1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Scilla [17]
3 years ago
14

Water can dissolve almost anything in the universe

Physics
1 answer:
Elodia [21]3 years ago
5 0

Answer:

No, its not possible for water to dissolve almost anything in the universe.

Explanation:

Solubility of a solute defines the ability of that solute to dissolve in a given solvent. It is defined as the maximum amount of solute dissolved in a solvent at equilibrium. The solution which results from dissolving this maximum amount is called a saturated solution, and one it has been reached, no more solute can be dissolved in it.

Different substances in the universe have diffferent solubilities in water, some very high (soluble) (eg. sugar and salt) and some very low (insoluble) (eg plastics). The substances that are able to form bonds with water (Hydrogen or Ionic) are more soluble than those who are not able to do so.

You might be interested in
An Earth satellite is orbiting at a distance from the Earth’s surface equal to one Earth radius (4 000 miles). At this location,
trasher [3.6K]

Answer:

The acceleration due to gravity is \dfrac{1}{4} times the value of g at the Earth’s surface.

(D) is correct option.

Explanation:

Given that,

Radius = 4000 miles

We need to calculate the gravitational force at surface

Gravitational force on the mass m on the surface of the earth

At r = R

F=mg

mg=\dfrac{GmM}{R^2}....(I)

We need to calculate the gravitational force at height

Gravitational force on a mass m from the center of the earth,

At r = R + R = 2 R

F'=mg'

mg'=\dfrac{GmM}{4R^2}....(II)

Dividing equation (II) by equation (I)

\dfrac{mg'}{mg}=\dfrac{\dfrac{GmM}{4R^2}}{\dfrac{GmM}{R^2}}

\dfrac{g'}{g}=\dfrac{1}{4}

Hence, The acceleration due to gravity is \dfrac{1}{4} times the value of g at the Earth’s surface.

4 0
3 years ago
A 44-cm-diameter water tank is filled with 35 cm of water. A 3.0-mm-diameter spigot at the very bottom of the tank is opened and
cricket20 [7]

Answer:

The frequency f = 521.59 Hz

The rate at which the frequency is changing = 186.9 Hz/s

Explanation:

Given that :

Diameter of the tank = 44 cm

Radius of the tank = \frac{d}{2} =\frac{44}{2} = 22 cm

Diameter of the spigot = 3.0 mm

Radius of the spigot = \frac{d}{2} =\frac{3.0}{2} = 1.5 mm

Diameter of the cylinder = 2.0 cm

Radius of the cylinder = \frac{d}{2} = \frac{2.0}{2} = 1.0 cm

Height of the cylinder = 40 cm = 0.40 m

The height of the water in the tank from the spigot = 35 cm = 0.35 m

Velocity at the top of the tank = 0 m/s

From the question given, we need to consider that  the question talks about movement of fluid through an open-closed pipe; as such it obeys Bernoulli's Equation and the constant discharge condition.

The expression for Bernoulli's Equation is as follows:

P_1+\frac{1}{2}pv_1^2+pgy_1=P_2+\frac{1}{2}pv^2_2+pgy_2

pgy_1=\frac{1}{2}pv^2_2 +pgy_2

v_2=\sqrt{2g(y_1-y_2)}

where;

P₁ and P₂ = initial and final pressure.

v₁ and v₂ = initial and final fluid velocity

y₁ and y₂ = initial and final height

p = density

g = acceleration due to gravity

So, from our given parameters; let's replace

v₁ = 0 m/s ; y₁ = 0.35 m ; y₂ = 0 m ; g = 9.8 m/s²

∴ we have:

v₂ = \sqrt{2*9.8*(0.35-0)}

v₂ = \sqrt {6.86}

v₂ = 2.61916

v₂ ≅ 2.62 m/s

Similarly, using the expression of the continuity for water flowing through the spigot into the cylinder; we have:

v₂A₂ = v₃A₃

v₂r₂² = v₃r₃²

where;

v₂r₂ = velocity of the fluid and radius at the spigot

v₃r₃ = velocity of the fluid and radius at the cylinder

v_3 = \frac{v_2r_2^2}{v_3^2}

where;

