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3 years ago

Describe how to prepare a 10.0% w/v solution of salt in water in a 100 mL volumetric flask

1 answer:

6 0

To prepare a 10.0% w/v solution of salt in water in a 100 mL volumetric flask, first you must weigh 10 g of salt because the 10 % 100 is 10 and the given should be 10 % w/v. place the 10 g salt to the volumetric flask then add water up until to mark of the volumetric flask then shake it.

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A ball is projected into the air with 100 j of kinetic energy which is transformed to gravitational potential energy at the top

raketka [301]

<span>when it returns to its original level after encountering air resistance, its kinetic energy is decreased.
In fact, part of the energy has been dissipated due to the air resistance.

The mechanical energy of the ball as it starts the motion is:
</span> $E=K = 100 J$
<span>where K is the kinetic energy, and where there is no potential energy since we use the initial height of the ball as reference level.
If there is no air resistance, this total energy is conserved, therefore when the ball returns to its original height, the kinetic energy will still be 100 J. However, because of the presence of the air resistance, the total mechanical energy is not conserved, and part of the total energy of the ball has been dissipated through the air. Therefore, when the ball returns to its original level, the kinetic energy will be less than 100 J.</span>

3 0

3 years ago

If a 2kg ball rolls down a ramp that is 15 meters long in 25 seconds, what is the

77julia77 [94]

Answer:

1.968504 ft/s

7 0

2 years ago

A spring-mass system has a spring constant of 3 Nm. A mass of 2 kg is attached to the spring, and the motion takes place in a vi

frosja888 [35]

Answer:

The answer to the question

The steady state response is u₂(t) = - $\frac{3\sqrt{2} }{2}$ cos(3t + π/4)

of the form R·cos(ωt−δ) with R = $-\frac{3\sqrt{2} }{2}$ , ω = 3 and δ = -π/4

Explanation:

To solve the question we note that the equation of motion is given by

m·u'' + γ·u' + k·u = F(t) where

m = mass = 2.00 kg

γ = Damping coefficient = 1

k = Spring constant = 3 N·m

F(t) = externally applied force = 27·cos(3·t)−18·sin(3·t)

Therefore we have 2·u'' + u' + 3·u = 27·cos(3·t)−18·sin(3·t)

The homogeneous equation 2·u'' + u' + 3·u is first solved as follows

2·u'' + u' + 3·u = 0 where putting the characteristic equation as

2·X² + X + 3 = 0 we have the solution given by $\frac{-1+/-\sqrt{23} }{4} \sqrt{-1}$ = $\frac{-1+/-\sqrt{23} }{4} i$

This gives the general solution of the homogeneous equation as

u₁(t) = $e^{(-1/4t)} (C_1cos(\frac{\sqrt{23} }{4}t) + C_2sin(\frac{\sqrt{23} }{4}t)$

For a particular equation of the form 2·u''+u'+3·u = 27·cos(3·t)−18·sin(3·t) which is in the form u₂(t) = A·cos(3·t) + B·sin(3·t)

Then u₂'(t) = -3·A·sin(3·t) + 3·B·cos(3·t) also u₂''(t) = -9·A·cos(3·t) - 9·B·sin(3·t) from which 2·u₂''(t)+u₂'(t)+3·u₂(t) = (3·B-15·A)·cos(3·t) + (-3·A-15·B)·sin(3·t). Comparing with the equation 27·cos(3·t)−18·sin(3·t) we have

3·B-15·A = 27

3·A +15·B = 18

Solving the above linear system of equations we have

A = -1.5, B = 1.5 and u₂(t) = A·cos(3·t) + B·sin(3·t) becomes 1.5·sin(3·t) - 1.5·cos(3·t)

u₂(t) = 1.5·(sin(3·t) - cos(3·t) = $-\frac{3\sqrt{2} }{2}$ ·cos(3·t + π/4)

The general solution is then u(t) = u₁(t) + u₂(t)

however since u₁(t) = $e^{(-1/4t)} (C_1cos(\frac{\sqrt{23} }{4}t) + C_2sin(\frac{\sqrt{23} }{4}t)$ ⇒ 0 as t → ∞ the steady state response = u₂(t) = $-\frac{3\sqrt{2} }{2}$ ·cos(3·t + π/4) which is of the form R·cos(ωt−δ) where

R = $-\frac{3\sqrt{2} }{2}$

ω = 3 and

δ = -π/4

8 0

3 years ago

What are examples of devices that use electromagnetic waves? Check all that apply.

yKpoI14uk [10]

Well, there aren't actually ANY that apply, because you haven't listed any. But I've lived a long time, and I remember hearing about electromagnetic waves and things that use them, so I can list a FEW of them for you:

-- radios

-- TVs

-- garage-door openers

-- TV remotes

-- cell phones

-- smart phones

-- GPS

-- walkie-talkies

-- car headlights

-- lava lamps

-- toasters

-- LEDs

-- light bulbs

-- fluorescent light tubes

-- police radios

-- Doppler weather radars

-- CB radios

-- ham radios

-- neon signs

-- eyeglasses

-- microscopes

-- telescopes

-- gas stoves

-- electric stoves

-- wood stoves

-- microwave ovens

-- tanning beds

-- cameras

-- lasers

-- CD recorders and players

-- DVD recorders and players

-- Bluray recorders and players

-- movie cameras

-- movie projectors

-- reading lamps

-- candles

-- whale-oil lamps

-- kerosene lanterns

-- flashlights

-- campfires

-- coffee percolators

-- heat lamps

-- cordless phones

These are just the ones I can think of right now off the top of my head. There are a lot of others.

5 0

3 years ago

Read 2 more answers

The emission of x rays can be described as an inverse photoelectric effect.

timama [110]

Answer:

The potential difference through which an electron accelerates to produce x rays is $1.24\times 10^4\ volts$ .

Explanation:

It is given that,

Wavelength of the x -rays, $\lambda=0.1\ nm=0.1\times 10^{-9}\ m$

The energy of the x- rays is given by :

$E=\dfrac{hc}{\lambda}$

The energy of an electron in terms of potential difference is given by :

$E=eV$

So,

$\dfrac{hc}{\lambda}=eV$

V is the potential difference

e is the charge on electron

$V=\dfrac{hc}{e\lambda}$

$V=\dfrac{6.63\times 10^{-34}\times 3\times 10^8}{1.6\times 10^{-19}\times 0.1\times 10^{-9}}$

V = 12431.25 volts

$V=1.24\times 10^4\ volts$

So, the potential difference through which an electron accelerates to produce x rays is $1.24\times 10^4\ volts$ . hence, this is the required solution.

4 0

2 years ago