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WINSTONCH [101]

3 years ago

14

A 0.43 m long and 0.43 m wide loop is moved at a constant velocity of 0.15 m/s into a perpendicular constant magnetic field of 0

.31 T. Calculate the magnitude of the induced voltage in the loop.

1 answer:

olya-2409 [2.1K]3 years ago

7 0

Answer:

The magnitude of the induced voltage in the loop is 20 mV.

Explanation:

given;

length of loop, L = 0.43 m

width of loop,w = 0.43 m

velocity of moved loop, v = 0.15m/s

magnetic field strength,B = 0.31 T

To determine the magnitude of the induced voltage in the loop, we apply Faraday's law;

magnitude induced E.M.F = BLv

magnitude induced E.M.F = 0.31 x 0.43 x 0.15 = 0.02 V = 20 mV

Therefore, the magnitude of the induced voltage in the loop is 20 mV.

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How do intermolecular forces differ from intramolecular forces

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Answer:

Explanation:

Intramolecular forces is a strong bond that helps to bond atoms together while intermolecular forces are weak bond that are present between molecules.

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Determine the acceleration due to gravity for low Earth orbit (LEO) given: MEarth = 6.00 x 1024 kg, rEarth = 6.40 x 106 m, G = 6

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Answer:

The answer to the question is as follows

The acceleration due to gravity for low for orbit is 9.231 m/s²

Explanation:

The gravitational force is given as

$F_{G}= \frac{Gm_{1} m_{2}}{r^{2} }$

Where $F_{G}$ = Gravitational force

G = Gravitational constant = 6.67×10⁻¹¹ $\frac{Nm^{2} }{kg^{2} }$

m₁ = mEarth = mass of Earth = 6×10²⁴ kg

m₂ = The other mass which is acted upon by $F_{G}$ and = 1 kg

rEarth = The distance between the two masses = 6.40 x 10⁶ m

therefore at a height of 400 km above the erth we have

r = 400 + rEarth = 400 + 6.40 x 10⁶ m = 6.80 x 10⁶ m

and $F_{G}$ = $\frac{6.67*10^{-11} *6.40*10^{24} *1}{(6.8*10^{6})^{2} }$ = 9.231 N

Therefore the acceleration due to gravity = $F_{G}$ /mass

9.231/1 or 9.231 m/s²

Therefore the acceleration due to gravity at 400 kn above the Earth's surface is 9.231 m/s²

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3 years ago

Read 2 more answers

Does current in Helmholtz coils flow in the same or opposite direction through each coil? Explain.

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To get a uniform field in the central region between the coils, current flows in the same direction in each.

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The greatest ocean depths on the Earth are found in the Marianas Trench near the Philippines. Calculate the pressure (in atm) du

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Answer:

$P = 103867260$ atm

Explanation:

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Thus, $P = \rho*g*h$

Substituting the given values, we get -

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8 0

3 years ago

The flywheel of a steam engine runs with a constant angular velocity of 190 rev/min. When steam is shut off, the friction of the

Ivanshal [37]

Answer:

(a) $\alpha = - 1.32\ rev/m^{2}$

(b) $\theta = 13674\ rev$

(c) $\alpha_{tan} = 8.75\times 10^{- 4}\ m/s^{2}$

(d) $a = 22.458\ m/s^{2}$

Solution:

As per the question:

Angular velocity, $\omega = 190\ rev/min$

Time taken by the wheel to stop, t = 2.4 h = $2.4\times 60 = 144\ min$

Distance from the axis, R = 38 cm = 0.38 m

Now,

(a) To calculate the constant angular velocity, suing Kinematic eqn for rotational motion:

$\omega' = \omega + \alpha t$

$omega'$ = final angular velocity

$\omega$ = initial angular velocity

$\alpha$ = angular acceleration

Now,

$0 = 190 + \alpha \times 144$

$\alpha = - 1.32\ rev/m^{2}$

Now,

(b) The no. of revolutions is given by:

$\omega'^{2} = \omega^{2} + 2\alpha \theta$

$0 = 190^{2} + 2\times (- 1.32) \theta$

$\theta = 13674\ rev$

(c) Tangential component does not depend on instantaneous angular velocity but depends on radius and angular acceleration:

$\alpha_{tan} = 0.38\times 1.32\times \frac{2\pi}{3600} = 8.75\times 10^{- 4}\ m/s^{2}$

(d) The radial acceleration is given by:

$\alpha_{R} = R\omega^{2} = 0.32(80\times \frac{2\pi}{60})^{2} = 22.45\ rad/s$

Linear acceleration is given by:

$a = \sqrt{\alpha_{R}^{2} + \alpha_{tan}^{2}}$

$a = \sqrt{22.45^{2} + (8.75\times 10^{- 4})^{2}} = 22.458\ m/s^{2}$

5 0

3 years ago