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3 years ago

Is this right? if not how do u do it?

Physics

1 answer:

Ronch [10]3 years ago

3 0

You've definitely got the right idea and your procedure is correct.

But I see one mistake for sure, plus a few other things that
I'm not sure of.

The mistake for sure: If the ending water is 12 ml,
then the beginning water is 9 ml, not 11 ml.

Things I'm not sure of:

-- I'm not sure that I'm remembering correctly at the moment,
but I have a sneaking hunch that you're supposed to read the
level at the BOTTOM of the meniscus, not the top.
It won't matter, because the DIFFERENCE will be the same
either way. But if the teacher wants you to read from the bottom,
then your numbers may be marked wrong.

-- The question at the bottom asks for the "volume of cherry",
but I seem to see TWO cherries in the water. Does than mean
they expect you to take only half of the displacement ?

-- They ask for "volume of cherry" twice.
Maybe that means they want you to end up with the volume in cm³
instead of ml. (The number is the same.)

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3 years ago

A 20.0-N weight slides down a rough inclined plane which makes an angle of 30 degree with the horizontal. The weight starts from

Ulleksa [173]

Answer:

$1270.64\ \text{J}$

Explanation:

m = Mass of object = $\dfrac{mg}{g}$

mg = Weight of object = 20 N

g = Acceleration due to gravity = $9.81\ \text{m/s}^2$

v = Final velocity = 15 m/s

u = Initial velocity = 0

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$\theta$ = Angle of slope = $30^{\circ}$

f = Force of friction

fd = Work done against friction

The force balance of the system is

$\dfrac{1}{2}m(v^2-u^2)=(mg\sin\theta-f)d\\\Rightarrow \dfrac{1}{2}mv^2=mg\sin\theta d-fd\\\Rightarrow fd=mg\sin\theta d-\dfrac{1}{2}mv^2\\\Rightarrow fd=20\times \sin 30^{\circ}\times 150-\dfrac{1}{2}\times \dfrac{20}{9.81}\times 15^2\\\Rightarrow fd=1270.64\ \text{J}$

The work done against friction is $1270.64\ \text{J}$ .

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3 years ago

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3 years ago

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MrMuchimi

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3 years ago

Read 2 more answers

A tank is full of water. Find the work W required to pump the water out of the spout. (Use 9.8 m/s2 for g. Use 1000 kg/m3 as the

Sergio039 [100]

Answer:

W = 1.06 MJ

Explanation:

- We will use differential calculus to solve this problem.

- Make a differential volume of water in the tank with thickness dx. We see as we traverse up or down the differential volume of water the side length is always constant, hence, its always 8.

- As for the width of the part w we see that it varies as we move up and down the differential element. We will draw a rectangle whose base axis is x and vertical axis is y. we will find the equation of the slant line that comes out to be y = 0.5*x. And the width spans towards both of the sides its going to be 2*y = x.

- Now develop and expression of Force required:

F = p*V*g

F = 1000*(2*0.5*x*8*dx)*g

F = 78480*x*dx

- Now, the work done is given by:

W = F.s

- Where, s is the distance from top of hose to the differential volume:

s = (5 - x)

- We have the work as follows:

dW = 78400*x*(5-x)dx

- Now integrate the following express from 0 to 3 till the tank is empty:

W = 78400*(2.5*x^2 - (1/3)*x^3)

W = 78400*(2.5*3^2 - (1/3)*3^3)

W = 78400*13.5 = 1058400 J

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3 years ago