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valkas [14]
4 years ago
14

A pulley system operates with 40% efficiency. if the work put in is 200 joules, how much useful work is produced?

Physics
2 answers:
QveST [7]4 years ago
8 0

<span>1.    </span><span>Efficiency is the measure of how efficient a process is. It is used to assess the ability of a process in avoiding waste energy, materials, money and time in doing a desirable output. It is calculated as;

Efficiency = useful energy ouput / total energy input</span>

<span>.40 = useful work / 200</span>

<span>useful work = 80 joules</span>

FinnZ [79.3K]4 years ago
5 0

Answer:

The useful work produced by a pulley system is 80 Joules.

Explanation:

Efficiency is a measure of how much work is preserved in a process. The formula for efficiency is the energy output, divided by the energy input, expressed as a percentage. A perfect process would have an efficiency of 100%.

η = efficiency (Greek letter "eta")

Wo = the work or energy produced by a process. Units are Joules (J).

Wi = the work or energy put in to a process. Units are Joules (J).

n = \frac{Wo}{Wi}

0.4 = \frac{Wo}{200J}

Wo = 0.4*200J

Wo = 80 J

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2 years ago
An archer fish launches a droplet of water from the surface of a small lake at an angle of 70° above the horizontal. He is aimin
Norma-Jean [14]

Answer:

a). v=2776 m/s

Explanation:

The speed of the water droplet for the fish be successful is

Taking the distance in axis 'x' and 'y'

x_{tx}=40cm\frac{1m}{100cm}=0.40m\\x_{ty}=23cm\frac{1m}{100cm}=0.23m

The time is the velocity in axis 'x' with the angle 70 so

t=\frac{0.40m}{v_{x}*cos(70)}

Now using the time in terms of velocity the motion in axis 'y' can find the velocity to be the fish successful

x_{yf}=x_{yo}+v_{o}*t+\frac{1}{2}*g*t^{2}\\0.23m=0.40m\frac{vo*sin(70)}{vo*cos(70)} +\frac{1}{2}*9.8*(\frac{0.40}{vo*cos(70)} )^{2}\\0.23m=0.40m*vo*tan(70)+4.9*(\frac{0.16m^{2} }{vo^{2} *cos(70)^{2} }) \\vo^{2}cos(70)^{2}=\frac{0.16m^{2} }{0.177}\\vo=\sqrt{\frac{0.9021}{cos(70)^{2}}} \\vo=2.776 \frac{m}{s}

4 0
3 years ago
If the density of a diamond is 3.5 g/cm", what would be the mass of a diamond whose
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Answer:

Explanation:

7 0
4 years ago
Light takes 8 minutes to reach the Earth, and the speed of light is 3.0×10^8 m/s. a) What is the orbital speed of the Earth arou
spin [16.1K]

Answer:

(a) 28690 m/s (b) 2.46x10^{33}J

Explanation:

The orbital speed is define as:

v = \frac{2 \pi r}{T}   (1)

Where r is the radius of the trajectory and T is the orbital period.

To determine the orbital speed of the Earth it is necessary to know the orbital period and the radius of the trajectory. That can be done by means of the Kepler's third law and average velocity equation.

The average velocity in a Uniform Rectilinear Motion is defined as:

v = \frac{d}{t}   (2)

Where v is the velocity, d is the covered distance and t is the time.

Equation 2 can be rewritten for d to get:

d = vt   (3)

In this case, v will be the speed of light and t, the 8 minutes that takes to reach the Earth.

The time will be converted to seconds so the units in equation 3 can match:

8min . \frac{60s}{1min} ⇒  480s

t = 480s

Replacing all those values in equation 3 it is gotten:

d = (3.0x10^{8}m/s)(480s)

d = 1.44x10^{11}m

Kepler’s third law is defined as:

T^{2} = r^{3}

Where T is orbital period and r is the radius of the trajectory.

T = \sqrt{r^{3}}

T = \sqrt{(1.44x10^{11}m)^{3}}

It is necessary to pass from meters to astronomical unit (AU), 1 AU is defined as the distance between the Earth and the Sun.

T = \sqrt{1AU}

T = 1AU

That can be expressed in units of years.

T = 1AU . \frac{1year}{1AU}

T = 1year

But there are 31536000 seconds in one year:

T = 1year . \frac{31536000s}{1year}

T = 31536000s

Finally, equation  1 can be used:

v = \frac{2 \pi (1.44x10^{11}m)}{(31536000s)}

v = 28690 m/s

<u>So Earth orbital speed around the Sun is 28690 m/s.</u>

<em>b) What is its kinetic energy?</em>

The kinetic energy is defined as:

E = \frac{1}{2}mv^{2}  (4)

Notice that it is necessary to found the mass of the Earth, that can be done combining the Universal law of gravity and Newton's second law:

F = \frac{GMm}{r^{2}}

ma = \frac{GMm}{r^{2}}  (5)

M will be isolated in equation 5:

M = \frac{r^{2}a}{G}

Where r is the radius of the Earth (6.38x10^{6}m)

M = \frac{(6.38x10^{6}m)^{2}(9.8m/s^{2})}{6.67x10^{-11}kg.m/s^{2}.m^{2}/Kg^{2}})

M = 5.98x10^{24} Kg

E = \frac{1}{2}( 5.98x10^{24} Kg))(28.690m/s)^{2}

E = 2.46x10^{33}Kg.m^{2}/s^{2}

E = 2.46x10^{33}J

<u>Hence, the kinetic energy of Earth is 2.46x10^{33}J.</u>

8 0
3 years ago
What is the answer??
kati45 [8]

Answer:

lol do you still need help?

Explanation:

8 0
3 years ago
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