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3 years ago

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

Physics

2 answers:

loris [4]3 years ago

6 0

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.

jek_recluse [69]3 years ago

3 0

Answer:

70%

Explanation:

I did the USA Test Prep.

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A point source at the origin emits sound of frequency 175 Hz uniformly in all directions. On the x-axis at x=100 m, the sound in

hammer [34]

Answer:

Multiple answers:

1. Power output P=17.59W

2.Intensity 160m I=17.6W/ $m^{2}$

3. dB = 77.3

4. f=178.5 Hz

Explanation:

First one comes from the expression

$I=\frac{P}{4\pi r^{2} }$

where I is the intensity, P is the power and r is the radio of the spherical wave, or in this case, the distance x. I solved for the Power by multiplying Intensity with the area (4 $\pi x^{2}$

Second one is done with:

$\frac{I_{2} }{I_{1} } =\frac{x^{2}_{1} }{x^{2} _{2}}$

Solving for Intensity 2, the result mentioned.

The third is simply computed with

$dB=10*log\frac{I}{10^{-12} }$

And finally the last one is done with doppler effect, taking into account the speed of the air as in 10ºC 337m/s.

$f=f_{initial} *(\frac{s+v_{receiver} }{s+v_{source} } )$

Where Finitial is the frequency emitted and s is the speed of the sound. The wind blowing in positive is, in principle, going away of the observer.

8 0

3 years ago

A block with a mass of 31.8 kg is pushed on a frictionless

OlgaM077 [116]

Answer: Total work done on the block is 3670.5 Joules.

Step by step:

Work done:

$W = F.d.\cos\theta$

With F the force, d the displacement, and theta the angle of action (which is 0 since the block is pushed along the direction of displacement, and cos 0 = 1)

$W = F.d$

Given:

F = 75 N

m = 31.8 kg

Final velocity $v_f = 15.3 \frac{m}{s}$

In order to calculate the Work we need to determine the displacement, or distance the block travels. We can use the information about F and m to first figure out the acceleration:

$F = ma\\\implies a=\frac{F}{m}=\frac{75 N }{31.8 kg}\approx 2.36 \frac{m}{s^2}\\$

Now we can determine the displacement from the following formula:

$d = \frac{1}{2}a^2+v_0t+d_0$

Here, the initial displacement is 0 and initial velocity is also 0 (at rest):

$d = \frac{1}{2}at^2\\$

Now we still have "t" as unknown. But we are given one more bit of information from which this can be determined:

$v_f = a\cdot t_f\\\implies t_f = \frac{v_f}{a} = \frac{15.2 \frac{m}{s}}{2.36 \frac{m}{s^2}}\approx 6.44 s$

(using vf as final velocity, and tf as final time)

So it takes about 6.44 seconds for the block to move. This allows us to finally calculate the displacement:

$d = \frac{1}{2}at^2=\frac{1}{2}2.36 \frac{m}{s^2}\cdot 6.44^2 s^2 \approx 48.94 m$

and the corresponding work:

$W = F\cdot d=75 N\cdot 48.94 m =3670.5J$

7 0

3 years ago

A sample of an ideal gas has a volume of 2.37 L at 2.80×102 K and 1.15 atm. Calculate the pressure when the volume is 1.68 L and

iogann1982 [59]

Answer:

$p_2 = 1.76 atm$

Explanation:

given data:

v_1 = 2.37 L

v_2 = 1.68 L

p_1 =1.15 atm

p_2 = ?

t_1 = 280 K

t_2 = 304 K

from Gas Law Equation

, WE HAVE

$\frac{p_1 v_1}{t_1} =\frac{p_2 v_2}{t_2}$

Putting the values

$\frac{1.15*2.37}{280} =\frac{p_2 *1.68}{304}$

$9.733*10^{-3} = \frac{p_2 *1.68}{304}$

$9.733*10^{-3}*304 = p_2*1.68$

$\frac{9.733*10^{-3}*304}{1.68} =p_2$

$p_2= 1.76 atm$

7 0

4 years ago

An automobile traveling 95 km/h overtakes a 1.30-km-long train traveling in the same direction on a track parallel to the road.

SCORPION-xisa [38]

Answer: 6.175 km

Explanation:

from the question, we have the following

velocity of the automobile = 95 km/g

velocity of the train = 75 km/h

length of the train = 1.30 km

since the automobile and the train are moving in the same direction, we need to find the velocity of the car relative to the train which will be their difference in speed = 95 - 75 = 20 km/h

we need to find the time it takes the automobile to overtake the train using the formula time = distance / speed , with the distance being the length of the train.

time (t) = 1.3 / 20

= 0.065 hour

now we can find the distance traveled by the automobile using the the time taken for it to overtake the train and the speed of the automobile.

therefore, distance = speed x time

distance = 95 x 0.065 =6.175 km

5 0

4 years ago

The siren on an ambulance is emitting a sound whose frequency is2550 Hz. The speed of sound is 343 m/s.(a) If the ambulance is s

Sati [7]

Answer:

0.13451 m

0.124196 m