Questions 8 out of 20

What are examples of a solution in solids, liquids, and gases

Nina [5.8K]

Answer:

Solid: metal alloy

Liquid: beer

Gas: Air

Explanation:

A solution is a type of mixture where the solvent and solute are homogeneously mixed. Homogeneous mixture means that the solute shouldn't be able to be seen with the naked eye, filtered and stable enough.

Metal alloy will be an example of a solution in solid-state. Beer is a solution made of liquid alcohol and liquid water. Air mostly composed of nitrogen, but it has oxygen, carbon dioxide, and many other substances in gaseous form.

3 0

2 years ago

The current produced in homes is ______ current.

Lelu [443]

If any current is PRODUCED in an individual home, like with
a windmill or solar panels, it is direct current (DC).

The current SUPPLIED to homes by the local utility company
is alternating current (AC).

8 0

2 years ago

Read 2 more answers

Which wave has a greater frequency

larisa86 [58]

Answer:

A I think

Explanation:

because what is the most frequency a because it has more frequency I think I'm not that sure

5 0

2 years ago

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X rays of wavelength 0.0169 nm are directed in the positive direction of an x axis onto a target containing loosely bound electr

mamaluj [8]

Answer:

a) 4.04*10^-12m

b) 0.0209nm

c) 0.253MeV

Explanation:

The formula for Compton's scattering is given by:

$\Delta \lambda=\lambda_f-\lambda_i=\frac{h}{m_oc}(1-cos\theta)$

where h is the Planck's constant, m is the mass of the electron and c is the speed of light.

a) by replacing in the formula you obtain the Compton shift:

$\Delta \lambda=\frac{6.62*10^{-34}Js}{(9.1*10^{-31}kg)(3*10^8m/s)}(1-cos132\°)=4.04*10^{-12}m$

b) The change in photon energy is given by:

$\Delta E=E_f-E_i=h\frac{c}{\lambda_f}-h\frac{c}{\lambda_i}=hc(\frac{1}{\lambda_f}-\frac{1}{\lambda_i})\\\\\lambda_f=4.04*10^{-12}m +\lambda_i=4.04*10^{-12}m+(0.0169*10^{-9}m)=2.09*10^{-11}m=0.0209nm$

c) The electron Compton wavelength is 2.43 × 10-12 m. Hence you can use the Broglie's relation to compute the momentum of the electron and then the kinetic energy.

$P=\frac{h}{\lambda_e}=\frac{6.62*10^{-34}Js}{2.43*10^{-12}m}=2.72*10^{-22}kgm\\$

$E_e=\frac{p^2}{2m_e}=\frac{(2.72*10^{-22}kgm)^2}{2(9.1*10^{-31}kg)}=4.06*10^{-14}J\\\\1J=6.242*10^{18}eV\\\\E_e=4.06*10^{-14}(6.242*10^{18}eV)=0.253MeV$

5 0

3 years ago

A 20-kg child running at 1.4 m/s jumps onto a playground merry-go-round that has mass 180 kg and radius 1.6m. She is moving tang

Dominik [7]

Answer:

ωf = 0.16 rad/s

Explanation:

Moment of inertia of the child = mr² = 20(1.6²) = 51.2 kg•m²

Moment of Inertia of the MGR = ½mr² = ½(180)1.6² = 230.4 kg•m²

(ASSUMING it is a uniform disk)

Initial angular momentum of the child = Iω = I(v/r) = 51.2(1.4/1.6) = 44.8 kg•m²/s

Conservation of angular momentum

44.8 = (51.2 + 230.4)ωf

ωf = 0.15909090...

4 0

2 years ago