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gizmo_the_mogwai [7]
3 years ago
8

Consider a 4-m-long, 4-m-wide, and 1.5-m-high above-the-ground swimming pool that is filled with water to the rim. (a) Determine

the hydrostatic force on each wall and the distance of the line of action of this force from the ground. (b) If the height of the walls of the pool is doubled, will the hydrostatic force on each wall double or quadruple? Why?
Physics
1 answer:
pashok25 [27]3 years ago
3 0

Answer:

44100 N

Explanation:

Each wall will have dimension of 4 m x 1.5 m

Whole force will act on central point of wall situated at a depth of 1.5 /2 = .75m

pressure at CM = h d g , h = .75 , d ( density of water = 10³ )

pressure at CM = .75 x 10³ x 9.8

= 7350 N / m²

Total force on each wall

= pressure x area

= 7350 x 4 x 1.5

= 44100 N Ans

b ) If h = 1.5 x 2 = 3

Pressure = hdg

1.5 x 10³ x 9.8

= 14700 N / m²

Force

= pressure x area

14700 x 3 x 4

= 176400 N

Which is 4 times 44100 N

So force will quadruple.

It is so because both area and height have become twice.

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A fully loaded, slow-moving freight elevator has a cab with a total mass of 1200 kg, which is required to travel upward 54 m in
Lana71 [14]

Answer:

The average power is calculated as 735.0 W

Solution:

As per the question:

Total mass, M = 1200 kg

Counter mass of the elevator, m = 950

Distance traveled by the elevator, d = 54 m

Time taken, t = 3 min = 180 s

Now,

To calculate the average power:

First, we find the force needed for lifting the weight:

Force, F = (M - m)g = (1200 - 950)\times 9.8 = 2450 N

Now, the work done by this force:

W = Fd = 2450\times 54 = 132300\ J = 132.3\ kJ

Average power is given as:

P_{avg} = \frac{W}{t} = \frac{132300}{180} = 735.0\ W

4 0
3 years ago
When a body of mass 0.25 kg is attached to a vertical massless spring, it is extended 5.0 cm from its unstretched length of 4.0
lora16 [44]

Answer:

d=0.165m

Explanation:

Given

m=0.25kg,x_{1}=5cm*\frac{1m}{100cm}=0.05m,x_{2}=4cm*\frac{1m}{100cm}=0.04m,v=2\frac{rev}{s}

The tension of the spring is

F_{k}=K*x_{1}=m*g

K=\frac{m*g}{x_{1}}

K=\frac{0.25kg*9.8m/s^2}{0.05m}=49N/m

The force in the spring is equal to centripetal force so

F_{c}=\frac{m*v^2}{r}

v=w*r=2\pi*r

But Fc is also

Fc=KxΔr

F_{c}=K*(r-x_{2})

Replacing

m*4\pi^2*r=K*(r-x_{2})

0.25kg*4\pi^2*r=49*(r-0.04m)

r=0.205m

total distance is

d=0.205-0.04=0.165m

3 0
3 years ago
1.
Cloud [144]

Answer:

broom

Explanation:

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3 years ago
What determines velocity rather than speed??
Stella [2.4K]
Velocity - <span><span>the speed of something in a given direction
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Velocity is the speed in a certain direction, whereas speed is just the rate of fastness.
Does that make sense?
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3 0
3 years ago
A nonconducting spherical shell, with an inner radius of 4 cm and an outer radius of 6 cm, has charge spread non uniformly throu
Aloiza [94]
In other words a infinitesimal segment dV caries the charge 
<span>dQ = ρ dV </span>

<span>Let dV be a spherical shell between between r and (r + dr): </span>
<span>dV = (4π/3)·( (r + dr)² - r³ ) </span>
<span>= (4π/3)·( r³ + 3·r²·dr + 3·r·(dr)² + /dr)³ - r³ ) </span>
<span>= (4π/3)·( 3·r²·dr + 3·r·(dr)² + /dr)³ ) </span>
<span>drop higher order terms </span>
<span>= 4·π·r²·dr </span>

<span>To get total charge integrate over the whole volume of your object, i.e. </span>
<span>from ri to ra: </span>
<span>Q = ∫ dQ = ∫ ρ dV </span>
<span>= ∫ri→ra { (b/r)·4·π·r² } dr </span>
<span>= ∫ri→ra { 4·π·b·r } dr </span>
<span>= 2·π·b·( ra² - ri² ) </span>

<span>With given parameters: </span>
<span>Q = 2·π · 3µC/m²·( (6cm)² - (4cm)² ) </span>
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<span>= 3.77×10⁻⁸C </span>
<span>= 37.7nC</span>
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3 years ago
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