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KatRina [158]
3 years ago
13

What is sieving? Give an example where this method is used. (2)​

Physics
1 answer:
Anna007 [38]3 years ago
8 0

Answer:

sieving is when you separate particles of different sizes.

Explanation:

separating sand mixtures

separating chaffs from local garri

You might be interested in
A tennis ball is dropped from a height of 10.0 m. It rebounds off the floor and comes up to a height of only 4.00 m on its first
Dominik [7]

Answer:

a) V=14.01 m/s

b) V=8.86 \, m/s

c)t = 2.33s

Explanation:

Our most valuable tool in solving this problem will be the conservation of mechanical energy:

E_m = E_k +E_p

That is, mechanical energy is equal to the sum of potential and kinetic energy, and  the value of this E_m mechanical energy will remain constant. (as long as there is no dissipation)

For a point particle, we have that kinetic energy is:

E_k = \frac{1}{2} m \, V^2

Where m is the mass, and V is the particle's velocity,

Potential energy on the other hand is:

E_p= m\, g\, h

where g is the acceleration due to gravity (g=9.81 \, m/s^2) and h is the height of the particle. How do we define the height? It's a bit of an arbitrary definition, but we just need to define a point for which h=0, a "floor". conveniently we pick the actual floor as our reference height, but it could be any point whatsoever.

Let's calculate  the mechanical energy just before the ball is dropped:

As we drop the ball, speed must be initially zero, and the height from which we drop it is 10 meters, therefore:

E_m = \frac{1}{2}m\,0^2+ mg\cdot 10 \,m\\E_m=mg\cdot10 \, m

That's it, the actual value of m is not important now, as we will see.

Now, what's the potential energy at the bottom? Let's see:

At the bottom, just before we hit the floor, the ball is no longer static, it has a velocity V that we want to calculate, on the other hand, it's height is zero! therefore we set h=0

E_m = \frac{1}{2}m\,V^2+ mg\cdot 0\\\\E_m = \frac{1}{2}m\,V^2

So, at the bottom, all the energy is kinetic, while at the top all the energy is potential, but these energies are the same! Because of conservation of mechanical energy. Thus we can set one equal to the other:

E_m = \frac{1}{2}m\,V^2 = mg\cdot 10m\\\\\\ \frac{1}{2}m\,V^2 = mg\cdot 10m\\\\V = \sqrt[]{2g\cdot 10m} \\

And so we have found the velocity of the ball as it hits the floor.

V = \sqrt[]{2g\cdot 10m}=14.01\, m/s

Now, after the ball has bounced, we can again do an energy analysis, and we will get the same result, namely:

V = \sqrt[]{2g\cdot h}

where h is the maximum height of the ball, and v is the maximum speed of the ball (which is always attained at the bottom). If we know that now the height the ball achieves is 4 meters, plugging that in:

V = \sqrt[]{2g\cdot 4m} =8.86 \, m/s

Now for C, we need to know for how long the ball will be in the air from the time we drop it from 10 meters, and how long it will take the ball to reach its new maximum height of 4 meters.

As the acceleration of gravity is a constant, that means that the velocity of the ball will change at a constant rate. When something changes at a constant rate, what is its average?  It's the average between initial and final velocity, look at diagram to understand. The area under the Velocity vs time curve is the displacement of the ball, and:

V_{avg}\cdot t=h\\t=h/V_{avg}

what's the average speed when the ball is descending?

V_{avg}=\frac{1}{2} (14.01\, m/s+0)=7 \, m/s

so the time it takes the ball to go down is:

t=h/V_{avg}=\frac{10m}{7m/s} =1.43s\\

Now, when it goes up, it's final and initial speeds are 0 and 8.86 meters per second, thus the average speed is:

V_{avg}=\frac{1}{2} (8.86\, m/s+0)=4.43 \, m/s

and the time it takes to go up is:t=h/V_{avg}=\frac{4m}{4.43m/s} =0.90s

When we add both times , we get:

t_{total}=t_{down}+t_{up}=1.43s+0.90s = 2.33s

6 0
3 years ago
Two extremely large nonconducting horizontal sheets each carry uniform charge density on the surfaces facing each other. The upp
sashaice [31]

This problem refers to a parallel plate capacitor. There is an electric field between the two plates. The working equation to be used is the Gauss’s Law which is

Electric field = Surface charge density / ε0

The answer is -2.52 μC/m2.

8 0
3 years ago
A uniform rod of length L rests on a frictionless horizontal surface. The rod pivots about a fixed frictionless axis at one end.
Veronika [31]

Answer:

A) ω = 6v/19L

B) K2/K1 = 3/19

Explanation:

Mr = Mass of rod

Mb = Mass of bullet = Mr/4

Ir = (1/3)(Mr)L²

Ib = MbRb²

Radius of rotation of bullet Rb = L/2

A) From conservation of angular momentum,

L1 = L2

(Mb)v(L/2) = (Ir+ Ib)ω2

Where Ir is moment of inertia of rod while Ib is moment of inertia of bullet.

(Mr/4)(vL/2) = [(1/3)(Mr)L² + (Mr/4)(L/2)²]ω2

(MrvL/8) = [((Mr)L²/3) + (MrL²/16)]ω2

Divide each term by Mr;

vL/8 = (L²/3 + L²/16)ω2

vL/8 = (19L²/48)ω2

Divide both sides by L to obtain;

v/8 = (19L/48)ω2

Thus;

ω2 = 48v/(19x8L) = 6v/19L

B) K1 = K1b + K1r

K1 = (1/2)(Mb)v² + Ir(w1²)

= (1/2)(Mr/4)v² + (1/3)(Mr)L²(0²)

= (1/8)(Mr)v²

K2 = (1/2)(Isys)(ω2²)

I(sys) is (Ir+ Ib). This gives us;

Isys = (19L²Mr/48)

K2 =(1/2)(19L²Mr/48)(6v/19L)²

= (1/2)(36v²Mr/(48x19)) = 3v²Mr/152

Thus, the ratio, K2/K1 =

[3v²Mr/152] / (1/8)(Mr)v² = 24/152 = 3/19

3 0
3 years ago
(1 point) A rectangular tank that is 3 feet long, 9 feet wide and 12 feet deep is filled with a heavy liquid that weighs 110 pou
Aloiza [94]

Answer:

Explanation:

Work in pumping water from the tank is given as

W = ∫ y dF. From a to b

Where dF is the differential weight of the thin layer of liquid in the tank, y is the height of the differential layer

a is the lower limit of the height

b is the upper limit of the height.

We know that, .

F = ρVg

Where F is the weight

ρ is the density of water

V is the volume of water in tank

g is the acceleration due to gravity

Then,

dF = ρg ( Ady)

We know that the density and the acceleration due to gravity is constant, also the base area of the tank is constant, only the height that changes.

Then,

ρg = 62.4 lbs/ft³

Area = L×B = 3 × 9 = 27ft²

dF = ρg ( Ady)

dF = 1684.8dy

The height reduces from 12ft to 0ft

Then,

W = ∫ y dF. From a to b

W = ∫ 1684.8y dy From 0 to 12

W = 1684.8y²/2 from 0 to 12

W = 842.4 [y²] from y = 0 to y = 12

W = 842.4 (12²-0²)

W = 121,305.6 lb-ft

3 0
3 years ago
A trolley at rest is pushed to accelerate at a
laila [671]
V^2 = 2a S
V = square root ( 2 x 0.5 . 4)
V = 2 m/s
Answer is B
6 0
3 years ago
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