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soldier1979 [14.2K]

2 years ago

5

A swimming pool has the shape of a box with a base that measures 30 m by 12 m and a uniform depth of 2.2 m. How much work is req

uired to pump the water out of the pool when it is full? Use 1000 kg divided by m cubed for the density of water and 9.8 m divided by s squared for the acceleration due to gravity.

1 answer:

Alex777 [14]2 years ago

5 0

Answer:

853776 J

Explanation:

The work-energy needs to pump water out of the pool is the product of the weight of water and distance h

E = Wh = mgh

Since water mass is a body of water we can treat it as the product of density 1000kg/m3 and volume, which is the product of base area and uniform height h

$m = \rho V = \rho \int\limits^{2.2}_0 {A} \, dh$

Therefore:

$E = mgh = g\rho A\int\limits^{2.2}_0 {h} \, dh\\E = 9.8*1000*30*12[h^2/2]^{2.2}_0 = 1764000(2.2^2 - 0^2) = 853776 J$

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A block (mass = 61.2 kg) is hanging from a massless cord that is wrapped around a pulley (moment of inertia = 1/2MR2 kg · m2, wh

kolezko [41]

Answer:

The angular velocity is $w = 53.35 \ rounds /minute$

Explanation:

From the question we are told that

The mass of the block is $m = 61.2kg$

The of the pulley is $M = 14.2 kg$

The radius of the pulley is $R = 1.5m$

The radius of the cord around the pulley is $r = 1.5 m$

The distance of the block to the floor is $d = 8.0 m$

From the question we are told that the moment of inertia of the pulley is

$I = \frac{1}{2} MR^2 kg \cdot m^2$

Substituting value

$I = \frac{1}{2} * 14.2 * (1.5)^2$

$I = 15.975 kg \cdot m^2$

Using the Newtons law we can express the force acting on the vertical axis as

$ma = mg -T$

=> $T = mg -ma$

Now when the pulley is rotated that torque generated on the massless cord as a r result of the tension T and the radius of the cord around the pulley is mathematically represented as

$\tau = I \alpha$

Here $\alpha$ is the angular acceleration

Here $\tau$ is the torque which can be equivalent to

$\tau = T r$

Substituting this above

$Tr = I \alpha$

Substituting for T

$(mg - ma ) r = I\ r \alpha$

Here $a$ is the linear acceleration which is mathematically represented as

$a = r\alpha$

$(mg - m(r\alpha ) ) r = I\ r \alpha$

$mgr = I\alpha + m(r\alpha ) r$

$mgr = \alpha [ I + mr^2]$

making $\alpha$ the subject

$\alpha = \frac{mgr}{I -mr ^2}$

Substituting values

$\alpha = \frac{61.2 * 1.5 * 9.8}{15.975 + (61.2 ) * (1.5)^2}$

$\alpha =5.854 rad /s^2$

Now substituting into the equation above to obtain the acceleration

$a = 5.854 * 1.5$

$a=8.78 m/s^2$

This acceleration is $a = \frac{v}{t}$

and v is the linear velocity with is mathematically represented as

$v = \frac{d}{t}$

Substituting this into the formula acceleration

$a = \frac{d}{t^2}$

making t the subject

$t = \sqrt{\frac{d}{a} }$

substituting value

$t = \sqrt{\frac{8}{8.78}}$

$t = 0.9545 \ s$

Now the linear velocity is

$v = \frac{8}{0.9545}$

$v = 8.38 m/s$

The angular velocity is

$w = \frac{v}{r}$

So

$w = \frac{8.38}{1.5}$

$w = 5.59 rad/s$

Generally 1 radian is equal to 0.159155 rounds or turns

So 5.59 radian is equal to x

Now x is mathematically obtained as

$x = \frac{5.59 * 0.159155}{1}$

$= 0.8892 \ rounds$

Also

60 second = 1 minute

So 1 second = z

Now z is mathematically obtained as

$z = \frac{ 1}{60}$

z $= 0.01667 \ minute$

Therefore

$w = \frac{0.8892}{0.01667}$

$w = 53.35 \ rounds /minute$

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