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Zina [86]
3 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]3 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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A flare is launched from a life raft with an initial velocity of 192 ft/sec. How many seconds will it take for the flare to retu
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What body systems help these cells get the energy they need?
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2 years ago
Now consider the statements in Part A that are inferred from models. A solar model is used to calculate interior conditions base
mafiozo [28]

Answer:

d. We can calculate it by applying Newton's version of Kepler's third law

Explanation:

The measurements of a Star like the Sun have several problems, the first one is distance, but the most important is the temperature since as we get closer all the instruments will melt. This is why all measurements must be indirect because of the effects that these variables create on nearby bodies.

Kepler's laws are deduced from Newton's law of universal gravitation, in these laws the mass of the Sun affects the orbit of the planets since it creates a force of attraction, if measured the orbit and the time it takes to travel it we can know the centripetal acceleration and with it knows the force, from where we clear the mass of the son.

Let's review the statements of the exercise

.a) False. We don't have good enough models for this calculation

.b) False. The size of the sun is very difficult to measure because it is a mass of gas, in addition the density changes strongly with depth

.c) False. The amount of light that comes out of the sun is not all the light produced and is due to quantum effects where the mass of the sun is not taken into account

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6 0
3 years ago
Scenario
Anvisha [2.4K]

Answer:

1) t = 23.26 s,  x = 8527 m, 2)   t = 97.145 s,  v₀ = 6.4 m / s

Explanation:

1) First Scenario.

After reading your extensive problem, we are going to solve it, for this exercise we must use the parabolic motion relationships. Let's carry out an analysis of the situation, for deliveries the planes fly horizontally and we assume that the wind speed is zero or very small.

Before starting, let's reduce the magnitudes to the SI system

         v₀ = 250 miles/h (5280 ft / 1 mile) (1h / 3600s) = 366.67 ft/s

         y = 2650 m

Let's start by looking for the time it takes for the load to reach the ground.

         y = y₀ + v_{oy} t - ½ g t²

in this case when it reaches the ground its height is zero and as the plane flies horizontally the vertical speed is zero

         0 = y₀ + 0 - ½ g t2

          t = \sqrt{ \frac{2y_o}{g} }

          t = √(2 2650/9.8)

          t = 23.26 s

this is the horizontal scrolling time

          x = v₀ t

          x = 366.67  23.26

          x = 8527 m

the speed at the point of arrival is

         v_y = v_{oy} - g t = 0 - gt

         v_y = - 9.8 23.26

         v_y = -227.95 m / s

Module and angle form

        v = \sqrt{v_x^2 + v_y^2}

         v = √(366.67² + 227.95²)

        v = 431.75 m / s

         θ = tan⁻¹ (v_y / vₓ)

         θ = tan⁻¹ (227.95 / 366.67)

         θ = - 31.97º

measured clockwise from x axis

We see that there must be a mechanism to reduce this speed and the merchandise is not damaged.

2) second scenario. A catapult located at the position x₀ = -400m y₀ = -50m with a launch angle of θ = 50º

we look for the components of speed

           cos θ = v₀ₓ / v₀

           sin θ = v_{oy} / v₀

            v₀ₓ = v₀ cos θ

            v_{oy} = v₀ sin θ

we look for the time for the arrival point that has coordinates x = 0, y = 0

            y = y₀ + v_{oy} t - ½ g t²

            0 = y₀ + vo sin θ t - ½ g t²

            0 = -50 + vo sin 50 t - ½ 9.8 t²

            x = x₀ + v₀ₓ t

            0 = x₀ + vo cos θ t

            0 = -400 + vo cos 50 t

podemos ver que tenemos un sistema de dos ecuación con dos incógnitas

          50 = 0,766 vo t – 4,9 t²

          400 =   0,643 vo t

resolved

          50 = 0,766 ( \frac{400}{0.643 \ t}) t – 4,9 t²

          50 = 476,52 t – 4,9 t²

          t² – 97,25 t + 10,2 = 0

we solve the quadratic equation

         t = [97.25 ± \sqrt{97.25^2 - 4 \ 10.2}] / 2

         t = 97.25 ±97.04] 2

         t₁ = 97.145 s

         t₂ = 0.1 s≈0

the correct time is t1 the other time is the time to the launch point,

         t = 97.145 s

let's find the initial velocity

         x = x₀ + v₀ cos 50 t

         0 = -400 + v₀ cos 50 97.145

         v₀ = 400 / 62.44

         v₀ = 6.4 m / s

5 0
2 years ago
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Answer:

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7 0
2 years ago
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