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3 years ago

1) What is the power in horsepower of a 100 watt light bulb? What is the power in kilowatts of a 200 horsepower engine?

Physics

1 answer:

Yuki888 [10]3 years ago

3 0

1). Use the definition 1 Horsepower = 746 watts

100 watts = (100/746) HP = 0.134 Horsepower

2). Use the definition Power = (work done) / (time to do the work)

and the definition Work = (force) x (distance moved)

(3,000 N x 5 m) / second

= (3,000 x 5 N-m) / second

= (15,000 joules) / second

= 15,000 watts

3). Use the definition Power = (work done) / (time to do the work) , and the definition Work = (force) x (distance moved) , and the 'customary' definition of 1 Horsepower = 550 foot-pounds per second.

Let the answer ... the number of gallons ... be called ' G ' until we know what it actually is.

Weight of water lifted in 1 minute = (G gallons) x (8.34 pounds/gallon)

The distance it's lifted = 30 feet

Work done in 1 minute = (force x distance) = (8.34G x 30) foot-pounds

In just a moment, we're going to need this in seconds:

Power = (8.34G x 30 foot-pounds/minute) x (1 minute / 60 seconds)

Power = (8.34G x 30/60) (foot-pound-minute/minute-sec)

Power = 4.17G foot-pound/sec

The power = 1/2 Horsepower = 225 foot-pounds / sec

So finally, 225 = 4.17G

Divide each side by 4.17, and we have

G = (225/4.17) gallons

G = 53.96 gallons

I really should check this over, but I've had it up to here with gallons, feet, horsepower, and pounds, so I'm not gonna do it. Read my lips. It is what it is. I stand or fall by my answer. What could go wrong ? !

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Answer:

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$A_M= 1.12 *10^{-3} T$ respectively

Explanation:

According to right hand rule, your finger (direction of electric field) would be pointing in the positive x-axis i.e towards your right let your palms be face toward the direction of the magnetic field i.e negative y-axis (toward the ground ) Then anywhere your thumb stretched out is facing is the direction of propagation of the wave here in this case is the negative z -axis

The Intensity of the wave is mathematically represented as

$I = \frac{1}{2} c \epsilon _O E_{rms}^2$

Given that $I = 7.43 \frac{kW}{cm^2} = 7.43 \frac{*10^3}{*10^-{4} }= 7.43*10^7 \frac{W}{m^2}$

Making $E_{rms}$ the subject we have

$E_{rms} = \sqrt{\frac{I}{0.5*c*\epsilon_o} }$

Substituting values as given on the question

$E_{rms} = \sqrt{\frac{7.43 *10^7[\frac{W}{m^2} ]}{0.5 * 3.08*10^8 *8.85*10^{-12}} }$

$= 2.37*10^5 \ V/m$

The amplitude of the electric field is mathematically represented as

$A_E = \sqrt{2} * E_{rms}$

$= \sqrt{2} * 2.37*10^5$

$A_E= 3.35*10^5 V/m$

The amplitude of the magnetic field is mathematically represented as

$A_M = \frac{A_E}{c}$

Substituting value

$A_M = \frac{3.35 *10^5}{3.0*10^8}$

$A_M= 1.12 *10^{-3} T$

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An isotope undergoes a radioactive decay to attain stability.

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Beta decay is defined as the decay process in which a neutron gets converted to a proton and an electron. In this decay process, beta particle is emitted. The emitted particle carries a charge of -1 units and has a mass of 0 units. The released beta particle is also known as electron.

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An Apple with the mass of 200g falls from the tree.what is the acceleration of the apple towards the earth.what is the accelerat

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Answer:

Assume that the earth is a sphere (with radius equal to its actual radius at the equator) of uniform density.

Acceleration of the apple towards the earth: approximately $9.79\; \rm m \cdot s^{-2}$ .

Acceleration of the earth towards the apple: approximately $3.28 \times 10^{-25}\; \rm m \cdot s^{-2}$ .

Both values were rounded to three significant figures. Air resistance is assumed to be negligible.

Explanation:

Convert the mass of the apple to standard units:

$m(\text{apple}) = 200\; \rm g = 0.200\; \rm kg$ .

Look up the mass and radius of the earth:

Mass of the earth: $m(\text{earth})\approx 5.97\times 10^{24}\; \rm kg$ .
Radius of the earth: $r \approx 6.3781\times 10^{6}\; \rm m$ (at the equator.)

Assume that the earth is a sphere of uniform density. The acceleration of an object moving towards the earth under free fall would be approximately equal to the gravitational field strength of the earth at that position.

For a sphere of mass $m$ (assuming uniform density,) the gravitational field strength $g$ at a distance of $r$ away from the center of the sphere would be:

$\displaystyle g = \frac{G \cdot m}{r^2}$ .

The height of the tree should be much smaller than the radius of the earth. Therefore, the distance between the apple and the center of the earth would be approximately equal to the radius of the earth.

The strength of the gravitational field of the earth at the position of the apple would be approximately:

$\begin{aligned} & \frac{G \cdot m(\text{earth})}{r^2} \\ &\approx \frac{6.67\times 10^{-11}\; \rm m^3\cdot kg^{-1}\cdot s^{-2} \times 5.97 \times 10^{24}\; \rm kg}{\left(6.3781\times 10^{6}\; \rm m\right)^2} \\ &\approx 9.79\; \rm m\cdot s^{-2} \end{aligned}$ .

This value should be approximately equal to the acceleration of the apple towards the earth.

On the other hand, assume that the apple acts like a point mass when compared to the earth. In other words, assume that the radius of the apple is much smaller than the distance between the apple and the earth.

It can be shown (using calculus) that if the earth is a sphere with uniform density, the earth will behave just like a point mass at the center of the earth when studying interactions between the earth and objects outside of it.

At the center of the earth, the strength of the gravitational field due to the apple would be:

$\begin{aligned} & \frac{G \cdot m(\text{apple})}{r^2} \\ &\approx \frac{6.67\times 10^{-11}\; \rm m^3\cdot kg^{-1}\cdot s^{-2} \times 0.200\; \rm kg}{\left(6.3781\times 10^{6}\; \rm m\right)^2} \\ &\approx 3.28 \times 10^{-25}\; \rm m\cdot s^{-2} \end{aligned}$ .

Under the assumption that the earth is a sphere of uniform density, this value should be equal to the acceleration of the earth towards the apple due to the gravitational attraction of the apple.

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