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Marina CMI [18]
3 years ago
15

A wooden log is displaced to a distance of 20m in 10 seconds by applying 500N effort . Calculate the workdone and power.​

Physics
2 answers:
nata0808 [166]3 years ago
8 0

A wooden log is displaced to a distance of 20m in 10 seconds by applying 500N effort . Calculate the workdone and power...

Solution,

displacement = 20 m

time = 10 sec

force = 500 N

work done = ?

power = ?

Now ,

work done = f × s

= 500 N × 20 m

= 10000 j

Now ,

power \:  =  \frac{w}{t}  \\  =  \frac{10000j}{10sec}  \\   = 1000W

~nightmare 5474~

andriy [413]3 years ago
7 0

The answer is 10000 j and 1000 W

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Jill applies a force of 250 N to a machine. The machine applies a force of 25 N to an object. What is the mechanical advantage o
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Mechanical advantage is defined as the ratio of output load to the input load. The mechanical advantage of the machine will be 0.1.

<h3>What is mechanical advantage?</h3>

Mechanical advantage is a measure of the ratio of output force to input force in a system,

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applied force = 250 N

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3 0
2 years ago
As shown in the diagram, two forces act on an object. The forces have magnitudes F1 = 5.7 N and F2 = 1.9 N. What third force wil
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Answer:

Second option 6.3 N at 162° counterclockwise from  

F1->

Explanation:

Observe the attached image. We must calculate the sum of all the forces in the direction x and in the direction y and equal the sum of the forces to 0.

For the address x we have:

-F_3sin(b) + F_1 = 0

For the address and we have:

-F_3cos(b) + F_2 = 0

The forces F_1 and F_2 are known

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We have 2 unknowns (F_3 and b) and we have 2 equations.

Now we clear F_3 from the second equation and introduce it into the first equation.

F_3 = \frac{F_2}{cos (b)}

Then

-\frac{F_2}{cos (b)}sin(b)+F_1 = 0\\\\F_1 = \frac{F_2}{cos (b)}sin(b)\\\\F_1 = F_2tan(b)\\\\tan(b) = \frac{F_1}{F_2}\\\\tan(b) = \frac{5.7}{1.9}\\\\tan^{-1}(\frac{5.7}{1.9}) = b\\\\b= 72\°\\\\m = b +90\\\\\m= 162\°

Then we find the value of F_3

F_3 = \frac{F_1}{sin(b)}\\\\F_3 =\frac{5.7}{sin(72\°)}\\\\F_3 = 6.01 N

Finally the answer is 6.3 N at 162° counterclockwise from  

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oksano4ka [1.4K]

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