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3 years ago

11

Which could result from untreated diabetes check all that apply A .Asthma B. Numbness in hands and feet C.Exhaustion D.Weight lo

ss E.Blurred vision

2 answers:

soldier1979 [14.2K]3 years ago

7 0

Answer:

B. Numbness In Hands and Feet

C. Exhaustion

D. Weight Loss

E. Blurred Vision

grigory [225]3 years ago

3 0

With the exception of Asthma, all of those things, and a lot more that you don't want, could result from untreated diabetes.

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A sound wave has a speed of 345 m/s and a wavelength of 4.15 meters. What is the frequency of this wave?

Debora [2.8K]

Answer:

83 hertz

OPTION A

Explanation:

$wavelength = \frac{speed}{frequency}$

Given

wavelength=4.15m

speed=345 m/s

$frequency = \frac{speed}{wavelength} = \frac{345}{4.15}$

$frequency = 83 \: hertz$

I hope it helped you

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2 years ago

The atmosphere protects us from all of the following except Meteorites The sun's rays Pollution Wind

denis23 [38]

Meteorites is the answer

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3 years ago

What is the energy in joules of a mole of photons associated with red light of wavelength 7.00 × 102 nm?

konstantin123 [22]

<span>The energy of a single photon is given by E = hc/lambda, where h is Planck's constant, c is the speed of light, and lambda is the wavelength. Plugging the values in gives E = 6.63E-34 x 3.00E8 / 700E-9 = 2.84E-19 Joules Now one mole of substance is equivalent to 6.02E23 particles, so one mole of these photons will be: 2.84E-19 x 6.02E23 = 1.71E5 Joules</span>

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3 years ago

Clarify this statement acceleration due to gravity of the Earth is 9.8m/s square

muminat

Answer:

it means that velocity of a body rises by 9.8m/s each second if the air resistance is nrelated

mark me

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3 years ago

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