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dlinn [17]
3 years ago
6

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 = <em>0.134 Horsepower</em>


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

= 1<em>5,000 watts</em>


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

<em>G = 53.96 gallons</em>

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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a penny

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How would radiation occur in space?
Marizza181 [45]

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3 years ago
A box with mass m is dragged across a level floor with coefficient of kinetic friction μk by a rope that is pulled upward at an
Komok [63]

Answer:

a) F =  μk  mg Cosθ

b) F = 279.78 N

Explanation:

a) F = μk R

Based on the description in the question, the horizontal reaction is:

R = mg Cosθ

The force required to move the box with constant speed in terms of m, μk, θ, and g is :

F =  μk  mg Cosθ

b) If m = 90 kg

g = 9.8 m/s²

μk=0.35

θ = 25⁰

Force required to slide the 90-kg patient across a floor at constant speed by pulling on him at an angle of 25∘ above the horizontal will be:

F =  μk  mg Cosθ

F = 0.35 * 90 * 9.8 * cos25

F = 279.78 N

5 0
4 years ago
A crystalline grain of nickel in a metal plate is situated so that a tensile load is oriented along the [111] crystal direction
hoa [83]

Given:

Applied stress, \sigma = 0.45 MPa

Critical Resolved Stress,  T_{cRss}= 0.242 MPa

Solution:

a). According to the question, orientation of tensile load is along [1 1 1],

\sigma = 0.45 MPa

Now, for resolved shear stress,  \t_{Rss} along [1 0 1] within (1 1 T)

let  '\theta' be the angle between [1 1 1] and [1 0 1], then by coordinate formula:

cos\theta = \frac{1\times 1 + 1\times 0 + 1\times 1}{\sqrt{1^{2}+1^{2}+1^{2}\sqrt{1^{2}+0^{2}+1^{2}}}}

cos\theta = \sqrt{\frac{2}{3}}

let  '\phi' be the angle between [1 1 1] and [1  1 1], then by coordinate formula:

cos\phi  = \frac{1\times 1 + 1\times 1 + 1\times 1}{\sqrt{1^{2}+1^{2}+1^{2}\sqrt{1^{2}+1^{2}+1^{2}}}}

cos\phi = 1

Now, for the resolved components along [1 0 1]

\t_{Rss}  = \sigma  cos\theta cos\phi

\t_{Rss} = 0.45\times 10^{6}\times \sqrt{\frac{2}{3}}\times 1 = 0.3673 MPa

b).  For required tensile stress to produce  T_{cRss}= 0.242 MPa:

\sigma _{1} = \frac{T_{cRss}}{cos\theta  cos\phi }

\sigma _{1} = \frac{0.242\times 10^{6}}{\sqrt{\frac{2}{3}}\times 1}

\sigma _{1} = 0.2964 MPa

5 0
3 years ago
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