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evablogger [386]

3 years ago

5

A classroom has 24 fluorescent bulbs, each of which is 32 W. how much energy does it take to light the room for a minute?(unit=J

)PLEASE HELP

1 answer:

Alina [70]3 years ago

5 0

The energy prdoduced by the 24 bulbs is 46,080 J

Explanation:

Power is defined as the ratio between the energy produced and the time taken:

$P=\frac{E}{t}$

where

E is the energy

t is the time

In this problem, the power of each bulb is

P = 32 W

Therefore, in a time of

t = 1 min = 60 s

The energy produced by each bulb is

$E=Pt=(32)(60)=1920 J$

Here we have 24 bulbs, so the total energy produced is

$E=24(1920)=46,080 J$

Learn more about power and energy here:

brainly.com/question/7956557

#LearnwithBrainly

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At what time of day would you be most likely to find that the air over water is significantly warmer than the air over land near

Brut [27]

This would happen later at night or early in the morning.

The reason being land becomes warm and cold quicker than the water because of the heat capacity. So during the day water warms up because of sunlight but at night the land becomes a lot cooler as compared to the water which is still war. So the air over water is significantly warmer than the air over land.

4 0

4 years ago

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What personal experiences have you had with watching a substance change from one phase to another?

Dmitry_Shevchenko [17]

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3 years ago

Wave traveling at 330 m/sec has a wavelength of 4.3 meters. What is the frequency of this wave?

Rasek [7]

Answer:

76.74 Hz

Explanation:

Given:

Wave velocity ( v ) = 330 m / sec

wavelength ( λ ) = 4.3 m

We have to calculate Frequency ( f ):

We know:

v = λ / t [ f = 1 / t ]

v = λ f

= > f = v / λ

Putting values here we get:

= > f = 330 / 4.3 Hz

= > f = 3300 / 43 Hz

= > f = 76.74 Hz

Hence, frequency of sound is 76.74 Hz.

6 0

3 years ago

A small meteorite with mass of 1 g strikes the outer wall of a communication satellite with a speed of 2Okm/s (relative to the s

strojnjashka [21]

Answer:

The energy coverted to heat is 200 kilojoules.

Explanation:

GIven the absence of external forces exerted both on the small meteorite and on the communication satellite, the Principle of Linear Momentum is considered and let suppose that collision is completely inelastic and that satellite is initially at rest. Hence, the expression for the satellite-meteorite system:

$m_{M}\cdot v_{M} + m_{S}\cdot v_{S} = (m_{M}+m_{S})\cdot v$

Where:

$m_{M}$ , $m_{S}$ - Masses of the small meteorite and the communication satellite, measured in kilograms.

$v_{M}$ , $v_{S}$ - Speeds of the small meteorite and the communication satellite, measured in meters per second.

$v$ - Final speed of the satellite-meteorite system, measured in meters per second.

The final speed of the satellite-meteorite system is cleared:

$v = \frac{m_{M}\cdot v_{M}+m_{S}\cdot v_{S}}{m_{M}+m_{S}}$

If $m_{M} = 1\times 10^{-3}\,kg$ , $m_{S} = 200\,kg$ , $v_{M} = 20000\,\frac{m}{s}$ and $v_{S} = 0\,\frac{m}{s}$ , the final speed is now calculated:

$v = \frac{(1\times 10^{-3}\,kg)\cdot \left(20000\,\frac{m}{s} \right)+(200\,kg)\cdot \left(0\,\frac{m}{s} \right)}{1\times 10^{-3}\,kg+200\,kg}$

$v = 0.1\,\frac{m}{s}$

Which means that the new system remains stationary and all mechanical energy from meteorite is dissipated in the form of heat. According to the Principle of Energy Conservation and the Work-Energy Theorem, the change in the kinetic energy is equal to the dissipated energy in the form of heat:

$K_{S} + K_{M} - K - Q_{disp} = 0$

$Q_{disp} = K_{S}+K_{M}-K$

Where:

$K_{S}$ , $K_{M}$ - Initial translational kinetic energies of the communication satellite and small meteorite, measured in joules.

$K$ - Kinetic energy of the satellite-meteorite system, measured in joules.

$Q_{disp}$ - Dissipated heat, measured in joules.

The previous expression is expanded by using the definition for the translational kinetic energy:

$Q_{disp} = \frac{1}{2}\cdot [m_{M}\cdot v_{M}^{2}+m_{S}\cdot v_{S}^{2}-(m_{M}+m_{S})\cdot v^{2}]$

Given that $m_{M} = 1\times 10^{-3}\,kg$ , $m_{S} = 200\,kg$ , $v_{M} = 20000\,\frac{m}{s}$ , $v_{S} = 0\,\frac{m}{s}$ and $v = 0.1\,\frac{m}{s}$ , the dissipated heat is:

$Q_{disp} = \frac{1}{2}\cdot \left[(1\times 10^{-3}\,kg)\cdot \left(20000\,\frac{m}{s} \right)^{2}+(200\,kg)\cdot \left(0\,\frac{m}{s} \right)^{2}-(200.001\,kg)\cdot \left(0.001\,\frac{m}{s} \right)^{2}\right]$ $Q_{disp} = 200000\,J$

$Q_{disp} = 200\,kJ$

The energy coverted to heat is 200 kilojoules.

4 0

3 years ago

Zolol [24]

Answer:

48

Explanation:

you basically divide 1200 into 25

8 0

4 years ago