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igor_vitrenko [27]
3 years ago
15

People have proposed driving motors with the earth's magnetic field. This is possible in principle, but the small field means th

at unrealistically large currents are needed to produce noticeable torques. Suppose a 20-cm-diameter loop of wire is oriented for maximum torque in the earth's field.What current would it need to carry in order to experience a very modest 1.0×10−3N⋅m torque?
Physics
1 answer:
NISA [10]3 years ago
3 0

Answer:

Explanation:

Torque on a loop in a magnetic field

Maximum torque = M B

M is magnetic moment  of loop and B is magnetic field

M = area x current

= 3.14 x .10² x i

Maximum torque = M B

1 x 10⁻³ = 3.14 x .10² x i x .65 x 10⁻⁴

i = 490 A

Current = 490 A.

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viva [34]
The correct answer is D.
8 0
3 years ago
A series of pulses, each of amplitude 0.1m , is sent down a string that is attached to a post at one end. The pulses are reflect
sp2606 [1]

The net displacement at a point on the string where the pulses cross is 0.2 m.

The term "displacement" refers to a shift in an object's position. It has a magnitude and a direction, making it a vector quantity. An arrow pointing from the starting point to the finishing point serves as its symbol.

A string that is connected to a post at one end is used to transmit a sequence of pulses, each measuring 0.1 meters in amplitude.

At the post, the pulses are reflected and return along the string without losing any of their amplitude.

Now, let's say the ends are free.

There is no inversion on reflection if the end is free. The amplitude at their intersection is 2A.

Now, since A = 0.1 m

Then, 2A = 2(0.1) = 0.2 m

As a result, the net displacement at the string's intersection of two pulses is 0.2 m.

The correct option is (c).

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4 0
2 years ago
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Vlad1618 [11]
1 every light in your household is parallel this is useful bc as one goes out not all do as they would in a series circuit 2 plastic and glass they are common bc it does not let electricity travel from one place to another outside of it
8 0
3 years ago
Read 2 more answers
g A 1.45-kg block is pushed against a vertical wall by means of a spring (k = 860 N/m). The coefficient of static friction betwe
olga_2 [115]

Answer:

The minimum compression is  x= 0.046m

Explanation:

From the question we are told that

              The mass of the block is m_b = 1.45 kg

               The spring constant is  k = 860 N/m

               The coefficient of static friction is  \mu = 0.36

For the the block not slip it mean the sum of forces acting on the  horizontal axis is equal to the forces acting on the vertical axis

     Now the force acting on the vertical axis is the force due to gravity which is mathematically given as

                   F_y = m_b*g

And the force acting on the horizontal axis is  force due to the spring which is mathematically represented as

                   F_x = k *x * \mu

where x is the minimum compression to keep the block from slipping

        Now equating this two formulas and making x the subject

                      x = \frac{m_b * g}{k * \mu}

substituting values we have

                     x = \frac{1.45 * 9.8}{860 *0.36}

                        x= 0.046m

 

3 0
3 years ago
The forces in (Figure 1) are acting on a 1.0 kg object.What is ax , the x -component of the object's acceleration
a_sh-v [17]

The x -component of the object's acceleration is 2 m/s².

<h3>What's the resultant force along x- direction?</h3>
  • Forces along x axis direction are as follows
  1. 4N along +x axis, so it's taken as +4 N
  2. 2N along -x axis , so it's taken as -2N.
  • Resultant force along x direction = 4N - 2N = 2 N which is along + ve x direction.

<h3>What's the acceleration along x axis direction?</h3>
  • As per Newton's second law, Force = mass × acceleration of the object
  • Force along x axis= mass × acceleration along x axis= 2N
  • Acceleration = 2/ mass = 2/1 = 2 m/s²

Thus, we can conclude that the acceleration along x axis is 2 m/s².

Disclaimer: The question was given incomplete on the portal. Here is the complete question.

Question: The forces in (Figure 1) are acting on a 1.0 kg object. What is ax, the x-component of the object's acceleration?

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3 0
2 years ago
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