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Molodets [167]
3 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]3 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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2. Four resistors of 2 ohm each are connected first in series and then in parallel with a battery.Find the ratio of electric cur
Serjik [45]

Answer:

Ratio of series current to parallel

= 1 : 8

Explanation:

Total resistance Rt

For series, Rt = 2+2+2+2 = 4ohms

For parallel, 1/Rt = 1/2 + 1/2 + 1/2 + 1/2

1/Rt = 4/2, Rt = 2/4 ohms.

If we use a 1V battery, then,

I = V/Rt

I = 1/4 = 0.25 ampere for series arrangement.

I = 1/0.5 = 2 ohms.

Ratio of current of series to parallel = 0.25 : 2

= 1 : 8

4 0
3 years ago
Read 2 more answers
Find the moments of inertia Ix, Iy, I0 for a lamina that occupies the part of the disk x2 y2 ≤ 36 in the first quadrant if the d
Tasya [4]

Answer:

I(x)  = 1444×k ×{\pi}

I(y)  = 1444×k ×{\pi}

I(o) = 3888×k ×{\pi}  

Explanation:

Given data

function =  x^2 + y^2 ≤ 36

function =  x^2 + y^2 ≤ 6^2

to find out

the moments of inertia Ix, Iy, Io

solution

first we consider the polar coordinate (a,θ)

and polar is directly proportional to a²

so p = k × a²

so that

x = a cosθ

y = a sinθ

dA = adθda

so

I(x) = ∫y²pdA

take limit 0 to 6 for a and o to \pi /2 for θ

I(x) = \int_{0}^{6}\int_{0}^{\pi/2} y²p dA

I(x) = \int_{0}^{6}\int_{0}^{\pi/2} (a sinθ)²(k × a²) adθda

I(x) = k  \int_{0}^{6}a^(5)  da ×  \int_{0}^{\pi/2}  (sin²θ)dθ

I(x) = k  \int_{0}^{6}a^(5)  da ×  \int_{0}^{\pi/2}  (1-cos2θ)/2 dθ

I(x)  = k ({r}^{6}/6)^(5)_0 ×  {θ/2 - sin2θ/4}^{\pi /2}_0

I(x)  = k × ({6}^{6}/6) × (  {\pi /4} - sin\pi /4)

I(x)  = k ×  ({6}^{5}) ×   {\pi /4}

I(x)  = 1444×k ×{\pi}    .....................1

and we can say I(x) = I(y)   by the symmetry rule

and here I(o) will be  I(x) + I(y) i.e

I(o) = 2 × 1444×k ×{\pi}

I(o) = 3888×k ×{\pi}   ......................2

3 0
3 years ago
Children's immune systems are very active and excludes them from being a high risk to foodbourne illness.
Alex17521 [72]

Answer:

TRUE...is ans........

8 0
3 years ago
What did you include in your description? Check all that apply
Alexeev081 [22]
I included things
Hope this helps!
7 0
3 years ago
A ball is thrown vertically upward from the top of a building 112 feet tall with an initial velocity of 96 feet per second. The
nordsb [41]

Answer:

t=6.96s

Explanation:

From this exercise, our knowable variables are <u>hight and initial velocity </u>

v_{oy}=96ft/s

y_{o}=112ft

To find how much time does the <u>ball strike the ground</u>, we need to know that the final position of the ball is y=0ft

y=y_{o}+v_{oy}t+\frac{1}{2}gt^{2}

0=112ft+(96ft/s)t-\frac{1}{2}(32.2ft/s^{2})t^{2}

Solving for t using quadratic formula

t=\frac{-b±\sqrt{b^{2}-4ac } }{2a}

a=-\frac{1}{2} (32.2)\\b=96\\c=112

t=-0.999s or t=6.96s

<u><em>Since time can't be negative the answer is t=6.96s</em></u>

7 0
3 years ago
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