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3 years ago

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A ten loop coil of area 0.23 m2 is in a 0.047 T uniform magnetic field oriented so that the maximum flux goes through the coil.

The coil is then rotated so that the flux through it goes to zero in 0.34 s. The average emf induced in the coil during the 0.34 s is:

1 answer:

Nana76 [90]3 years ago

6 0

Answer:

0.32 V

Explanation:

N = 10, A = 0.23 m^2, B = 0.47 T, t = 0.34 s

The average induced emf is given by

e = - N dФ / dt

Where, dФ be the change in magnetic flux in time dt.

dФ / dt = d / dt (B A) = A dB/dt

So,

e = - 10 x 0.23 x 0.047 / 0.34 = - 0.32 V

The negative sign shows the direction of induced emf.

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What is the net torque on the square plate, with sides 0.2 m, from each of the three forces? F1=18 N, F2=26 N, and F3=14 N. Use

marshall27 [118]

Answer:

The net torque on the square plate is 2.72 N-m.

Explanation:

Given that,

Side = 0.2 m

Force $F_{1}=18\ N$

Force $F_{2}=26\ N$

Force $F_{3}=14\ N$

We need to calculate the torque due to force F₁

Using formula of torque

$\tau_{1}=-F_{1}d_{1}$

$\tau_{1}=-F_{1}\times\dfrac{a}{2}$

Put the value into the formula

$\tau_{1}=-18\times\dfrac{0.2}{2}$

$\tau_{1}=-1.8\ N-m$

We need to calculate the torque due to force F₂

Using formula of torque

$\tau_{2}=F_{2}d_{2}$

$\tau_{2}=F_{2}\times\dfrac{a}{2}$

Put the value into the formula

$\tau_{2}=26\times\dfrac{0.2}{2}$

$\tau_{2}=2.6\ N-m$

We need to calculate the torque due to force F₃

Using formula of torque

$\tau_{3}=F_{3}d_{3}$

$\tau_{3}=(F_{3}\sin45+F_{3}\cos45)\times\dfrac{a}{2}$

Put the value into the formula

$\tau_{3}=0.1(14\sin45+14\cos45)$

$\tau_{3}=1.92\ N-m$

We need to calculate the net torque on the square plate

$\tau=\tau_{1}+\tau_{2}+\tau_{3}$

$\tau=-1.8+2.6+1.92$

$\tau=2.72\ N-m$

Hence, The net torque on the square plate is 2.72 N-m.

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Leviafan [203]

Answer:x=23.4 cm

Explanation:

Given

mass of block $m=0.5 kg$

inclination $\theta =30$

coefficient of static friction $\mu =0.35$

coefficient of kinetic friction $\mu _k=0.25$

distance traveled $d=77.3 cm$

spring constant $k=35 N/m$

work done by gravity+work done by friction=Energy stored in Spring

$mg\sin \theta d-\mu _kmg\cos \theta d=\frac{kx^2}{2}$

$mgd\left ( \sin \theta -\mu _k\cos \theta \right )=\frac{kx^2}{2}$

$0.5\times 9.8\times 0.773\left ( \sin 30-0.25\cos 30\right )=\frac{35\times x^2}{2}$

$x=\sqrt{\frac{2\times 0.5\times 9.8\times 0.773(\sin 30-0.25\times \cos 30)}{35}}$

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3 years ago

A 1.15 kg book is at rest on the table. What is the magnitude of the normal force that the table is exerting on the book?

Agata [3.3K]

Answer:

11.27N

Explanation:

Given parameters:

Mass of the book = 1.15kg

Unknown:

Magnitude of the normal force = ?

Solution:

The normal force is the vertical force exerted by a body on an object.

It can be described as the weight of an object.

Normal force = mass x acceleration due to gravity

Normal force = 1.15 x 9.8 = 11.27N

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Novosadov [1.4K]

Answer:

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Explanation:

Capacitance is directly proportional to dielectric constant

Aluminium and zinc are highly reactive and have high dielectric contact.

Copper has less dielectric constant hence capacitance will decrease

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