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2 years ago

A solution is oversaturated with solute. which could be done to decrease the oversaturation?

Physics

2 answers:

Rzqust [24]2 years ago

5 0

Answer: The correct awnser is B add more solute

Explanation:

Grace [21]2 years ago

4 0

A solution is oversaturated with solute. The thing that could be done to decrease the oversaturation is to add more solvent in order to decrease the concentration of the solute. You can also increase the temperature to increase solubility of the solute. Hope this answers the question.

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A classical estimate of the vibrational frequency is ff = 7.0×10137.0×1013 HzHz. The mass of a hydrogen atom differs little from

vlada-n [284]

Answer:

The force constant of the spring is 317.8 N/m.

Explanation:

Given that,

Frequency $f=7.0\times10^{13}\ Hz$

We need to calculate the reduced mass

Using formula of reduced mass

$\mu=\dfrac{m_{H}m_{I}}{m_{H}+m_{I}}$

Where, $m_{H}$ = atomic mass of H

$m_{I}$ = atomic mass of I

Put the value into the formula

$\mu=\dfrac{1\times126.9}{1+126.9}$

$\mu=0.99\ u$

$\mu=0.99\times1.66\times10^{-27}\ Kg$

$\mu=1.643\times10^{-27}\ kg$

We need to calculate the force constant of the spring

Using formula of frequency

$f=\dfrac{1}{2\pi}\times\sqrt{\dfrac{k}{\mu}}$

$k=f^2\times 4\pi^2\times\mu$

Put the value into the formula

$k=(7.0\times10^{13})^2\times4\pi^2\times1.643\times10^{-27}$

$k=317.8\ N/m$

Hence, The force constant of the spring is 317.8 N/m.

0 0

2 years ago

A boy is pulling a load of 150N with a string inclined at an angle 30 to the horizontal if the tension of string is 105N the for

Lorico [155]

The force tending to lift the load (vertical force) is equal to 22.5N.

Why?

Since the boy is pulling a load (150N) with a string inclined at an angle of 30° to the horizontal, the total force will have two components (horizontal and vertical component), but we need to consider the given information about the tension of the string which is equal to 105N.

We can calculate the vertical force using the following formula:

$VerticalForce=Force*Sin(30\° )=(BoysForce-StringForce)*\frac{1}{2}\\\\VerticalForce=(150N-105N)*\frac{1}{2}=VerticalForce=45N*\frac{1}{2}=22.5N$

Hence, we can see that the force tending to lift the load off the ground (vertical force) is equal to 22.5N.

Have a nice day!

8 0

2 years ago

A huge (essentially infinite) horizontal nonconducting sheet 10.0 cm thick has charge uniformly spread over both faces. The uppe

Nonamiya [84]

Answer:

6.78 X 10³ N/C

Explanation:

Electric field near a charged infinite plate

= surface charge density / 2ε₀

Field will be perpendicular to the surface of the plate for both the charge density and direction of field will be same so they will add up.

Field due to charge density of +95.0 nC/m2

E₁ = 95 x 10⁻⁹ / 2 ε₀

Field due to charge density of -25.0 nC/m2

E₂ = 25 x 10⁻⁹ / 2ε₀

Total field

E = E₁ + E₂

= 95 x 10⁻⁹ / 2 ε₀ + 25 x 10⁻⁹ / 2ε₀

= 6.78 X 10³ N/C

4 0

2 years ago

You set a tuning fork into vibration at a frequency of 723 Hz and then drop it off the roof of the Physics building where the ac

zaharov [31]

Answer:

Explanation:

Given

Original Frequency $f=723\ Hz$

apparent Frequency $f'=697\ Hz$

There is change in frequency whenever source move relative to the observer.

From Doppler effect we can write as

$f'=f\cdot \frac{v-v_o}{v+v_s}$

where

$f'=$ apparent frequency

v=velocity of sound in the given media

$v_s=$ velocity of source

$v_0=$ velocity of observer

here $v_0=0$

$697=723\cdot (\frac{343-0}{343+v_s})$

$v_s=(\frac{f}{f'}-1)v$

$v_s=(\frac{723}{697}-1)\cdot 343$

$v_s=12.79\approx 12.8\ m/s$

i.e.fork acquired a velocity of $12.8 m/s$

distance traveled by fork is given by

$v^2-u^2=2as$

where v=final velocity

u=initial velocity

a=acceleration

s=displacement

$v_s^2-0=2\times 9.8\times s$

$s=\frac{12.8^2}{2\times 9.8}$

$s=8.35\ m$

5 0

2 years ago

Adam is teeing off on hole number two. The hole is 390 yards away. It is a par four hole. What club should he use to tee off? Ex

kkurt [141]

Answer:

A driver.

Explanation:

Using a driver while at least 350 yds away is better than using a iron, because it will be a waste of the par 4 as it is not as powerful as the driver.

3 0

2 years ago