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Molodets [167]
4 years ago
14

A heavy rope, 80 ft long and weighing 32 lbs, hangs over the edge of a building 100 ft high. how much work w is done in pulling

the rope up 20 ft?
Physics
1 answer:
allochka39001 [22]4 years ago
6 0
The first thing you should know for this case is that work is defined as the product of force by the distance traveled in the direction of force.
 We have then:
 W = Fd
 The distance varies, so we must integrate:
 from 0 to 20:
 W = ∫F (x) dx
 W = ∫32xdx
 W = 32∫xdx
 W = 32 (x ^ 2/2) = (16) (20 ^ 2) = 6400 ft * lbs
 answer:
 6400 ft * lbs is work done pulling the rope up 20 ft
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Answer:

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3 years ago
a spotted lizard runs at 3m/s at top speed. a girl wants to catch the lizard to keep as a pet. where should the girl place her c
Ann [662]

The acceleration due to earth's gravity is -9.8 m/s [dn] I thought... I'm assuming this is a projectile motion question asking for the range.

Break each kinematic quantity into their x and y components.

          x           y

v₁ = 3 m/s        0

v₂ = 3 m/s        ?

Δd = ?            -1.5

Δt  = ?              ?

a = 0           -10 m/s²

So the variable we are trying to find is Δdx (x component of displacement). We need to use a kinematic equation to do so. However, we obviously don't have enough given to find Δdx. This means we need to find something first, something we can use. How about Δt? Δt can be applied to both the x and y components. We have enough information in the y component list to find Δt. We can use this formula and solve for Δt.

Δdy = v₁y ( Δt ) + 1/2 ( ay ) ( Δt )²

Δdy = 1/2 ( ay ) ( Δt )²   <- the first term cancels out since v₁y = 0.

2Δdy = ay ( Δt )²

2Δdy / ay = ( Δt )²

√ 2Δdy / ay = Δt

√ 2(-1.5 m/s) / -9.8 m/s² = Δt

√ -3.0 m/s / -9.8 m/s² = Δt

√ 0.3<u>0</u>6122449 s² = Δt

0.5<u>5</u>32833352 s = Δt

Now, we can use this newly found quantity to solve for Δdx using the x component values using the appropriate kinematic equation.

Δdx = ( v₁x + v₂x / 2) ( Δt )

Δdx = ( ( 3.0 m/s + 3.0 m/s ) / 2 ) ( 0.5<u>5</u>32833352 s )

Δdx = ( 6.0 m/s / 2 ) ( 0.5<u>5</u>32833352 s )

Δdx = ( 3.0 m / s )( 0.5<u>5</u>32833352 s )

Δdx = 1.<u>6</u>59850006 m

Therefore, the girl should place her cage 1.7 m away from the platform to catch the lizard.

This solution assumes that the acceleration due to gravity is -10 m/s² [dn] and not -9.8 m/s² [dn]. If you need -9.8 m/s² [dn], then just substitute it into my solution. This was a pain to type lol





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Naddik [55]

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Yosef is playing with different kinds of rubber bands. Some are very narrow and some are quite wide. Yosef is curious about the
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Answer:

He could have many different hypothesis, but here is one.

Explanation:

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dezoksy [38]
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Therefore the correct answer is "all of the above".

b) The planets closer to the Sun were (and still are) subject to higher temperatures, due to their close distance to the Sun. In these conditions, rocky materials undergo condensation, while iced gaseous materials undergo vaporization. In the outer parts of the Solar System temperatures are too low to allow these transformations.
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