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3 years ago

A ______ is a tiny packet of energy

Physics

1 answer:

Nataliya [291]3 years ago

5 0

<span>Individual packets or quanta of light energy are called Photons.
The answer is photons.
</span>

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What formed the lunar maria?

Greeley [361]

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3 years ago

What is the equation for horizontal velocity

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3 years ago

The clothes washer in your house consumes 470 kWh of energy per year. Price of the washer is $360 and the lifetime of the washer

VashaNatasha [74]

Answer:

$893

Explanation: the complete question should be

The clothes washer in your house consumes 470 kWh of energy per year. Price of the washer is $360 and the lifetime of the washer is 10 yrs. Energy price in your city is 9 cents per kWh. What is the lifecycle cost of the clothes washer? (assume a maintenance cost of $11 per year)

SOLUTION

Given:

The clothes washe power consumption (PC) is 470 kWh

Price of the washer (P) is $360

lifetime of the washer (L) is 10 yrs

Energy price in the city (E) is 9 cents per kWh (Covert to $ by dividing 100)

maintenance cost (M) is $11 per year

Lifecycle cost = P + (PC × L × E) +M + L

Lifecycle cost = $360 + (470kWh × 10years × 9cents/100) + ($11 × 10years)

=$893

7 0

3 years ago

A 4 cm diameter "bobber" with a mass of 3 grams floats on a pond. A thin, light fishing line is tied to the bottom of the bobber

Tasya [4]

Answer:

Explanation:

Calculate the volume of the lead

$V=\frac{m}{d}\\\\=\frac{10g}{11.3g'cm^3}$

Now calculate the bouyant force acting on the lead

$F_L = Vpg$

$F_L=(\frac{10g}{11.3g/cm^3} )(1g/cm^3)(9.8m/s^2)\\\\=8.673\times 10^{-3}N$

This force will act in upward direction

Gravitational force on the lead due to its mass will act in downward direction

Hence the difference of this two force

$T=mg-F_L\\\\=(10\times10^{-3}kg(9.8m/s^2)-8.673\times 10^{-3}\\\\=8.933\times10^{-3}N$

If V is the volume submerged in the water then bouyant force on the bobber is

$F_B=V'pg$

Equate bouyant force with the tension and gravitational force

$F_B=T_mg\\\\V'pg=\frac{(8.933\times10^{-2}N)+mg}{pg} \\\\V'=\frac{(8.933\times10^{-2}N)+mg}{pg}$

Now Total volume of bobble is

$\frac{V'}{V^B} =\frac{\frac{(8.933\times10^{-2})+Mg}{pg} }{\frac{4}{3} \pi R^3 }\times100\\\\=\frac{\frac{(8.933\times10^{-2})+(3)(9.8)}{(1000)(9.8)} }{\frac{4}{3} \pi (4.0\times10^{-2})^3 }\times100\\\\$

= $\large\boxed{4.52 \%}$

7 0

3 years ago