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3 years ago

If 478 watts of power are used in 14 seconds,how much work was done

Physics

1 answer:

zepelin [54]3 years ago

6 0

Answer:

6692J

Explanation:

Power is defined as the rate at which work is being done.

So,

Power = $\frac{workdone}{time }$

Work done = Power x time

Given parameters:

Power = 478watts

Time = 14s

So;

Work done = 478 x 14 = 6692J

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3 years ago

What is specific heat? List the formula

sladkih [1.3K]

An example of a high specific heat is water’s specific heat, which requires 4.184 joules of heat to increase the temperature of 1 gram of water 1 degree Celsius. Scientifically, water’s specific heat is written as: 1 calorie/gm °C = 4.186 J/gm °C.

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3 years ago

7.22 Ignoring reflection at the air–water boundary, if the amplitude of a 1 GHz incident wave in air is 20 V/m at the water surf

Serga [27]

Answer:

z = 0.8 (approx)

Explanation:

given,

Amplitude of 1 GHz incident wave in air = 20 V/m

Water has,

μr = 1

at 1 GHz, r = 80 and σ = 1 S/m.

depth of water when amplitude is down to 1 μV/m

Intrinsic impedance of air = 120 π Ω

Intrinsic impedance of water = $\dfrac{120\pi}{\epsilon_r}$

Using equation to solve the problem

$E(z) = E_0 e^{-\alpha\ z}$

E(z) is the amplitude under water at z depth

E_o is the amplitude of wave on the surface of water

z is the depth under water

$\alpha = \dfrac{\sigma}{2}\sqrt{\dfrac{(120\pi)^2}{\Epsilon_r}}$

$\alpha = \dfrac{1}{2}\sqrt{\dfrac{(120\pi)^2}{80}}$

$\alpha =21.07\ Np/m$

now ,

$1 \times 10^{-6} = 20 e^{-21.07\times z}$

$e^{21.07\times z}= 20\times 10^{6}$

taking ln both side

21.07 x z = 16.81

z = 0.797

z = 0.8 (approx)

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3 years ago

What kind of weather is signaled by cumulus clouds

Novay_Z [31]

Cumulonimbus clouds are associated by severe weather. They are thunderstorm clouds that bring heavy rain, snow, hail, lightning and even tornadoes.

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3 years ago

an athlete whirls an 8.71 kg hammer tied to the end of a 1.5 m chain in a simple horizontal circle where you should ignore any v

-BARSIC- [3]

Answer:

T = 692.42 N

Explanation:

Given that,

Mass of hammer, m = 8.71 kg

Length of the chain to which an athlete whirls the hammer, r = 1.5 m

The angular sped of the hammer, $\omega=1.16\ rev/s=7.28\ rad/s$

We need to find the tension in the chain. The tension acting in the chain is balanced by the required centripetal force. It is given by the formula as follows :

$F=m\omega^2r\\\\=8.71\times (7.28)^2\times 1.5\\\\=692.42\ N$

So, the tension in the chain is 692.42 N.

5 0

3 years ago