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lesantik [10]
3 years ago
9

Polonium, the Period 6 member of Group 6A(16), is a rare radioactive metal that is the only element with a crystal structure bas

ed on the simple cubic unit cell. If its density is 9.232 g/cm3, calculate an approximate atomic radius for polonium―209.
Physics
1 answer:
Burka [1]3 years ago
7 0

Answer:

Approximate atomic radius for polonium-209 is 167.5 pm .

Explanation:

Number of atom in simple cubic unit cell = Z = 1

Density of platinum = 9.232 g/cm^3

Edge length of cubic unit cell= a = ?

Atomic mass of Po (M) = 209 g/mol

Formula used :  

\rho=\frac{Z\times M}{N_{A}\times a^{3}}

where,

ρ = density

Z = number of atom in unit cell

M = atomic mass

(N_{A}) = Avogadro's number  

a = edge length of unit cell

On substituting all the given values , we will get the value of 'a'.

9.232 g/cm3=\frac{1 \times 209 g/mol}{6.022\times 10^{23} mol^{-1}\times (a)^{3}}

a = 3.35\times 10^{-8} cm

Atomic radius of the polonium in unit cell = r

r = 0.5a

r=0.5\times 3.35\times 10^{-8} cm=1.675\times 10^{-8} cm

1 cm = 10^{10} pm

1.675\times 10^{-8} cm=1.675\times 10^{-8}\times 10^{10}=167.5 pm

Approximate atomic radius for polonium-209 is 167.5 pm.

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6 0
2 years ago
Gibbons, small Asian apes, move by brachiation, swinging below a handhold to move forward to the next handhold. A 9.4 kg gibbon
Katarina [22]

Answer:

upward force acting = 261.6 N

Explanation:

given,

mass of gibbon = 9.4 kg

arm length = 0.6 m

speed of the swing

net force must provide

F_{branch} + F_{gravity}=F_{centripetal}

force of gravity = - mg

F_{branch}=F_{centripetal}-F_{gravity}

                        = \dfrac{mv^2}{r} + mg

                        = m(\dfrac{3.4^2}{0.6} +9.8)

                        =9 x 29.067

                        = 261.6 N

upward force acting = 261.6 N

7 0
2 years ago
Please help me with my physics! I do not understand and please provide work.
Marrrta [24]
26) A) 400J as GPE = mgh
27) C) Both obj has same PE
28) B) It decreases as energy is not being provided.
29) C) (42 - 0)/7 = 6m/s2
5 0
3 years ago
Read 2 more answers
A rugby player sits on a scrum machine that weighs 200 Newtons. Given that the coefficient of static friction is 0.67, the coeff
Trava [24]

a. 850 N is the minimum force needed to get the machine/player system moving, which means this is the maximum magnitude of static friction between the system and the surface they stand on.

By Newton's second law, at the moment right before the system starts to move,

• net horizontal force

∑ F[h] = F[push] - F[s. friction] = 0

• net vertical force

∑ F[v] = F[normal] - F[weight] = 0

and we have

F[s. friction] = µ[s] F[normal]

It follows that

F[weight] = F[normal] = (850 N) / (0.67) = 1268.66 N

where F[weight] is the combined weight of the player and machine. We're given the machine's weight is 200 N, so the player weighs 1068.66 N and hence has a mass of

(1068.66 N) / g ≈ 110 kg

b. To keep the system moving at a constant speed, the second-law equations from part (a) change only slightly to

∑ F[h] = F[push] - F[k. friction] = 0

∑ F[v] = F[normal] - F[weight] = 0

so that

F[k. friction] = µ[k] F[normal] = 0.56 (1268.66 N) = 710.45 N

and so the minimum force needed to keep the system moving is

F[push] = 710.45 N ≈ 710 N

4 0
1 year ago
Can someone answer this 2 correctly please
ahrayia [7]

Answer:

I know the first one is C.) 4J. I don't know of the answer for the second oneis suppose to be in N/m form? but I got

2,500N/m

5 0
3 years ago
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