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Daniel [21]
3 years ago
15

When water diffuses across a semi permeable membrane the process is called __________.

Physics
1 answer:
expeople1 [14]3 years ago
6 0
Osmosis !!!!!!!!!!!!!!!!!!
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An olympic high diver has gravitational potential energy because of her height. as she dives, what becomes of her energy just be
sammy [17]

An Olympic high diver has gravitational potential energy because of her height. As she dives, kinetic energy becomes of her energy just before she hits the water.

Gravitational potential energy is the energy possessed or acquired by an object due to a change in its position when it is present in a gravitational field. In simple terms, it can be said that gravitational potential energy is an energy that is related to gravitational force or to gravity.

Kinetic energy is the energy of motion, observable as the movement of an object, particle, or set of particles.

When the high diver is standing stable and not moving , that diver has a gravitational potential energy because of the height . The moment she dives , before hitting the water , from being stationary she gained some momentum and come in motion , due to motion her gravitational potential energy will change to kinetic energy before hitting the ground.

To learn more about Gravitational potential energy  here

brainly.com/question/15978356

#SPJ4

6 0
1 year ago
Two identical cylindrical vessels with their bases at the same level each contain a liquid of density 1.23 g/cm3. The area of ea
motikmotik

Explanation:

Work done by gravity is given by the formula,

           W = \rho A (h_{1} - h)g (h - h_{2}) ......... (1)

It is known that when levels are same then height of the liquid is as follows.

           h = \frac{h_{1} + h_{2}}{2} ......... (2)

Putting value of equation (2) in equation (1) the overall formula will be as follows.

       W = \frac{1}{4} \rho gA(h_{1} - h_{2})^{2})

           = \frac{1}{4} \times 1.23 g/cm^{3} \times 9.80 m/s^{2} \times 3.89 \times 10^{-4} m^{2}(1.76 m - 0.993 m)^{2})

           = 0.689 J

Thus, we can conclude that the work done by the gravitational force in equalizing the levels when the two vessels are connected is 0.689 J.

3 0
3 years ago
Which particle of the atom has a negative charge?
lilavasa [31]
D, electron, the nucleus is not a single particle to begin with, the proton has a positive charge, a neutron has a neutral charge or no charge, and an electron has a negative charge
3 0
3 years ago
A photon ionizes a hydrogen atom from the ground state. The liberated electron 11. recombines with a proton into the first excit
anygoal [31]

Answer:

a) 23.2 e V

b) energy of the original photon is 36.8 eV

Explanation:

given,

energy at ground level = -13.6 e V

energy at first exited state = - 3.4 e V

A photon of energy ionized from ground state and electron of energy K is released.

h ν₁ - 13.6 = K

K combine with photon in first exited state giving out photon of energy

h\nu_2 =\dfrac{hc}{\lambda}=\dfrac{12400}{466}

            = 26.6 e V

h c = 6.626 ×  10⁻³⁴ ×  3  × 10⁸  = 12400 e V A°

K + ( 3.4 ) = 26.6 e V

a) energy of free electron

K = 26.6 - 3.4 = 23.2 e V

b) energy of the original photon

h ν₁ - 13.6 = K

h ν₁  = 23.2 + 13.6

       = 36.8 e V

energy of the original photon is 36.8 eV

3 0
3 years ago
A ball of mass M collides with a stick with moment of inertia I = βml2 (relative to its center, which is its center of mass). Th
ZanzabumX [31]

Answer:

Part a)

v_2 = \frac{\frac{2\beta mL^2v_o}{d}}{(md + \frac{\beta mL^2}{d}(1 + \frac{m}{M})}

Part b)

v_1 = v_0 - \frac{m}{M}(\frac{\frac{2\beta mL^2v_o}{d}}{(md + \frac{\beta mL^2}{d}(1 + \frac{m}{M})})

Explanation:

Since the ball and rod is an isolated system and there is no external force on it so by momentum conservation we will have

Mv_o = M v_1 + m v_2

here we also use angular momentum conservation

so we have

M v_o d = M v_1 d + \beta mL^2 \omega

also we know that the collision is elastic collision so we have

v_o = (v_2 + d\omega) - v_1

so we have

\omega = \frac{v_o + v_1 - v_2}{d}

also we know

M v_o d - M v_1 d = \beta mL^2(\frac{v_o + v_1 - v_2}{d})

also we know

v_1 = v_o - \frac{m}{M}v_2

so we have

M v_o d - M(v_o - \frac{m}{M}v_2)d = \beta mL^2(\frac{v_o + v_o - \frac{m}{M}v_2 - v_2}{d})

mv_2 d = \beta mL^2\frac{2v_o}{d} - \beta mL^2(1 + \frac{m}{M})\frac{v_2}{d}

now we have

(md + \frac{\beta mL^2}{d}(1 + \frac{m}{M})v_2 = \frac{2\beta mL^2v_o}{d}

v_2 = \frac{\frac{2\beta mL^2v_o}{d}}{(md + \frac{\beta mL^2}{d}(1 + \frac{m}{M})}

Part b)

Now we know that speed of the ball after collision is given as

v_1 = v_o - \frac{m}{M}v_2

so it is given as

v_1 = v_0 - \frac{m}{M}(\frac{\frac{2\beta mL^2v_o}{d}}{(md + \frac{\beta mL^2}{d}(1 + \frac{m}{M})})

3 0
3 years ago
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