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3 years ago

State three ways in which a machines are useful to us.

1 answer:

3 0

1: a machine makes our work easier

2: a machine can help us find important information

3: a machine can help us transport from a place to another

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Twopucksofequalmasscollideonafrictionlesssurface,asillustratedinthefigure.Immediatelyafterthe collision, the speed of the black

seropon [69]

Answer:

The speed of the white puck immediately after the collision is 2.6 m/s.

Explanation:

Given that,

Two pucks are equal masses.

Speed of black puck = 1.5 m/s

According to given figure,

We need to calculate the speed of the white puck immediately after the collision

Using law of conservation of momentum

$mv=m_{1}v_{1}\cos\theta+m_{2}v_{2}\cos\theta$

Put the value into the formula according to figure

$m\times3=m\times v_{1}\times\cos30+m\times1.5\times\cos60$

$3m=0.866m v_{1}+0.75m$

$v_{1}=\dfrac{3-0.75}{0.866}$

$v_{1}=2.6\ m/s$

Hence, The speed of the white puck immediately after the collision is 2.6 m/s.

8 0

3 years ago

Using 15 percent as machine energy efficiency, what is the actual work output if the total work input is 7500 kilojoules?

g100num [7]

Answer:

The actual work output is 1,125 kilojoules.

Explanation:

Energy efficiency is defined as the efficient use of energy. An appliance, process, or facility is energy efficient when it consumes less than the average amount of energy to perform an activity. Then, energy efficiency is the ratio between the amount of energy used in an activity and the amount expected to be carried out.

Efficiency is calculated as:

efficiency = output / input

Where output is the amount of mechanical work (in watts) or energy consumed by the process (in joules), and input (input) is the amount of work or energy that is used as input to carry out the process.

In this case:

Efficiency always has a value between 0 and 1. In this case, efficiency=0.15
output=?
input= 7500 kilojoules

Replacing:

0.15=output/7500 kilojoules

Solving:

Output=0.15* 7500 kilojoules

output=1,125 kilojoules

<u><em>The actual work output is 1,125 kilojoules.</em></u>

7 0

3 years ago

I need help finding the solution

omeli [17]

Answer:

objects with greater mass should have greater inertia

6 0

3 years ago

Air is a solution of 78% nitrogen, 21% oxygen and 1% other gases. Which part of this solution is the solvent?

Tom [10]

The answer is 78%. i got it right

6 0

3 years ago

Read 2 more answers

Unless indicated otherwise, assume the speed of sound in air to be v = 344 m/s. A stationary police car emits a sound of frequen

bixtya [17]

Answer:

a) The velocity of the car is 7.02 m/s and the car is approaching to the police car as the frequency of the police car is increasing.

b) The frequency is 1404.08 Hz

Explanation:

If the police car is a stationary source, the frequency is:

$f_{a} =(\frac{v+v_{c} }{v} )f_{s}$ (eq. 1)

fs = frequency of police car = 1200 Hz

fa = frequency of moving car as listener

v = speed of sound of air

vc = speed of moving car

If the police car is a stationary observer, the frequency is:

$f_{L} =f_{a} (\frac{v}{v-v_{c} } )=(\frac{v+v_{c} }{v-v_{c} } )f_{s}$ (eq. 2)

Now,

fL = frequecy police car receives

fs = frequency police car as observer

a) The velocity of car is from eq. 2:

$1250=1200(\frac{v+v_{c} }{v-v_{c} } )\\1250(v-v_{c} )=1200(v+v_{c} )\\v_{c} =\frac{50*344}{2450} =7.02m/s$

b) Substitute eq. 1 in eq. 2:

$f_{L} =(\frac{v+v_{p} }{v-v_{c} } )(\frac{v+v_{c} }{v-v_{p} } )f_{s} =(\frac{344+20}{344-7.02} )(\frac{344+7.02}{344-20} )*1200=1404.08Hz$

7 0

4 years ago