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1 year ago

A machine has an efficiency of 15%. If the energy input is 300 joules, how much useful energy is generated?(1 point).

Engineering

1 answer:

Crazy boy [7]1 year ago

3 0

With an energy input of 300 J and an efficiency of 15%, the machine produces 45 J of useable energy.

<h3>What is energy?</h3>

Energy is characterized as a quantitative characteristic that is transmitted to a person or a physical system and can be observed in the manner in which labor is accomplished as well as in the generation of heat and light. The sun is one of the most significant energy sources.

The machine's useful energy output, or work output, can be calculated as follows:

Efficiency = work out / work in x 100

0.15 = work out / 300

Multiply by cross

Production = 0.15 x 300

Production: 45 J

Thus, with an energy input of 300 J and an efficiency of 15%, the machine produces 45 J of useable energy.

To learn more about energy, refer to the link below:

brainly.com/question/1932868

#SPJ1

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vaieri [72.5K]

Answer:21.3%

Explanation:

Given

80 % reduction in tool life

According to Taylor's tool life

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where V is cutting velocity

T=tool life of tool

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Thus

$VT^{0.12}=V'\left ( 0.2T\right )^{0.12}$

$V'=\frac{V}{0.2^{0.12}}$

$V'=\frac{V}{0.824}$ =1.213V

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6 0

3 years ago

The primary winding of a 120 volt transformer has 200 turns. The secondary winding has 20 turns. What is the secondary winding v

Katyanochek1 [597]

Answer:

12 volts

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The voltage ratio is equal to the turns ratio.

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The secondary winding voltage is 12 volts.

6 0

2 years ago

A mechanism is a device that transmits movement so the output movement is different from the input movement. identify three ways

igomit [66]

Answer:

• Gear ratios compare the output (or driven gear)

to the input (or drive gear)

• Gear Ratios can be determined using number

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3 years ago

Two steel plates are to be held together by means of 16-mm-diameter high-strength steel bolts fitting snugly inside cylindrical

Vilka [71]

Answer:

24.87 mm

Explanation:

The area of the bolt is given by

$A_b=\pi r^{2}$

Since diameter is 16mm, the radius is 16/2= 8 mm= 0.008 m

Area, $A_b=\pi\times 0.008^{2}=0.000201062 m^{2}$

For safe design of the bolt, we use stress of 201 Mpa

$\sigma_b=\frac {P}{A_b}$ where P is the load and $\sigma_b$ is normal stress.

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$P=A_b \sigma_b$

Substituting the figures given and already calculated area of bolt

$P=201\times 10^{6}\times 0.000201062 m^{2}=40413.4479 N$

The area of spacer is given by

$\pi (r_o^{2}- r_i^{2})$ where r is radius and the subscripts o and I denote inner and outer respectively

The value of 142 Mpa by default becomes the stress on spacer hence

$\sigma_s=\frac {P}{A_s}$ and making $A_s$ the subject then

$A_s=\frac {P}{\sigma_s}$

$\pi (r_o^{2}- r_i^{2})=\frac {P}{\sigma_s}$

Since we already calculated the value of P as 40413.4479 N and inner radius is 8mm= 0.008 m then

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5 0

3 years ago

Water is flowing steadily through a 3-m long, 20 mm innerdiameter cast iron pipe. The water enters at a uniform velocity U and e

devlian [24]

Answer:

Hello your question is incomplete attached below is the complete question

answer :

U/r = R = $U_{max}$ [ 1 - $(\frac{R}{R} )^2$ ] = 0

∴ No slip condition is satisfied

Explanation:

Given that

Outlet velocity U(r) = $U_{max}$ [ 1 - $(\frac{r}{R} )^2$ ]

<u>prove that the output velocity profile satisfies the no slip condition</u>

at no slip u = 0 ( i.e. at the pipe's inner surface )

at , r = R ( inference is at center )

hence U/r = R = $U_{max}$ [ 1 - $(\frac{R}{R} )^2$ ] = 0

∴ No slip condition is satisfied

8 0

2 years ago