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2 years ago

How many calories are equal to one BTU? (One calorie = 4.186 J, one BTU = 1 054 J.)

Physics

1 answer:

I am Lyosha [343]2 years ago

3 0

<h2>Option C is the correct answer.</h2>

Explanation:

We need to find how many calories is 1 BTU.

Given

1 BTU = 1054 J

1 calorie = 4.186 J

So we have

1 BTU = 4.186 x 251.79 J

1 BTU =251.79 calorie

1 BTU = 252 calorie.

Option C is the correct answer.

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Calculate the minimum average power output necessary for a person to run up a 12.0 m long hillside, which is inclined at 25.0° a

Viktor [21]

Answer:

Power, P = 924.15 watts

Explanation:

Given that,

Length of the ramp, l = 12 m

Mass of the person, m = 55.8 kg

Angle between the inclined plane and the horizontal, $\theta=25^{\circ}$

Time, t = 3 s

Let h is the height of the hill from the horizontal,

$h=l\ sin\theta$

$h=12\times \ sin(25)$

h = 5.07 m

Let P is the power output necessary for a person to run up long hill side as :

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

$P=\dfrac{mgh}{t}$

$P=\dfrac{55.8\times 9.8\times 5.07}{3}$

P = 924.15 watts

So, the minimum average power output necessary for a person to run up is 924.15 watts. Hence, this is the required solution.

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2 years ago

Stu wanted to calculate the resistance of a light bulb connected to a 4.0 V battery, with a resulting current of 0.5 A. He used

Dimas [21]

His answer was incorrect because according to ohm's law the formula used should have been R=V/I instead of multiplying and the answer should be 8ohms

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2 years ago

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A spring with a spring constant of 2500 n/m. is stretched 4.00 cm. what is the work required to stretch the spring?

Yuri [45]

W = 1/2k*x^2.

k = spring constant = 2500 n/m.
x = distance = 4 cm = 0.04m (convert to same units).

W = 1/2(2500)(0.04)^2 = 2J.

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2 years ago

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Sonrisa owns a 300 W television. If the total energy usage for February is 32.4 kWh, how many hours per week does Sonrisa watch

Zielflug [23.3K]

the correct answer is 27 hours per week :) hope this helps

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2 years ago

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A 0.200-kg mass attached to the end of a spring causes it to stretch 5.0 cm. If another 0.200-kg mass is added to the spring, th

ziro4ka [17]

Answer:

A: 4 times as much

B: 200 N/m

C: 5000 N

D: 84,8 J

Explanation:

In the first question, we have to caculate the constant of the spring with this equation:

$m*g=k*x$

Getting the k:

$k=\frac{m*g}{x} =\frac{0,2[kg]*9,81[\frac{m}{s^{2} } ]}{0,05[m]} =39,24[\frac{N}{m}]$

Then we can calculate how much the spring stretch whith the another mass of 0,2kg:

$x=\frac{m*g}{k} =\frac{0,4[kg]*9,81[\frac{m}{s^{2} } ]}{39,24[\frac{N}{m}]} =0,1[m]\\$

The energy of a spring:

$E=\frac{1}{2}*k*x^{2}$

For the first case:

$E=\frac{1}{2} *39,24[\frac{N}{m}]*(0,05[m])^{2} =0,049 [J]$

For the second case:

$E=\frac{1}{2} *39,24[\frac{N}{m}]*(0,1[m])^{2} =0,0196 [J]$

If you take the relation E2/E1 = 4.

We have the next facts:

x=0,005 m

E = 0,0025 J

Using the energy equation for a spring:

$E=\frac{1}{2}*k*x^{2}$ ⇒ $k=\frac{E*2}{x^{2} } =\frac{0,0025[J]*2}{(0,005[m])^{2} } =200[\frac{N}{m} ]$

The potential energy of the diver will be equal to the kinetic energy in the moment befover hitting the watter.

$E=W*h=500[N]*10[m]=5000[J]$

Watch out the units in this case, the 500 N reffer to the weighs of the diver almost relative to the earth, thats equal to m*g.

The work is equal to the force acting in the direction of the motion. so we have to do the diference beetwen angles to obtain the effective angle where the force is acting: 47-15=32 degree.

The force acting in the direction of the ramp will be the projection of the force in the ramp, equal to F*cos(32). The work will be:

W=F*d=F*cos(32)*d=10N*cos(32)*10m=84,8J

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2 years ago