in which phase change does a substance change from a solid to a gas vapor without changing to a liquid first?

The efficiency of a device such as a lamp can be calculated using this equation:

loris [4]

efficiency = (useful energy transferred ÷ energy supplied) × 100

It's easy to use this formula, but we have to know both the useful energy and the energy supplied. The drawing doesn't tell us the useful energy, so we have to find a clever way to figure it out. I see two ways to do it:

Way #1:

We all know about the law of conservation of energy. So we know that the total energy coming out must be 250J, because that's how much energy is going in. The wasted energy is 75J, so the rest of the 250J must be the useful energy . . . (250J - 75J) = 175J useful energy.

(useful energy) / (energy supplied) = (175J) / (250J) = 70% efficiency

================================

Way #2:

How much of the energy is wasted ? . . . 75J wasted

What percentage of the Input is that 75J ? . . . 75/250 = 30% wasted

30% of the input energy is wasted. That leaves the other 70% to be useful energy.

6 0

2 years ago

Read 2 more answers

A material that transmits nearly all the light in a ray because it offers little resistance to the light is ________.

Nina [5.8K]

Answer:

D. transparent.

Explanation:

A material that transmits nearly all the light in a ray because it offers little resistance to the light is transparent.

A transparent material allows light to pass through them with little or no resistance enabling them see-through. A material that transmits nearly all the light rays that pass through it because it offers little resistance to the light is TRANSPARENT. Examples of transparent materials are water, glass (flint and crown), air, and diamond.

4 0

2 years ago

A hockey stick strikes a hockey puck of mass 0.17 kg. If the force exterted on the hockey puck is 35.0 N and there is a force of

GREYUIT [131]

Answer:

$a=190\ m/s^2$

Explanation:

Mass of a hockey puck, m = 0.17 kg

Force exerted by the hockey puck, F' = 35 N

The force of friction, f = 2.7 N

We need to find the acceleration of the hockey puck.

Net force, F=F'-f

F=35-2.7

F=32.3 N

Now, using second law of motion,

F = ma

a is the acceleration of the hockey puck

$a=\dfrac{F}{m}\\\\a=\dfrac{32.3}{0.17}\\\\a=190\ m/s^2$

So, the acceleration of the hockey puck is $190\ m/s^2$ .

5 0

2 years ago

If the wavelength of an s-wave is 23,000 m, and its speed 4500 m/s, what is its frequency?

wolverine [178]

Answer:

5 metre.

Explanation:

Wavelength = Velocity / Frequency

= 23,000/ 4,500

= 5 metre.

3 0

2 years ago

The engine starter and a headlight of a car are connected in parallel to the 12.0-V car battery. In this situation, the headligh

stepladder [879]

Answer:

The total power they will consume in series is approximately 2.257 W

Explanation:

The connection arrangement of the headlight and the engine starter = Parallel to the battery

The voltage of the battery, V = 12.0 V

The power at which the headlight operates in parallel, $P_{headlight}$ = 38 W

The power at which the kick starter operates in parallel, $P_{kick \ starter}$ = 2.40 kW

We have;

P = V²/R

Where;

R = The resistance

V = The voltage = 12 V (The voltage is the same in parallel circuit)

For the headlight, we have;

R₁ = V²/ $P_{headlight}$ = 12²/38 = 72/19

R₁ = 72/19 Ω

For the kick starter, we have;

R₂ = V²/ $P_{kick \ starter}$ = 12²/2.4 = 60

R₂ = 60 Ω

When the headlight and kick starter are rewired to be in series, we have;

Total resistance, R = R₁ + R₂

Therefore;

R = ((72/19) + 60) Ω = (1212/19) Ω

The current flowing, I = V/R

∴ I = 12 V/(1212/19) Ω = (19/101) A

We note that power, P = I²R

In the series connection, we have;

$P_{headlight}$ = I² × R₁

∴ $P_{headlight}$ = ((19/101) A)² × 72/19 Ω = 1368/10201 W ≈ 0.134 W

The power at which the headlight operates in series, $P_{headlight, S}$ ≈ 0.134 W

$P_{kick \ starter}$ = ((19/101) A)² × 60 Ω = 21660/10201 W ≈ 2.123 W

The power at which the kick starter operates in series, $P_{kick \ starter, S}$ ≈ 2.123 W

The total power they will consume, $P_{Total}$ = $P_{headlight, S}$ + $P_{kick \ starter, S}$

Therefore;

$P_{Total}$ ≈ 0.134 W + 2.123 W = 2.257 W

4 0

2 years ago