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Alexeev081 [22]
2 years ago
5

A wire carrying a current of 10 A and 2 m in length is placed in a field of flux density 0.15 T. What’s the force on the wire if

it is placed (a) at right angles to the field, (b) at 450 to the field, (c) along the field​
Physics
1 answer:
saul85 [17]2 years ago
7 0

Explanation:

I = 10A

l = 2m

B = 0.15T

F = ?

a) ¶ = 90

F = BILsin¶

F = 0.15×10×2×sin90

F = 3N

b) ¶ = 45 degree

F = BILsin¶

F = 0.15×10×2×sin45

F = 2.12N

c) ¶ = 0 degree

F = BILsin¶

F = 0.15×10×2×sin0

F = 0

Goodluck

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A basketball rolls off a deck that is 3.2 m from the pavement. The basketball lands 0.75 m from the edge of the deck. How fast w
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The table is almost perfect. BUT ... since she called ay negative, that means she's calling the upward direction positive-y and the downward direction negative-y. In that case, since the ball moves downward from the deck to the pavement, the change in y should be negative 3.2 m. Everything else in her table is fine. Choice-D is the good one.

Now, regarding the speed of the ball ...
How long does it take to fall 3.2 m ?
Use the formula. D = 1/2 g T^2 .

3.2 = 4.9 T^2.

T^2 = 3.2/4.9

T = √(3.2/4.9) = 0.808 second.

The ball hit the pavement 0.808 second after it rolled off the deck. So that's also the time it took to move the 0.75 m horizontally.

Speed = distance / time

Speed = 0.75 m / 0.808 second

Speed = 0.928 meter/second .
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2 years ago
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In this chapter, we treated force as a push or pull, now we say it is part of an interaction. Is a force a push or pull, or part
mafiozo [28]

Answer:

A force can be described as both.

A force is a push or pull upon an item coming about because of the object's connection with another object.

8 0
2 years ago
In an electricity experiment, a 1.0g plastic ball is suspended on a60-cm-long string and given an electric charge. A charged rod
alisha [4.7K]

a) Magnitude of the electric force: 3.57\cdot 10^{-3} N

b) Tension in the string: 0.010 N

Explanation:

a)

When the charged rod is brought near the ball, then the ball remains "suspended" in an inclined position. Therefore, we can analzye the forces acting in two perpendicular directions:

- Along the horizontal direction, we have the electric force F_E, pushing in one direction, and the component of the tension in the string acting in the opposite direction, T sin \theta, where T is the tension and \theta=20^{\circ} is the angle with the vertical

- Along the vertical direction, we have the weight of the ball, mg, acting downward (where m=1.0 g = 0.001 kg is the mass of the ball and g=9.8 m/s^2 is the acceleration due to gravity), and the component of the tension acting in the upward direction, T cos \theta

Therefore, since the ball is in equilibrium, we have the two equations:

T sin \theta =F_E\\Tcos \theta = mg

By dividing the two equations, we get

tan \theta=\frac{F_E}{mg}

an solving for the electric force, we find

F_E=mg tan \theta=(0.001)(9.8)tan 20^{\circ}=3.57\cdot 10^{-3} N

b)

The tension in the string can now be found by using either of the two equations above; for instance, by using the equation along the horizontal direction,

T sin \theta =F_E

Where

F_E=3.57\cdot 10^{-3} N is the electric force

\theta=20^{\circ} is the angle with the vertical

We find the tension in the string:

T=\frac{F_E}{sin \theta}=\frac{3.57\cdot 10^{-3} N}{sin 20^{\circ}}=0.010 N

Learn more about electric force:

brainly.com/question/8960054

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6 0
2 years ago
A wedge with an inclination of angle θ rests next to a wall. A block of mass m is sliding down the plane. There is no friction b
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Answer:

  The net force on the block  F(net)  = mgsinθ).

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Explanation:

(a)

Here, m is the mass of the block, n is the normal force, \thetaθ is the wedge angle, and Fw  is the force exerted by the wall on the wedge.

Since the block sliding down, the net force on the block is along the plane of the wedge that is equal to horizontal component of weight of the block.

                    F(net)  = mgsinθ

The net force on the block  F(net)  = mgsinθ).

The direction of motion of the block is along the direction of net force acting on the block. Since there is no frictional force between the wedge and block, the only force acting on the block along the direction of motion is mgsinθ.

(b)

From the free body diagram, the normal force n is equal to mgcosθ .

                           n=mgcosθ

The horizontal component of normal force on the block is equal to force

                           Fw=n*sin(θ) that exerted by the wall on the wedge.

Substitute mgcosθ for n in the above equation;

                           Fw =mg(cosθ)(sinθ)

Since, there is no friction between the wedge and the wall, there is component force acting on the wall to restrict the motion of the wedge on the surface and that force is arises from the horizontal component for normal force on the block.

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2 years ago
A city uses a water tower to store water for times of high demand. When demand is light, water is pumped into the tower. When de
love history [14]

Answer:

The height of the tower will be 35.714 m

Explanation:

We have given gauge pressure P=350kPa=350\times 10^3Pa

Density of water \rho =1000kg/m^3

We have to find the height of the tower h

We know that gauge pressure is given by P=\rho gh

350\times 10^3=100 0\times 9.8\times h

h=35.714m

So the height of the tower will be 35.714 m

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2 years ago
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