All categories

3 years ago

When you turn on a flashlight, you are transforming ____ into radiant energy

Physics

2 answers:

Gnom [1K]3 years ago

4 0

You are transforming the light into radiant energy

lilavasa [31]3 years ago

3 0

Stored form of potential in the form of chemical energy into the radiant energy that's light

You might be interested in

A bullet of mass 0.107 kg traveling horizontally at a speed of 300 m/s embeds itself in a block of mass 3 kg that is sitting at

melisa1 [442]

Answer:

165.77J

Explanation:

M₁ = 0.107kg

u₁ = 300m/s

m₂ = 3kg

u₂ = 0

v =

m₁u₁ + m₂u₂ = (m₁ + m₂)V

(0.107*300) + 0 = (0.107 + 3)V

V = 32.1 / 3.107 = 10.33m/s

kinetic energy of the system after collision =

½m1v² + ½m2v²

K.E = ½(m1 + m2)v²

K.E = ½(0.107+3) * 10.33²

K.E = 165.77J

5 0

3 years ago

Estimate how many raindrops would fall on 1 acre plot during a 1 inch rainfall

Tom [10]

Answer:

1,580,088,782,700

Explanation:

There are 90,021 drops of water in one gallon, so multiplying all together. Roughly 1,580,088,782,700 drops of water will fall. Or about 1 1/2 trillion drops.

5 0

3 years ago

A heat pump is used to maintain a house at 22°C by extracting heat from the outside air on a day when the outside air temperatur

Ilia_Sergeevich [38]

Answer:

It is enough

Explanation:

To develop the problem it is necessary to take into account the concepts related to the coefficient of performance of a pump.

The two ways in which the performance coefficient can be expressed are given by:

$COP_p = \frac{T_H}{T_H-T_L}$

Where,

$T_H =$ High Temperature

$T_L =$ Low Temperature

And the other way is,

$COP_p = \frac{\dot{Q}}{W}$

Where $\dot{Q}$ is heat rate and W the power consumed.

We have all our terms in Celsius, so we calculate the temperature in Kelvin

$T_H = 22+273 = 295K$

$T_L = 2+273 = 275k$

The rate at which heat is lost is:

$\dot{Q} = 110000kJ/h$

The power consumed by the heat pump is

$\dot{W} = 5kW$

And the coefficient of performance is

$COP_p = \frac{T_H}{T_H-T_L}$

$COP_p = \frac{295}{295-275}$

$COP_p = 14.75$

With this value we can calculate the Power required,

$COP_p = \frac{\dot{Q}}{W}$

$14.75 = \frac{110000}{W}$

$W = \frac{110000}{3600*14.75}$

$W = 2.07kW$

<em>The power consumed is consumed is 5kW which is more than 2.07kW so this heat pump powerful enough.</em>

4 0

3 years ago

Why don’t we feel the gravitational pull of objects around us?

spayn [35]

It is because gravity is not an independent force. Gravity comes as result of warped space time according to the theory of general relativity. Therefore we can’t feel gravity because it’s not a force that pulls or pushes.

8 0

3 years ago

A 55 kg cheerleader uses an oil-filled hydraulic lift to hold four 110 kg football players at a height of 1.0 m. If her piston i

AVprozaik [17]

Answer:

$D = 55.2 cm$