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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?

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 charge of Q is fixed in space. A second charge of q was first placed at a distance r1 away from Q. Then it was moved along a s

topjm [15]

Answer:

$\Delta U = \frac{Qq}{4\pi\epsilon_0}(\frac{1}{r_2^2}-\frac{1}{r_1^2})$

Explanation:

The electrostatic potential energy is given by the following formula

$U = \frac{1}{4\pi\epsilon_0}\frac{q_1q_2}{r^2}$

Now, we will apply this formula to both cases:

$U_1 = \frac{1}{4\pi\epsilon_0}\frac{Qq}{r_1^2}\\U_2 = \frac{1}{4\pi\epsilon_0}\frac{Qq}{r_2^2}$

So, the change in the potential energy is

$\Delta U = U_2 - U_1 = \frac{Qq}{4\pi\epsilon_0}(\frac{1}{r_2^2}-\frac{1}{r_1^2})$

7 0

3 years ago

A soccer player takes a corner kick, lofting a stationary ball 30.0° above the horizon at 18.0 m/s. If the soccer ball has a mas

Radda [10]

Answer:

change in momentum, $\Delta p=7.65 \,kg.m.s^{-1}$

$\Delta p_x= 6.6251 \,kg.m.s^{-1}$

$\Delta p_y= 3.825 \,kg.m.s^{-1}$

Average Force, $F=144.3396\,N$

$F_x=125.0018\,N$

$F_y=72.1698\,N$

Explanation:

Given:

angle of kicking from the horizon, $\theta= 30^{\circ}$

velocity of the ball after being kicked, $v=18 m.s^{-1}$

mass of the ball, $m=0.425\, kg$

time of application of force, $t=5.3\times 10^{-2}\,s$

We know, since body is starting from the rest

$\Delta p=m.v$ .....................(1)

$\Delta p=0.425\times 18$

$\Delta p=7.65 \,kg.m.s^{-1}$

Now the components:

$\Delta p_x= 7.65\times cos 30^{\circ}$

$\Delta p_x= 6.6251 \,kg.m.s^{-1}$

similarly

$\Delta p_y= 7.65\times sin 30^{\circ}$

$\Delta p_y= 3.825 \,kg.m.s^{-1}$

also, impulse

$I=F\times t$ .........................(2)

where F is the force applied for t time.

Then from eq. (1) & (2)

$F\times t=m.v$

$F\times 5.3\times 10^{-2}= 7.65$

$F=144.3396\,N$

Now, the components

$F_x=144.3396\times cos 30^{\circ}$

$F_x=125.0018\,N$

&

$F_y=144.3396\times sin 30^{\circ}$

$F_y=72.1698\,N$

6 0

3 years ago

A truck moves 60 kilometers east from point A to point B. At point B, it turns back west and stops 15 kilometers away from point

iogann1982 [59]

Answer:

Distance: 75 km

Displacement: 45 km

Explanation:

- Distance is a scalar quantity that refers to the total space covered by an object. It is calculated as the sum of the distances covered in each motion, regardless of their direction. therefore in this case:

distance = 60 km + 15 km = 75 km

- Displacement is a vector quantity whose magnitude is equal to the difference between the final point and the starting point of the motion, so it also takes into account the direction of each motion. In this case, the truck moves 60 km east, and then 15 km west: if we call '0' the starting point, the final point will be then

$p_B = 60 km - 15 km=45 km$

And so the displacement is

$d=p_B - p_A=45 km - 0 = 45 km$

6 0

3 years ago

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Damm [24]

Answer:

P_(pump) = 98,000 Pa

Explanation:

We are given;

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P_(tank)+ P_(pump) = P_(nozzle)

Now, the pressure would be given by;

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So,

ρgh_1 + P_(pump) = ρgh_2

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Plugging in the relevant values to obtain;

P_(pump) = 1000•9.8(30 - 20)

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