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3 years ago

an alarm clock draws 0.5 a of current when connected to a 120 volt circuit. calculate its resistance.

Physics

1 answer:

Digiron [165]3 years ago

7 0

Resistance = (potential difference) / (current) = 120V / 0.5A =  240 ohms .

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Using the image provided, which moon phase is next?

ella [17]

Answer:

Waning Crescent

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2 years ago

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Which players are usually the tallest on their team, and stay close to the basket so they can shoot and rebound the ball?

alexdok [17]

Answer:

Center

Explanation:

The center is the tallest player on each team, playing near the basket. On offense, the center tries to score on close shots and rebound. But on defense, the center tries to block opponents' shots and rebound their misses.

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2 years ago

An object of mass 3.4 kg is moving in a straight line with kinetic energy 59.177 J. A force is applied in the direction of its m

rusak2 [61]

Answer:

Its momentum is multiplied by a factor of 1.25

Explanation:

First, we calculate the initial velocity of the object:

K = 0.5 * m * v₁²

59.177 J = 0.5 * 3.4 kg * v₁²

v₁ = 5.9 m/s

With that velocity we can calculate the initial momentum of the object:

p₁ = v₁ * m

p₁ = 20.06 kg·m/s

Then we calculate the velocity of the object once its kinetic energy has increased:

(59.177 J) * 1.57 = 0.5 * 3.4 kg * v₂²

v₂ = 7.4 m/s

And calculate the second momentum of the object:

p₂ = v₂ * m

p₂ = 25.16 kg·m/s

Finally we calculate the factor:

p₂/p₁ = 1.25

3 0

2 years ago

A house of mass 500 kg is lifted 25 m by balloons.

Evgen [1.6K]

Gpe = mgh
Gpe = 500*10*25
Gpe = 125000J

7 0

2 years ago

A cylindrical conductor with a circular cross section has a radius a and a resistivity p and carries a constant current I. (Take

Lady bird [3.3K]

Answer:

Explanation:

Using Ohms Law

$R= \rho \frac{l}{\pi a^2 }
\\V = IR = E l = I \rho \frac{l}{\pi a^2 }
\\E = \frac{I \rho}{\pi a^2 }
\\E = J \rho
$

Where J is the current density $J = \frac{I}{\pi a^2 }
$

and the direction of E is the same as the direction of the current. Since J is uniform throughout the conductor $E = \rhoJ$ just inside at a radius a (and anywhere else).

Since we have no changing electric fields we can use Ampere’s law in it’s simplest form without displacement current

$\oint B .dl = B 2 \pi a = \mu_{o} I
$

such that

$B = \frac{\mu_{o} I}{2 \pi a }
$

and by the right hand rule, since the current is going to the right, the magnetic field is circling around the conductor such that it’s pointing out of the page at the top and into the page at the bottom.

The Poynting vector is given by

$S = \frac{1}{ \mu_o} |E \textrm{x}B| = \frac{\rho I^2}{2 \pi^2 a^3}
$

and by the right hand rule it’s always pointing in towards the center of the conductor.

Note: directions of these three vectors are mentioned along with their magnitudes in above 3 parts a , b and c

7 0

2 years ago