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1 year ago

- A straight wire, 0.20 m long, moves at a constant speed of 7.0 m/s

Physics

1 answer:

Elis [28]1 year ago

7 0

(a) The emf induced in the wire is 0.112 V.

(b) The direction of the current will into the page.

(d) The current flowing through the wire is 0.224 A.

<h3>EMF induced in the wire</h3>

The emf induced in the wire is calculated as follows;

EMF = BLv

EMF = (8 x 10⁻²) x 0.2 x 7

EMF = 0.112 V

<h3>Direction of the current</h3>

Since, the magnetic field is out of the page, the current will follow counter clockwise and will point into the page.

<h3>Polarity of point A and point L</h3>

The current is flowing counter currently, hence the polarity of A will be positive and L will be negative.

<h3>Current through the wire</h3>

V = IR

I = V/R

I = (0.112)/0.5

I = 0.224 A

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If the current through a 20-Ω resistor is 8.0 A , how much energy is dissipated by the resistor in 1.0 h ? Express your answer w

marshall27 [118]

Answer:

$P(3600)=593.247W$

Explanation:

First, let's find the voltage through the resistor using ohm's law:

$V=IR=20*8=160V$

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$P(t)= V*I*cos(\phi)-V*I*cos(2 \omega t-\phi)$ (1)

Where:

$\phi=Phase\hspace{3}angle\\\omega= Angular\hspace{3}frequency$

Because of the problem doesn't give us additional information, let's assume:

$\phi=0\\\omega=2 \pi f=2*\pi *(60)=120\pi$

Evaluating the equation (1) in t=3600 (Because 1h equal to 3600s):

$P(3600)=160*8*cos(0)-160*8*cos(2*120\pi*3600-0)\\P(3600)=1280-1280*cos(2714336.053)\\P(3600)=1280-1280*0.5365255751\\P(3600)=1280-686.7527361=593.2472639\approx=593.247W$

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2 years ago

a scale model of the solar system where 50 cm represents 1.0x10 to the fifth km is actual distance what would be the dimension o

Fofino [41]

The distance between Mars and the Sun in the scale model would be 1140 m

Explanation:

In this scale model, we have:

$x_1 = 50 cm$ represents an actual distance of

$d_1 = 1.0\cdot 10^5 km$

The actual distance between Mars and the Sun is 228 million km, therefore

$d_2=228\cdot 10^6 km$

On the scale model, this would corresponds to a distance of $x_2$ .

Therefore, we can write the following proportion:

$\frac{x_1}{d_1}=\frac{x_2}{d_2}$

And solving for $x_2$ , we find:

$x_2=\frac{x_1 d_2}{d_1}=\frac{(50)(228\cdot 10^6)}{1\cdot 10^5}=1.14\cdot 10^5 cm = 1140 m$

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A 0.560 kg snowball is fired from a cliff 14.2 m high with an initial velocity of 13.3 m/s, directed 26.0° above the horizontal.

enot [183]

Answer:

a) v = 21.34 m/s

b) v = 21.34 m/s

c) v = 21.34 m/s

Explanation:

Mass of the snowball, m = 0.560 kg

Height of the cliff, h = 14.2 m

Initial velocity of the ball, u = 13.3 m/s

θ = 26°

The speed of the slow ball as it reaches the ground, v = ?

The initial Kinetic energy of the snow ball, $KE_{0} = 0.5 mu^{2}$

Potential energy of the snow ball at the given height, PE = mgh

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a) Using the principle of energy conservation,

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b) The speed remains v = 21.34 m/s since it is not a function of the angle of launch

c)The principle of energy conservation used cancels out the mass of the object, therefore the speed is not dependent on mass

v = 21. 34 m/s

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leonid [27]

Answer:

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