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soldier1979 [14.2K]
3 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.
Physics
1 answer:
Alex777 [14]3 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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Evgesh-ka [11]

Answer:

Both objects travel the same distance.

(c) is correct option

Explanation:

Given that,

Mass of first object = 4.0 kg

Speed of first object = 2.0 m/s

Mass of second object = 1.0 kg

Speed of second object = 4.0 m/s

We need to calculate the stopping distance

For first particle

Using equation of motion

v^2=u^2+2as

Where, v = final velocity

u = initial velocity

s = distance

Put the value in the equation

0= u^2-2as_{1}

s_{1}=\dfrac{u^2}{2a}....(I)

Using newton law

a=\dfrac{F}{m}

Now, put the value of a in equation (I)

s_{1}=\dfrac{8}{F}

Now, For second object

Using equation of motion

v^2=u^2+2as

Put the value in the equation

0= u^2-2as_{2}

s_{2}=\dfrac{u^2}{2a}....(I)

Using newton law

F = ma

a=\dfrac{F}{m}

Now, put the value of a in equation (I)

s_{2}=\dfrac{8}{F}

Hence, Both objects travel the same distance.

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3 years ago
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Answer:

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A friend asks you how much pressure is in your car tires. You know that the tire manufacturer recommends 30 psi, but it's been a
Bezzdna [24]

Answer:

25 psi

Explanation:

The weight of the car is:

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