v₂ = 2.62 m/s

r₂ = 1.5 mm

r₃ = 1.0 cm

we have;

v₃ = (2.62  m/s)* (\frac{1.5mm^2}{1.0mm^2} )

v₃ = 0.0589 m/s

∴ velocity  of the fluid in the cylinder =  0.0589 m/s

So, in an open-closed system we are dealing with; the frequency can be calculated by using the expression;

f=\frac{v_s}{4(h-v_3t)}

where;

v_s = velocity of sound

h = height of the fluid

v₃ = velocity  of the fluid in the cylinder

f=\frac{343}{4(0.40-(0.0589)(0.4)}

f= \frac{343}{0.6576}

f = 521.59 Hz

∴ The frequency f = 521.59 Hz

b)

What are the rate at which the frequency is changing (Hz/s) when the cylinder has been filling for 4.0 s?

The rate at which the frequency is changing is related to the function of time (t) and as such:

\frac{df}{dt}= \frac{d}{dt}(\frac{v_s}{4}(h-v_3t)^{-1})

\frac{df}{dt}= -\frac{v_s}{4}(h-v_3t)^2(-v_3)

\frac{df}{dt}= \frac{v_sv_3}{4(h-v_3t)^2}

where;

v_s (velocity of sound) = 343 m/s

v₃ (velocity  of the fluid in the cylinder) = 0.0589 m/s

h (height of the cylinder) = 0.40 m

t (time) = 4.0 s

Substituting our values; we have ;

\frac{df}{dt}= \frac{343*0.0589}{4(0.4-(0.0589*4.0))^2}

= 186.873

≅ 186.9 Hz/s

∴ The rate at which the frequency is changing = 186.9 Hz/s  when the cylinder has been filling for 4.0 s.

8 0
3 years ago
The ______ and _______ are used to calculate magnitude and direction of a resultant vector.
S_A_V [24]
Here are the correct answers that would complete the given statement above. The vector quantity and the vector arrow are used to calculate magnitude and direction of a resultant vector. Vector quantity has both magnitude and direction, whereas vector arrow represents<span> the magnitude of a quantity and the direction represents the direction of that quantity. </span>Hope this is the answer that you are looking for. 
3 0
3 years ago
1. boiling point of water
melomori [17]

Answer:

what is the question. . .

5 0
3 years ago
Starting from one oasis, a camel walks 25 km in a direction 30° south of west and then walks 30 km toward the north to a second
shepuryov [24]

Answer:

distance between both oasis ( 1 and 2) is  27.83 Km

Explanation:

let d is the distance between oasis1 and oasis 2

from figure

OC  = 25cos 30

OE = 25sin30

OE = CD

Therefore BC =  30-25sin30

distance between both oasis ( 1 and 2) is calculated by using phytogoras theorem

in\Delta BCO

OB^2 = BC^2 + OC^2

PUTTING ALL VALUE IN ABOVE EQUATION

d^2 = 930-25sin30)^2 + (25cos30)^2

d^2 = 775

d = 27.83 Km

distance between both oasis ( 1 and 2) is  27.83 Km

3 0
3 years ago
Other questions:
  • What is the difference b/w distance and displacement?
    10·1 answer
  • Would the solar panel work under a fluorescent or halogen light? explain your response being sure to relate your observations to
    15·1 answer
  • Do you wont to be friends
    9·1 answer
  • Explain a situation in which you can have a positive velocity but a negative acceleration
    11·1 answer
  • Who discovered the law of conservation of energy?
    11·2 answers
  • (PLEASE HELP ASAP) Which chemical equation models the law of conservation of mass?
    5·2 answers
  • A copper wire loop has a circular shape, with a radius a (see below). The loop is put perpendicularly to the uniform magnetic fi
    15·1 answer
  • GIVING BRAINLIEST PLEASE HELP!!
    9·1 answer
  • A box is at rest on a slope
    11·1 answer
  • Examine table b then read the statement and choose the best responses. as the resistance increased under 25 v, the current . com
    14·2 answers
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!