<span>The choices can be found elsewhere and as follows:
</span><span>a. they are so small that they stay close to the ground due to the attractive properties of charged soil particles.
b. they are easily carried by the wind.
c. they easily dissolve in liquid droplets.
d. it is easier for then to roll along the small crevices in the ground.</span><span>
</span>I think the correct answer from the choices listed above is option B. Only the smallest particles of soil can be displaced by suspension because they are so small that they are easily carried by the wind. Hope this answers the question. Have a nice day. Feel free to ask more questions.
Answer: 86.47 g of carbon-14 must have been present in the sample 11,430 years ago.
Explanation:
Half-life of sample of carbon -14= 5,730 days

Let the sample present 11,430 years(t) ago = 
Sample left till today ,N= 0.060 g

![ln[N]=ln[N]_o-\lambda t](https://tex.z-dn.net/?f=ln%5BN%5D%3Dln%5BN%5D_o-%5Clambda%20t)
![\log[0.060 g]=\log[N_o]-2.303\times 0.00012 day^{-1}\times 11,430 days](https://tex.z-dn.net/?f=%5Clog%5B0.060%20g%5D%3D%5Clog%5BN_o%5D-2.303%5Ctimes%200.00012%20day%5E%7B-1%7D%5Ctimes%2011%2C430%20days)
![\log[N_o]=1.9369](https://tex.z-dn.net/?f=%5Clog%5BN_o%5D%3D1.9369)

86.47 g of carbon-14 must have been present in the sample 11,430 years ago.
Explanation:
Geocentric model said that the Earth is at the center of the universe and everything revolves around it. It was considered to be stationary. Galileo proved this model incorrect with the help of his astronomical observations. Some of the key observation that he used to support the heliocentric model were:
1. He proposed the theory that the tides on the Earth occur because of its motion.
2. He observed the phases of the Venus which meant that the Venus revolved around the Sun and not the Earth.
3. He observed other planets and thus noted that they also move around the Sun and not Earth.
4. He discovered the Moons of other planets.
Answer:
12mph in 2hrs and 3mph in 0.5hrs the total distance would be 12*2 and 3*0.5 which would be 24 and 1.5 so we add those 24+1.5= 25.5. The answer would be 25.5
Answer:
magnification is - 159
objective distance is 3.85 cm
Explanation:
Given data
focal length f1 = 1.40 cm
focal length f2 = 2.20 cm
separated d = 19.6 cm
to find out
angular magnification and How far from the objective
solution
we know magnification formula that is
magnification = ( - L / f1 ) (D/f2)
here D = 25 cm put all value
magnification = ( - 19.6 / 1.40 ) (25/2.20)
magnification = - 159
and
now we apply lens formula
i/f = 1/q + 1/p
p = f2 = 2.20
so
q = f2 p / p -f2
q = 1.4(2.20) / ( 2.2 - 1.4 )
q = 3.85 cm
so objective distance is 3.85 cm