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lidiya [134]
3 years ago
14

Which is more work, pushing with 115 N over 15m or lifting 20N over 10 m?

Physics
2 answers:
Harrizon [31]3 years ago
6 0
Work is simply the force applied over a distance.

Work = Force × Displacement (m) ...In some cases where an angle is involved cosФ (theta) is used 
Work = F × D(m) × Cos(an angle measure)

So for lifting the object cos(90°) would be used...while pushing cos(0°)
When you work out the formula for each it should be clear what the correct answer is. 


Nikitich [7]3 years ago
3 0

Work = (force) x (distance)

You could look at the two cases, and see right away that
the first one has more force acting through more distance,
so it must be more work.  But since I just gave you the formula
for Work, let's calculate the amount of it for both cases:

First case:  Work = (115 N) x (15 m) =   1,725 joules

Second case:  Work = (20 N) x (10 m)  =  200 joules

The first case involves 8.625 times as much work as the second case.

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5 points
olga2289 [7]

Answer:

d. 5 ohms

Explanation:

For resistors in parallel, the equivalent resistance is found with:

1/Req = ∑(1/R)

1/R = 1/15 + 1/15 + 1/15

1/R = 3/15

R = 15/3

R = 5

8 0
3 years ago
A hot air balloon rising vertically is tracked by an observer located 3 miles from the lift-off point. At a certain moment, the
yuradex [85]

Answer:

\frac{dy}{dt}=1.2\frac{mi}{min}

Explanation:

We know that the tangent function relates the angle of the right triangle that forms the hot air balloon rising:

tan\theta=\frac{y}{x}\\y=xtan\theta(1)

Differentiating (1) with respect to time, we get:

\frac{dy}{dt}=tan\theta\frac{dx}{dt}+xsec^{2}\theta\frac{d\theta}{dt}\\

\frac{dx}{dt}=0 since x is a constant value. Replacing:

\frac{dy}{dt}=3mi(sec^{2}\frac{\pi}{3})0.1\frac{rad}{min}\\\frac{dy}{dt}=1.2\frac{mi}{min}

5 0
3 years ago
50 N Vertical 10 N Horizontal Force X can you find force x?<br>​
malfutka [58]

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45

Explanation:

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Morgarella [4.7K]

Answer:

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two

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Explanation:

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