All categories

2 years ago

Why is it generally easy to filter out particles in suspension?

1 answer:

7 0

Answer: A suspension is a heterogeneous mixture in which some of the particles settle out of the mixture upon standing. The particles in a suspension are far larger than those of a solution, so gravity is able to pull them down out of the dispersion medium (water).

Explanation:

You might be interested in

In a football game, a 90 kg receiver leaps straight up in the air to catch the 0.42 kg ball the quarterback threw to him at a vi

irinina [24]

To solve this problem it is necessary to apply the equations related to the conservation of momentum. Mathematically this can be expressed as

$m_1v_1+m_2v_2 = (m_1+m_2)v_f$

Where,

$m_{1,2}$ = Mass of each object

$v_{1,2}$ = Initial velocity of each object

$v_f$ = Final Velocity

Since the receiver's body is static for the initial velocity we have that the equation would become

$m_2v_2 = (m_1+m_2)v_f$

$(0.42)(21) = (90+0.42)v_f$

$v_f = 0.0975m/s$

Therefore the velocity right after catching the ball is 0.0975m/s

8 0

3 years ago

You are designing a 108 cm3 right circular cylindrical can whose manufacture will take waste into account. There is no waste in

FinnZ [79.3K]

Explanation:

It is given that,

The volume of a right circular cylindrical, $V=108\ cm^3$

We know that the volume of the cylinder is given by :

$V=\pi r^2 h$

$108=\pi r^2 h$

$h=\dfrac{108}{\pi r^2}$ ............(1)

The upper area is given by :

$A=32r^2+2\pi rh$

$A=32r^2+2\pi r\times \dfrac{108}{\pi r^2}$

$A=32r^2+\dfrac{216}{r}$

For maximum area, differentiate above equation wrt r such that, we get :

$\dfrac{dA}{dr}=64r-\dfrac{216}{r^2}$

$64r-\dfrac{216}{r^2}=0$

$r^3=\dfrac{216}{64}$

r = 1.83 m

Dividing equation (1) with r such that,

$\dfrac{h}{r}=\dfrac{108}{\pi r}$

$\dfrac{h}{r}=\dfrac{108}{\pi 1.83}$

$\dfrac{h}{r}=59 \pi$

Hence, this is the required solution.

8 0

3 years ago

Suppose the gas resulting from the sublimation of 1.00 g carbon dioxide is collected over water at 25.0◦c into a 1.00 l containe

AlexFokin [52]

Answer:

0.56 atm

Explanation:

First of all, we need to find the number of moles of the gas.

We know that

m = 1.00 g is the mass of the gas

$Mm=44.0 g/mol$ is the molar mass of the carbon dioxide

So, the number of moles of the gas is

$n=\frac{m}{M_m}=\frac{1.00 g}{44.0 g/mol}=0.023 mol$

Now we can find the pressure of the gas by using the ideal gas equation:

$pV=nRT$

where

p is the pressure

$V=1.00 L = 0.001 m^3$ is the volume

n = 0.023 mol is the number of moles

$R=8.314 J/mol K$ is the gas constant

$T=25.0^{\circ}+273=298 K$ is the temperature of the gas

Solving the equation for p, we find

$p=\frac{nRT}{V}=\frac{(0.023 mol)(8.314 J/mol K)(298 K)}{0.001 m^3}=5.7 \cdot 10^4 Pa$

And since we have

$1 atm = 1.01\cdot 10^5 Pa$

the pressure in atmospheres is

$p=\frac{5.7\cdot 10^4 Pa}{1.01\cdot 10^5 Pa/atm}=0.56 atm$

5 0

3 years ago

(a) The electric potential due to a point charge is given by V = kq⁄r where q is the charge, r is the distance from q and k = 8.

LiRa [457]

Answer:

a) [volts] = [N m / C],

b) The lines or surface that has the same potential are called equipotential

c) the equipotential lines must also be perpendicular to the electric field lines

Explanation:

a) find the units of the volt

the electric potential energy is

V = k q / r

V = [N m² / C²] C / m

V = [N m / C]

The electric potential is defined as

V = E .s

V = [N / C] [m]

V = [N m / C] = [volt]

we see that in the two expressions the same result is obtained therefore the volt is

[volts] = [N m / C]

b) The lines or surface that has the same potential are called equipotential surfaces, the great utility of these lines or surfaces is that a face can be displaced on it without doing work.

c) The electric potential is defined as the gradient of the electric field

v = $- \frac{dE}{dx} i^$

therefore the equipotential lines must also be perpendicular to the electric field lines

4 0

3 years ago

Two identical stones are thrown from the top of a tall building. Stone 1 is thrown vertically downward with an initial speed v,

Molodets [167]

If the resistance of the Air is ignored, we can use the theory given by Galileo in which he warned that the thermal velocity of a body in free fall was given by

$v= \frac{1}{2}gt$

Where

g = Gravitational acceleration

t = time

As we can see the speed of objects in free fall is indifferent to the position that is launched (as long as the resistance of the air is ignored) or its mass.

Both bodies will end with the same thermal speed.

5 0

3 years ago