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Zina [86]
4 years ago
11

Suppose you wish to construct a motor that produces a maximum torque whose magnitude is 1.7 × 10-2 N·m. The coil of the motor ha

s an area of 9.0 × 10-4 m2 , consists of N turns, and contains a current of 1.1 A. The coil is placed in a uniform magnetic field of magnitude 0.20 T. What must N be?
Physics
1 answer:
Len [333]4 years ago
3 0

Answer:

86 turns

Explanation:

Parameters given:

Magnetic torque, τ = 1.7 * 10^(-2) Nm

Area of coil, A = 9 * 10^(-4) m²

Current in coil, I = 1.1 A

Magnetic field, B = 0.2 T

The magnetic toque is given mathematically as:

τ = N * I * A * B

Where N = number of turns

To find the number of turns, we make N subject of formula:

N = τ/(I * A * B)

Therefore:

N = (1.7 * 10^(-2)) / (1.1 * 9 * 10^(-4) * 0.2)

N = 85.85 = 86 turns (whole number)

The number of turns must be 86.

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Answer:

lamda=10.5m

Explanation:

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3 years ago
A tennis ball is dropped from a height of 10.0 m. It rebounds off the floor and comes up to a height of only 4.00 m on its first
Dominik [7]

Answer:

a) V=14.01 m/s

b) V=8.86 \, m/s

c)t = 2.33s

Explanation:

Our most valuable tool in solving this problem will be the conservation of mechanical energy:

E_m = E_k +E_p

That is, mechanical energy is equal to the sum of potential and kinetic energy, and  the value of this E_m mechanical energy will remain constant. (as long as there is no dissipation)

For a point particle, we have that kinetic energy is:

E_k = \frac{1}{2} m \, V^2

Where m is the mass, and V is the particle's velocity,

Potential energy on the other hand is:

E_p= m\, g\, h

where g is the acceleration due to gravity (g=9.81 \, m/s^2) and h is the height of the particle. How do we define the height? It's a bit of an arbitrary definition, but we just need to define a point for which h=0, a "floor". conveniently we pick the actual floor as our reference height, but it could be any point whatsoever.

Let's calculate  the mechanical energy just before the ball is dropped:

As we drop the ball, speed must be initially zero, and the height from which we drop it is 10 meters, therefore:

E_m = \frac{1}{2}m\,0^2+ mg\cdot 10 \,m\\E_m=mg\cdot10 \, m

That's it, the actual value of m is not important now, as we will see.

Now, what's the potential energy at the bottom? Let's see:

At the bottom, just before we hit the floor, the ball is no longer static, it has a velocity V that we want to calculate, on the other hand, it's height is zero! therefore we set h=0

E_m = \frac{1}{2}m\,V^2+ mg\cdot 0\\\\E_m = \frac{1}{2}m\,V^2

So, at the bottom, all the energy is kinetic, while at the top all the energy is potential, but these energies are the same! Because of conservation of mechanical energy. Thus we can set one equal to the other:

E_m = \frac{1}{2}m\,V^2 = mg\cdot 10m\\\\\\ \frac{1}{2}m\,V^2 = mg\cdot 10m\\\\V = \sqrt[]{2g\cdot 10m} \\

And so we have found the velocity of the ball as it hits the floor.

V = \sqrt[]{2g\cdot 10m}=14.01\, m/s

Now, after the ball has bounced, we can again do an energy analysis, and we will get the same result, namely:

V = \sqrt[]{2g\cdot h}

where h is the maximum height of the ball, and v is the maximum speed of the ball (which is always attained at the bottom). If we know that now the height the ball achieves is 4 meters, plugging that in:

V = \sqrt[]{2g\cdot 4m} =8.86 \, m/s

Now for C, we need to know for how long the ball will be in the air from the time we drop it from 10 meters, and how long it will take the ball to reach its new maximum height of 4 meters.

As the acceleration of gravity is a constant, that means that the velocity of the ball will change at a constant rate. When something changes at a constant rate, what is its average?  It's the average between initial and final velocity, look at diagram to understand. The area under the Velocity vs time curve is the displacement of the ball, and:

V_{avg}\cdot t=h\\t=h/V_{avg}

what's the average speed when the ball is descending?

V_{avg}=\frac{1}{2} (14.01\, m/s+0)=7 \, m/s

so the time it takes the ball to go down is:

t=h/V_{avg}=\frac{10m}{7m/s} =1.43s\\

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V_{avg}=\frac{1}{2} (8.86\, m/s+0)=4.43 \, m/s

and the time it takes to go up is:t=h/V_{avg}=\frac{4m}{4.43m/s} =0.90s

When we add both times , we get:

t_{total}=t_{down}+t_{up}=1.43s+0.90s = 2.33s

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If it's volume changes when you move it to the new container it would be a solid

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Answer:

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Explanation:

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