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Romashka-Z-Leto [24]
4 years ago
5

How does solubility change with temperature changes?

Physics
1 answer:
matrenka [14]4 years ago
8 0

Answer:

Solubility increases with temperature.

Explanation:

-Increasing temperatures, increases the kinetic energy that allows solvent molecules to more effectively break apart the solute molecules.

-This solute molecules are held together by intermolecular attractions which weakens in higher temperatures.

-Solubility therefore increases with increasing temperatures.

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In an RLC series circuit that includes a source of alternating current operating at fixed frequency and voltage, the resistance
maw [93]

Answer:

Capacitive Reactance is 4 times of resistance

Solution:

As per the question:

R = X_{L} = j\omega L = 2\pi fL

where

R = resistance

X_{L} = Inductive Reactance

f = fixed frequency

Now,

For a parallel plate capacitor, capacitance, C:

C = \frac{\epsilon_{o}A}{x}

where

x = separation between the parallel plates

Thus

C ∝ \frac{1}{x}

Now, if the distance reduces to one-third:

Capacitance becomes 3 times of the initial capacitace, i.e., x' = 3x, then C' = 3C and hence Current, I becomes 3I.

Also,

Z = \sqrt{R^{2} + (X_{L} - X_{C})^{2}}

Also,

Z ∝ I

Therefore,

\frac{Z}{I} = \frac{Z'}{I'}

\frac{\sqrt{R^{2} + (R - X_{C})^{2}}}{3I} = \frac{\sqrt{R^{2} + (R - \frac{X_{C}}{3})^{2}}}{I}

{R^{2} + (R - X_{C})^{2}} = 9({R^{2} + (R - \frac{X_{C}}{3})^{2}})

{R^{2} + R^{2} + X_{C}^{2} - 2RX_{C} = 9({R^{2} + R^{2} + \frac{X_{C}^{2}}{9} - 2RX_{C})

Solving the above eqn:

X_{C} = 4R

6 0
3 years ago
A diver leaves the end of a 4.0 m high diving board and strikes the water 1.3s later, 3.0m beyond the end of the board. Consider
shutvik [7]

Answer:

4.0 m/s

Explanation:

The motion of the diver is the motion of a projectile: so we need to find the horizontal and the vertical component of the initial velocity.

Let's consider the horizontal motion first. This motion occurs with constant speed, so the distance covered in a time t is

d=v_x t

where here we have

d = 3.0 m is the horizontal distance covered

vx is the horizontal velocity

t = 1.3 s is the duration of the fall

Solving for vx,

v_x = \frac{d}{t}=\frac{3.0 m}{1.3 s}=2.3 m/s

Now let's consider the vertical motion: this is an accelerated motion with constant acceleration g=9.8 m/s^2 towards the ground. The vertical position at time t is given by

y(t) = h + v_y t - \frac{1}{2}gt^2

where

h = 4.0 m is the initial height

vy is the initial vertical velocity

We know that at t = 1.3 s, the vertical position is zero: y = 0. Substituting these numbers, we can find vy

0=h+v_y t - \frac{1}{2}gt^2\\v_y = \frac{0.5gt^2-h}{t}=\frac{0.5(9.8 m/s^2)(1.3 s)^2-4.0 m}{1.3 s}=3.3 m/s

So now we can find the magnitude of the initial velocity:

v=\sqrt{v_x^2+v_y^2}=\sqrt{(2.3 m/s)^2+(3.3 m/s)^2}=4.0 m/s

4 0
4 years ago
The work function for tungsten metal is 4.52eV a. What is the cutoff (threshold) wavelength for tungsten? b. What is the maximum
Tanya [424]

Answer: a) 274.34 nm; b) 1.74 eV c) 1.74 V

Explanation: In order to solve this problem we have to consider the energy balance for the photoelectric effect on tungsten:

h*ν = Ek+W ; where h is the Planck constant, ek the kinetic energy of electrons and W the work funcion of the metal catode.

In order to calculate the cutoff wavelength we have to consider that Ek=0

in this case  h*ν=W

(h*c)/λ=4.52 eV

λ= (h*c)/4.52 eV

λ= (1240 eV*nm)/(4.52 eV)=274.34 nm

From this h*ν = Ek+W;  we can calculate the kinetic energy for a radiation wavelength of 198 nm

then we have

(h*c)/(λ)-W= Ek

Ek=(1240 eV*nm)/(198 nm)-4.52 eV=1.74 eV

Finally, if we want to stop these electrons we have to applied a stop potental equal to 1.74 V . At this potential the photo-current drop to zero. This potential is lower to the catode, so this  acts to slow down the ejected electrons from the catode.

5 0
3 years ago
According to Hooke's law
SOVA2 [1]

Answer:

According to Hook's law, we know,

strain/stress =Constant

Explanation: So, the ratio between stress and strain is always constant.

So, if stress is increased, then strain changes in that way so that this ratio always remains constant.

7 0
3 years ago
Describe the motion of an automobile on an east-west
Tanya [424]

Answer:

1. The automobile is traveling due east and is speeding up.

2. The car is traveling due east and is is slowing down.

3. The automobile is traveling due east at a constant speed.

4. The car is traveling due west and is slowing down.

5. The automobile is traveling due west and is speeding up.

6. The automobile is traveling due west at a constant speed.

7. The automobile is accelerating due east from rest.

8. The automobile is accelerating due west from rest.

Explanation:

The key to understanding this is:

When the acceleration and initial velocity of the automobile have the same sign (positive or negative) then the automobile is speeding up. Explained further, if acceleration and the initial velocity are both positive or they are both negative the automobile is speeding up but whenever they have opposite signs (that is acceleration is positive and initial velocity is negative or vice versa) the automobile is slowing down. When the acceleration is zero the automobile is maintaining a unform motion at a constant speed (the speed is not changing with time). The + or - sign indicates the direction of travel. In this case east is + and west is -. It is my pleasure answering this question. I hope you find it helpful. Thank you.

4 0
3 years ago
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