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Bezzdna [24]
2 years ago
9

A dam holds back a lake that is 85m deep at the dam. If the lake is 20km long, how much thicker should the dam be if the lake we

re smaller, only 1.0km long?​
Physics
1 answer:
evablogger [386]2 years ago
7 0

Answer:

the thickness of the dam should be the same if the level of water does not change.

Explanation:

the thickness of dam depends on the pressure of water that has to withstand.

Since the pressure exerted by the water depends only on the depth of the lake and not the length of the lake

P =1/2 *ρ*g*h ,

where P= pressure exerted by the water on the dam, ρ= density of water , g= gravity , h= depth of the lake

Then if the lake were smaller and the depth remains the same, the thickness of the dam should be the same

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If a force of 65 N is exerted on a 45 kg sofa and the sofa is moved 6.0 meters, how much work is done in moving the sofa? 17,550
abruzzese [7]

Answer:

The answer to your question is:   W = 390 J

Explanation:

Work is the transfer of energy when a body is moved from one place to another.

Data

Force = 65 N

mass = 45 kg

distance = 6 meters

work = ? J

Formula

W = F x d

Process

                   W = 65 N x 6 m

                   W = 390 J

5 0
2 years ago
Read 2 more answers
In at least 150 words, discuss how Chang's use of personification, metaphor, or connotation express one his themes in "Garden of
Kay [80]
<span>Poet Kuangchi Chang did not remain in China long enough to be "re-educated." Following the Communist takeover he fled to the United States. His poem "Garden of My Childhood" describes China before the revolution as a peaceful, idyllic garden with a violent horde rapidly approaching. A vine, the wind, and the sea are each personified, and each beckons for him to run. It is not until "eons later," when he is "worlds away," that his "running is all done," and he finds himself at his destination: another garden, just like the one he had left behind.</span>
6 0
2 years ago
A woman fires a rifle with barrel length of 0.5400 m. Let (0, 0) be where the 125 g bullet begins to move, and the bullet travel
dybincka [34]

Answer:

Explanation:

Given that,

Length of barrel =0.54m

Mass of bullet=125g=0.125kg

Force extend

F=16,000+10,000x-26,000x²

a. Work done is given as

W= ∫Fdx

W= ∫(16,000+10,000x-26,000x² dx from x=0 to x=0.54m

W=16,000x+10,000x²/2 -26,000x³/3 from x=0 to x=0.54m

W=16,000x+5,000x²- 8666.667x³ from x=0 to x=0.54m

W= 16,000(0.54) + 5000(0.54²) - 8666.667(0.54³) +0+0-0

W=8640+1458-1364.69

W=8733.31J

The workdone by the gas on the bullet is 8733.31J

b. Work done is given as

Work done when the length=1.05m

W= ∫Fdx

W= ∫(16,000+10,000x-26,000x² dx from x=0 to x=1.05m

W=16,000x+10,000x²/2 -26,000x³/3 from x=0 to x=1.05m

W=16,000x+5,000x²- 8666.667x³ from x=0 to x=1.05mm

W= 16,000(1.05) + 5000(1.05²) - 8666.667(1.05³) +0+0-0

W=16800+5512.5-10032.75

W=12,279.75J

The workdone by the gas on the bullet is 12,279.75J

3 0
2 years ago
A bulldozer does 4,500 J of work to push a mound of soil to the top of a ramp that is 15 m high. The ramp is at an angle of 35°
spin [16.1K]

<em>Answer</em>


Force = 170 N



<em>Explanation</em>

First find the distance (d) travelled by the bulldozer.


Sin 35 = 15/d

d = 15/(sin 35)

= 26.15m


Now;

work done = force × distance.


4500 J = force × 26.15


dividing both sides by 26.15,


Force = 4500/26.15

= 172.07 N


Answer to two significant figures = 170 N

3 0
2 years ago
Read 2 more answers
What is the best approximate value for the elastic potential energy (EPE) of the spring elongated by 3.0 meters?
DIA [1.3K]

The elastic potential energy of the spring is 6.8 J

Explanation:

The elastic potential energy of a compressed/stretched spring is given by the equation:

E=\frac{1}{2}kx^2

where

k is the spring constant

x is the elongation of the spring

The spring constant of the spring in this problem can be found by keeping in mind the relationship between force (F) and elongation (x) (Hooke's law):

F=kx

By looking at the graph and comparing it with the formula, we realize that the slope of the force-elongation graph corresponds to the spring constant. Therefore in this case,

k=\frac{15.0-0}{10.0-0}=1.5 N/m

Therefore when the spring has a elongation of x=3.0 m, its potential energy is

E=\frac{1}{2}(1.5)(3.0)^2=6.8 J

Learn more about potential energy:

brainly.com/question/1198647

brainly.com/question/10770261

#LearnwithBrainly

3 0
2 years ago
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