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Novosadov [1.4K]
3 years ago
13

The half life of a radioactive element is 8 hours. A sample of the element is tested and found to contain 6g of the element. How

much of the element was present 32 hours before the element was tested? 0.375g 6g 48g 96g
Physics
1 answer:
Kay [80]3 years ago
4 0

Answer:

m₀ = 96 g

Explanation:

First, we will calculate the no. of half-lives passed:

n = \frac{T}{T_{1/2}}

where,

n = no. of half-lives passed = ?

T = Total time elapsed = 32 h

T_{1/2} = Half-Life = 8 h

Therefore,

n = \frac{32\ h}{8\ h}\\\\n = 4

Now, for the initial amount of element:

m = \frac{m_o}{2^n}\\\\m_o = m(2^n)

where,

m₀ = initial amount of element = ?

m = current amount of element = 6 g

Therefore,

m_o = (6\ g)(2^4)

m₀ = (6 g)(16)

<u>m₀ = 96 g</u>

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Answer:

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Explanation:

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Frequency of a wave is inversely proportional to period of oscillation of the wave. The higher the frequency of a wave, the shorter the period of oscillation.  Gamma ray has the highest wave frequency in electromagnetic spectrum and shorter period of oscillation, thereby causing it to have the highest penetration power.

7 0
3 years ago
The strength of the force of friction depends on which two factors?
Stolb23 [73]

Options A and D are correct. The strength of the force of friction depends on the objects' sizes and weights and the heat generated by the friction and the types of surfaces involved.

<h3 /><h3>What is the friction force?</h3>

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Mathematically, it is defined as the product of the coefficient of friction and normal reaction.

On resolving the given force and acceleration in the different components and balancing the equation gets. Components in the x-direction.

The strength of the force of friction depends on the two factors, as;

A. The objects' sizes and weights.

D. The heat generated by the friction and the types of surfaces involved.

Hence, options A and D are correct.

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3 0
2 years ago
Find the potential inside and outside a uniformly charged solid sphere whose radius is R and whose total charge is q. Use infini
amid [387]

Answer:

Recall that the electric field outside  a uniformly charged solid sphere  is exactly the same as if the charge were all at a point in the centre of the  sphere:

E_{outside} =\frac{1}{4\pi(e_{0})}\frac{Q}{r^{2} } r^{'}

lnside the sphere, the electric field also acts like a point charge, but only for the proportion of the charge further inside than the point r:

E_{inside} =\frac{1}{4\pi(e_{0})}\frac{Q}{R^{2} } \frac{r}{R} r^{'}

To find the potential, we integrate the electric field on a path from infinity (where of course, we take the direct path so that we can write the it as a 1 D integral):

V(r>R)=\int\limits^r_\infty {\frac{1}{4\pi(e_{0)} }\frac{Q}{r^2}  } \, dr=\frac{q}{4\pi(e_{0)} } \frac{1}{r} \\V(r

=\frac{q}{4\pi e_{0} } [\frac{1}{R} -\frac{r^{2}-R^{2}  }{2R^{3} } ]

∴NOTE: Graph is attached

8 0
3 years ago
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Answer:

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4 0
3 years ago
Sort the statements about heat transfer into the correct columns.
Hoochie [10]

Answer: Conduction- Touch transfer heat and Earth warms air

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Explanation:

5 0
3 years ago
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