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2 years ago

A pump is used to lift 100 KG of water from a wel 60 m deep,in 20 S If force of gravity on 1 KG is 10 N,find

Physics

1 answer:

Harlamova29_29 [7]2 years ago

6 0

Explanation:

Given,

m = 100 kg
g = 10 N/kg¹
h = 60 m
t = 20 s

To Find:

a) Work done by the pump

b) Potential energy stored in the water

c)Power spent by the pump

d)Power rating of the pump.

Solution:

a) Work done by the pump

We know that,

$\rm \: Work \: done = Force * Distance \: moved$

f = 100 kg * 10N/kg
d = 60 m

$\rm \: Work\; Done =(100 \: kg \times \cfrac{10N}{kg} ) \times 60 \: m$

$\rm \: Work\; Done =1000 \times 60 \: joule$

$\boxed{\rm \: Work\; Done =60000 \: joule}$

[The unit'll be joule since N×M = J]

b) Potential energy stored in the water

$\rm \: P.E = m \cdot g \cdot h$

m = 100 kg
g = 10N/kg
h = 60

$\rm \: P.E =100 \:kg \: \times \cfrac{10 \: N}{kg} \times 60$

$\boxed{\rm \: P.E =60000 \: joule}$

same condition here as well, N×M = J
c) Power of the Pump

$\rm \: P = W/T$

where P = Power; W = Work done & T = Time taken
As we got the value of work done on question (a),& ATQ time taken is 20 S.

$\rm \: P = \cfrac{60000 \: joule}{20 \: seconds} =\boxed{\rm { 3000 \: Watts }}\: or \: \boxed{\rm 3 \: kW}$

d) Power rating of the pump = 3 kW

Assumption: The pump is 100% efficient & works well.

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Answer:

v = 0.42m/s

Explanation:

In order o calculate the linear speed of the point at the border of the wheel, you first take into account that the total acceleration of such a point is given by:

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atotal: total acceleration = 1.2m/s^2

ar: radial acceleration of the wheel

at: tangential acceleration

The tangential acceleration is also given by:

$a_t=r\alpha$ (2)

r: radius of the wheel = (40cm/2 )= 20cm = 0.2m

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You replace the expression (2) into the expression (1) and solve for the radial acceleration:

$a_{total}^2=a_r^2+(r\alpha)^2\\\\a_r=\sqrt{(a_{total})^2-(r\alpha)^2}\\\\a_r=\sqrt{(1.2m/s^2)^2-((0.2m)(4.0rad/s^2))^2}=0.894\frac{m}{s^2}$

Next, you use the following formula for the radial acceleration and solve for the linear speed:

$a_r=\frac{v^2}{r}\\\\v=\sqrt{ra_r}=\sqrt{(0.2m)(0.894m/s^2)}=0.42\frac{m}{s}$

The linear speed of the point at the border of the wheel is 0.42m/s

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3 years ago

If a ball is rolled down a frictionless inclined plane from a height of 3 feet, to what height would it roll up to on the other

mojhsa [17]

Answer:

It will roll up to its initial height from which it rolled from on the first hill

Explanation:

The given information are;

The height from which the ball is rolled = 3 feet[

The equation of the potential energy of the ball as it rolled down the frictionless inclined plane is given by the following equation;

m × g × h₁ = 1/2 × m × v² + 1/2×[2/5× m × r²) × [v/r]

Where;

m = The mass of the ball

h = The from which the ball is rolled down

v = The velocity of the ball

r = The radius of the ball

Therefore, all the potential energy, m × g × h, is converted into kinetic energy when the ball reached the ground

Similarly, as the ball rolls up on the other side, by the principle of energy conservation, all the kinetic energy is transformed into potential energy as follows

1/2 × m × v² + 1/2×[2/5× m × r²) × [v/r] = m × g × h₂, therefore, we have;

1/2 × m × v² + 1/2×[2/5× m × r²) × [v/r] = m × g × h₁ = m × g × h₂

Therefore since m × g × h₁ = m × g × h₂, then h₁ = h₂, the ball rises to the same height it started rolling from on the first hill.

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3 years ago

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Answer:

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3 years ago

Light with a single wavelength falls on two slits separated by 0.510 mm. In the resulting interference pattern on a screen 2.24

scZoUnD [109]

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Explanation:

Given

The slits are separated by $d=0.510\ mm$

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