 44/9.81 = 4.485kg

b) F(tension) = ma + mg ( where a is the acceleration of the body and g is the acceleration of the gravity)

44 = m (a +g)

44 = m (1.37 + 9.81)

44/11.18 = m

m = 3.94 kg

3 0
3 years ago
When two waves collide and create a taller wave, what interference occurs?
oksano4ka [1.4K]

It would be Constructive interference. It occurs when two waves collide and make a taller wave. 
3 0
3 years ago
A thin rod of length 0.64 m and mass 120 g is suspended freely from one end. It is pulled to one side and then allowed to swing
valina [46]

Answer:

1. Kinetic Energy = 0.0161 Joules

2. Height = 0.0137m

Explanation:

Given

Length of Rod, l = 0.64m

Mass, m = 120g = 0.12kg

Angular speed, w = 1.40 rad/s

a.

Calculating the Rod's kinetic energy

This is calculated by

Kinetic Energy = ½Iw²

Where I = rotational inertia of the rod about an axis.

This is calculated as follows;

I = Icm + mh²

I = ImL² + m(L/2)²

I = 1/12 * 0.12 * 0.64² + 0.12 * (0.64/2)²

I = 0.016384 kgm²

By substituton

KE = ½Iw² becomes

KE = ½ * 0.016384 * 1.40²

KE = 0.01605632J

KE = 0.0161 Joules

2. Using the total conservation of momentum;

K + U = Kf + V

Where K = Initial Kinetic Energy of the rod at lowest point.

U = Initial gravitational potential energy of the rod at lowest point

Kf = Final Kinetic Energy of the rod at maximum height = 0 J

V = Final gravitational potential energy of the rod at maximum height

So, K + U = Kf + V become

K + U = 0 + V

K + U = V

K = V - U = mgh

substitute 0.01605632J for K

0.01605632J = mgh

h = 0.01605632J/mg

h = 0.01605632J/(0.12 * 9.8)

h = 0.013653333333333

h = 0.0137m

4 0
3 years ago
A 295-kg object and a 595-kg object are separated by 4.10 m.
kodGreya [7K]

Answer:

a)F=3 x 10⁻⁷ N

b)x=2.405 m

Explanation:

Given that

m₁=295 kg

m₂=595 kg

d= 4.1 m

a)

m₃=63 kg

r=d/2 = 2.05 m

The force between the mass m₁ and m₃

F_{13}=\dfrac{Gm_1m_3}{r^2}

by putting the values

F_{13}=\dfrac{Gm_1m_3}{r^2}

F_{13}=\dfrac{6.67\times 10^{-11}\times 295\times 63 }{2.05^2}

F₁₃=2.94 x 10⁻⁷ N

The force  between the mass m₂ and m₃

by putting the values

F_{23}=\dfrac{Gm_2m_3}{r^2}

F_{23}=\dfrac{6.67\times 10^{-11}\times 595\times 63 }{2.05^2}

F₂₃=5.94 x 10⁻⁷ N

The net force F

F= F₂₃- F₁₃

F=5.94 x 10⁻⁷ N-2.94 x 10⁻⁷ N

F=3 x 10⁻⁷ N

b)

Lest take at distance x from mass m₂ net force is zero.

F_{23}=\dfrac{Gm_2m_3}{x^2}

F_{13}=\dfrac{Gm_1m_3}{(4.1-x)^2}

Form above two equation

\dfrac{Gm_1m_3}{(4.1-x)^2}=\dfrac{Gm_2m_3}{x^2}

\dfrac{m_1}{(4.1-x)^2}=\dfrac{m_2}{x^2}

\dfrac{295}{(4.1-x)^2}=\dfrac{595}{x^2}

x²=2.01(4.1-x)²

x=1.42 (4.1-x)

x=5.82 - 1.42x

x=2.405 m

4 0
3 years ago
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