How many atoms are in each product silver sulfide

If a nucleus was as big as a nonpareil, an atom would be ____

Mamont248 [21]

Answer:

small marble

Explanation:

A nonpareil are confectionery tiny ball items that are made up of starch and sugar. It is very small of the size of sugar crystal or sand grains. They were the miniature version of the comfits. They are generally opaque white but bow available in all colors.

In the context, if the nucleus is compared to the size of a nonpareil then its atom would be of the size of small size marble. An atom is bigger in size than that of nucleus as the nucleus is located inside the atoms.

6 0

2 years ago

Explain how all the organelles in a cell model work together to provide nutrition and energy for the cell.

slega [8]

These organelles are like the organs in a human and they help the cell stay alive. Each organelle has it's own specific function to help the cell survive. The nucleus of a eukaryotic cell directs the cell's activities and stores DNA. Eukaryotes also have a golgi apparatus that packages and distributes proteins.

6 0

3 years ago

estimate the force that a cyclist travelling at 6m/s needs to apply to the brakes to bring the bicycle to a stop in a distance o

RSB [31]

v1 = 6m/s

v2 = 0

∆v = v1 - v2 = 6m\s

s = t * v = 15m

t = s\v1 = 15(m) \ 6(m\s) = 2.5s

a = ∆v\t = 6(m\s) \ 2.5s = 2.4m\s2

a = F\m = 2.4m\s2

F = a * m = 2.4m\s2 * ?kg

I can't tell you this because I don't know the mass of this cyclist

4 0

3 years ago

What is the kinetic energy of a 12 kg ball moving at a speed of 15 m/s?

vivado [14]

Answer:

1350 N

Explanation:

KE = ½mv²

KE = ½(12)(15²)

KE = 1350 N

3 0

2 years ago

8-14 A Cu-30% Zn alloy tensile bar has a strain-hardening coefficient of 0.50. The bar, which has an initial diameter of 1 cm an

77julia77 [94]

Answer:

The true stress at true strain 0.05cm/cm is 80MPa

Explanation:

Given that

the strength coefficient is K

true strain is ε

strain hardening exponent is n

initial diameter of bar is d = 1cm, (10mm)

tensile force is F

engineering stress(S) = 120

the engineering stress(S) = $\frac{F}{\frac{\pi }{4}(d^2) }$

To find force (F) =

120 = $\frac{F}{\frac{\pi }{4}(100^{2} )}$

F = 120 * (π/4) * (100)

F = 9425N

Calculate the true strain (ε) = In (l₀ / l₁)

where

l₀ = initial length of the metallic bar = 3cm

l₁ = final length of metallic bar = 3.5cm

ε = In (3.5 / 3)

= In 1.1667

= 0.154cm/cm

Calculate the true stress (σ) at fracture point

= $\frac{F}{\frac{\pi }{4}(d^2) }}$

tensile force is F and final diameter of bar is d₁ (d in the eqn)

Substitute 9425 N for F and 0.926 cm (9.26mm) for d₁ (d in the eqn)

σ = $\frac{9425}{\frac{\pi }{4}(9.26^2) }}$

= 140MPa

To find the strength coefficient (K) of the material bar

K = $\frac{140}{\sqrt{0.154} }$

K = $\frac{140}{0.3925}$

= 356.75MPa

To calculate the true stress σ true strain of 0.05cm/cm

K = 356.75MPa

σ = $356.75(0.05)^0^.^5$

= $356.75 ( 0.2236)$

= 80MPa

The true stress at true strain 0.05cm/cm is 80MPa

6 0

3 years ago

How many atoms are in each product silver sulfide ​

How many atoms are in each product silver sulfide