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cupoosta [38]
3 years ago
7

A flat loop of wire consisting of a single turn ofcross-sectional area 7.90 cm2is perpendicular to a magnetic field that increas

es uniformly inmagnitude from 0.500 T to 3.50 T in1.00 s. What is the resulting inducedcurrent if the loop has a resistance of 1.30 Ω?
Physics
1 answer:
Westkost [7]3 years ago
6 0

To develop this problem, it is necessary to apply the concepts related to Faraday's law and Magnetic Flow, which is defined as the change that the magnetic field has in a given area. In other words

\Phi = BA Cos\theta

Where

B= Magnetic Field

A = Area

\theta = Angle between magnetic field lines and normal to the area

The differentiation of this value allows us to obtain in turn the induced emf or electromotive force.

In this case we have that the flat loop of wire is perpendicular to the magnetic field, therefore the angle is 0 degrees, since its magnetic field acts parallel to the area:

\theta = 0 then our expression can be written as

\Phi = BA

From the same value of the electromotive force we have to

\epsilon = -\frac{d\Phi}{dt}

Replacing we have

\epsilon = -A\frac{B}{dt}

Replacing with our values we have that

\epsilon = -(7.9*10^{-4}m)\frac{(3.5-0.5)}{0.1}

\epsilon = -0.0237V

Therefore the magnitude of the induced emf in the loop is 0.0237V

On the other hand we have that the current by Ohm's Law can be defined as

I = \frac{\epsilon}{R}

For the given value of the resistance and the previously found potential we have to

I = \frac{0.237}{1.3}

I= 0.0182A

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A brick with a mass of 20 kg and a weight of 196 N is released on a frictionless plane at an angle of 30°. What is the accelerat
Veronika [31]

Answer:4.9m/s^2

Basically answer A

Explanation:

6 0
3 years ago
Fellow student of mathematical bent tells you that the wave function of a traveling wave on a thin rope is y(x,t)= 2.20 mm cos[(
bezimeni [28]

Answer

given,

y(x,t)= 2.20 mm cos[( 7.02 rad/m )x+( 743 rad/s )t]

length of the rope = 1.33 m

mass of the rope = 3.31 g

comparing the given equation from the general wave equation

y(x,t)= A cos[k x+ω t]

A is amplitude

now on comparing

a) Amplitude  = 2.20 mm

b) frequency =

     f = \dfrac{\omega}{2\pi}

     f = \dfrac{743}{2\pi}

          f = 118.25 Hz

c) wavelength

        k= \dfrac{2\pi}{\lambda}

        \lambda= \dfrac{2\pi}{k}

        \lambda= \dfrac{2\pi}{7.02}

        \lambda= 0.895\ m

d) speed

         v = \dfrac{\omega}{k}

         v = \dfrac{743}{7.02}

                v = 105.84 m/s

e) direction of the motion will be in negative x-direction

f) tension

  T = \dfrac{v^2\ m}{L}

  T = \dfrac{(105.84)^2\times 3.31 \times 10^{-3}}{1.33}

      T = 27.87 N

g) Power transmitted by the wave

  P = \dfrac{1}{2}m\ v \omega^2\ A^2

  P = \dfrac{1}{2}\times 0.00331\times 105.84\times 743^2\ 0.0022^2

      P = 0.438 W

5 0
3 years ago
In a closed energy system, energy is?​
Alisiya [41]

Answer:

Explanation:

A closed system can exchange energy but not matter, with its surroundings. An isolated system cannot exchange any heat, work, or matter with the surroundings, while an open system can exchange energy and matter.

Hope this helped you!

3 0
2 years ago
Example 3 :
kherson [118]

The friction factor and head loss when velocity is 1m/s is 0.289 and 1.80 × 10^8 respectively. Also, the friction factor and head loss when velocity is 3m/s is 0.096 and 5.3 × 10^8 respectively.

<h3>How to determine the friction factor</h3>

Using the formula

μ = viscosity = 0. 06 Pas

d =  diameter = 120mm = 0. 12m

V =  velocity = 1m/s and 3m/s

ρ = density = 0.9

a. Velocity = 1m/s

friction factor = 0. 52 × \frac{0. 06}{0. 12* 1* 0. 9}

friction factor = 0. 52 × \frac{0. 06}{0. 108}

friction factor = 0. 52 × 0. 55

friction factor = 0. 289

b. When V = 3mls

Friction factor = 0. 52 × \frac{0. 06}{0. 12 * 3* 0. 9}

Friction factor = 0. 52 × \frac{0. 06}{0. 324}

Friction factor = 0. 52 × 0. 185

Friction factor = 0.096

Loss When V = 1m/s

Head loss/ length = friction factor × 1/ 2g × velocity^2/ diameter

Head loss = 0. 289 × \frac{1}{2*6. 6743 * 10^-11} × \frac{1^2}{0. 120} × \frac{1}{100}

Head loss =  1. 80 × 10^8

Head loss When V = 3m/s

Head loss = 0. 096 × \frac{1}{1. 334 *10^-10} × \frac{3^2}{0. 120} × \frac{1}{100}

Head loss = 5. 3× 10^8

Thus, the friction factor and head loss when velocity is 1m/s is 0.289 and 1.80 ×10^8 respectively also, the friction factor and head loss  when velocity is 3m/s is 0.096 and 5.3 ×10^8 respectively.

Learn more about friction here:

brainly.com/question/24338873

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4 0
1 year ago
An egg is thrown nearly vertically upward from a point near the cornice of a tall building. It just misses the cornice on the wa
Lyrx [107]

Answer:

A) 17.7 m/s

B) 15.98 m

C) Zero

E) 9.8 m/s²

Explanation:

given information

distance, h = - 34 m

time, t = 5 s

A) What is the initial speed of the egg?

h - h₀ = v₀t - \frac{1}{2} gt², h₀ = 0

- 34 = v₀ 5 - \frac{1}{2} 9.8 5²

- 34 = 5 v₀ - 122.5

v₀ = 122.5 - 34/5

    = 17.7 m/s

B) How high does it rise above its starting point?

v² = v₀² - 2gh

v = 0 (highest point)

2gh = v₀²

h = v₀²/2g

  = 17.7²/2 (9.8)

  = 15.98 m

C) What is the magnitude of its velocity at the highest point?

v = 0 (at highest point)

E) What are the magnitude and direction of its acceleration at the highest point?

g= 9.8 m/s², since the egg is moved vertically, the acceleration is the same as the gravitational acceleration.

7 0
3 years ago
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