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3 years ago

What is the power of an electric with a current of 0.5 A and a voltage of 120 V ? The formula for power is P =IV

Physics

2 answers:

erica [24]3 years ago

8 0

Answer:

60 W

Explanation: I took the quiz

zaharov [31]3 years ago

7 0

Answer:

60Watts

Explanation:

Given parameters:

Current = 0.5A

Voltage = 120V

Unknown

Power = ?

Solution:

The power in the electric circuit is the product of current and voltage;

P = IV

Insert the given parameters and solve;

P = 0.5 x 120

P = 60Watts

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Vinil7 [7]

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Explanation:

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3 years ago

Europa, a satellite of Jupiter, is believed to have a liquid ocean of water (with a possibility of life) beneath its icy surface

Goryan [66]

Answer:

so maximum velocity for walk on the surface of europa is 0.950999 m/s

Explanation:

Given data

legs of length r = 0.68 m

diameter = 3100 km

mass = 4.8×10^22 kg

to find out

maximum velocity for walk on the surface of europa

solution

first we calculate radius that is

radius = d/2 = 3100 /2 = 1550 km

radius = 1550 × 10³ m

so we calculate no maximum velocity that is

max velocity = √(gr) ...............1

here r is length of leg

we know g = GM/r² from universal gravitational law

so G we know 6.67 × $10^{-11}$ N-m²/kg²

g = 6.67 × $10^{-11}$ ( 4.8×10^22 ) / ( 1550 × 10³ )

g = 1.33 m/s²

now

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max velocity = √(1.33 × 0.68)

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3 years ago

A fan at a rock concert is 50.0 m from the stage, and at this point the sound intensity level is 114 dB. Sound is detected when

Marianna [84]

Answer:

A) $P=13.92\ J.s^{-1}$

B) $v=3730.9912\ m.s^{-1}$

C) $v=74.44\ mm.s^{-1}$

D) mosquitoes speed in part B is very much larger than that of part C.

Explanation:

Given:

Distance form the sound source, $s=50\ m$

sound intensity level at the given location, $\beta=114\ dB$

diameter of the eardrum membrane in humans, $d=8.4 \times 10^{-3}\ m$

We have the minimum detectable intensity to the human ears, $I_0=10^{-12}\ W.m^{-2}$

(A)

Now the intensity of the sound at the given location is related mathematically as:

$\beta=10\ log(\frac{I}{I_0} )$ ..........................................(1)

$114=10\ log\ (\frac{I}{10^{-12}} )$

$11.4=log\ I+12\ log\ 10$

$I=0.2512\ W.m^{-2}$

As we know :

$I=\frac{P}{A}$

$0.2512=\frac{P}{\pi\times \frac{8.4^2}{4} }$

$P=13.92\ J.s^{-1}$ is the energy transferred to the eardrums per second.

(B)

mass of mosquito, $m=2\times 10^{-6}\ kg$

Now the velocity of mosquito for the same kinetic energy:

$KE=\frac{1}{2} m.v^2$

$13.92=\frac{1}{2}\times 2\times 10^{-6}\times v^2$

$v=3730.9912\ m.s^{-1}$

(C)

Given:

Sound intensity, $\beta = 20\ dB$

Using eq. (1)

$20=10\ log\ (\frac{I}{10^{-12}} )$

$2=log\ I+12\ log\ 10$

$I=10^{-10}\ W.m^{-2}$

Now, power:

$P=I.A$

$P=10^{-10}\times \pi\times \frac{8.4^2}{4}$

$P=5.54\times 10^{-9}\ J.s^{-1}$

Hence:

$KE=\frac{1}{2} m.v^2$

$5.54\times 10^{-9}=0.5\times 2\times 10^{-6}\times v^2$

$v=0.07444\ m.s^{-1}$

$v=74.44\ mm.s^{-1}$

(D)

mosquitoes speed in part B is very much larger than that of part C.

7 0

3 years ago