All categories

2 years ago

How can a scientist assess whether a pure niobium sample is responsible for contaminating the lab with radioactivity

Physics

2 answers:

soldier1979 [14.2K]2 years ago

8 0

Answer:

Test the niobium Sample to see whether it now contains other elements

Explanation:

d1i1m1o1n [39]2 years ago

6 0

A scientist can assess whether a pure niobium sample is responsible for contaminating the lab with radioactivity by testing the sample. By testing the niobium sample, a scientist can determine whether it has any other element.

You might be interested in

Thermopane window is constructed, using two layers of glass 4.0 mm thick, separated by an air space of 5.0 mm.

Bond [772]

To solve this problem it is necessary to apply the concepts related to rate of thermal conduction

$\frac{Q}{t} = \frac{kA\Delta T}{d}$

The letter Q represents the amount of heat transferred in a time t, k is the thermal conductivity constant for the material, A is the cross sectional area of the material transferring heat, $\Delta T$ , T is the difference in temperature between one side of the material and the other, and d is the thickness of the material.

The change made between glass and air would be determined by:

$(\frac{Q}{t})_{glass} = (\frac{Q}{t})_{air}$

$k_{glass}(\frac{A}{L})_{glass} \Delta T_{glass} = k_{air}(A/L)_{air} \Delta T_{air}$

$\Delta T_{air} = (\frac{k_{glass}}{k_{air}})(\frac{L_{air}}{L_{glass}}) \Delta T_{glass}$

$\Delta T_{air} = (\frac{0.84}{0.0234})(\frac{5}{4}) \Delta T_{glass}$

$\Delta T_{air} = 44.9 \Delta T_{glass}$

There are two layers of Glass and one layer of Air so the total temperature would be given as,

$\Delta T = \Delta T_{glass} +\Delta T_{air} +\Delta T_{glass}$

$\Delta T = 2\Delta T_{glass} +\Delta T_{air}$

$20\°C = 46.9\Delta T_{glass}$

$\Delta T_{glass} = 0.426\°C$

Finally the rate of heat flow through this windows is given as,

$\Delta {Q}{t} = k_{glass}\frac{A}{L_{glass}}\Delta T_{glass}$

$\Delta {Q}{t} = 0.84*24*10 -3*0.426$

$\Delta {Q}{t} = 179W$

Therefore the correct answer is D. 180W.

3 0

2 years ago

An apple weighs 1.00 N. When you hang it from the end of a long spring of force constant 1.50 N/m and negligible mass, it bounce

Ierofanga [76]

Answer:

2.67 m

Explanation:

k = Spring constant = 1.5 N/m

g = Acceleration due to gravity = 9.81 m/s²

l = Unstretched length

Frequency of SHM motion is given by

$f_s=\dfrac{1}{2\pi}\sqrt{\dfrac{k}{m}}$

Frequency of pendulum is given by

$f_p=\dfrac{1}{2\pi}\sqrt{\dfrac{g}{l}}$

Given in the question

$f_p=\dfrac{1}{2}f_s$

$\dfrac{1}{2\pi}\sqrt{\dfrac{g}{l}}=\dfrac{1}{2}\dfrac{1}{2\pi}\sqrt{\dfrac{k}{m}}\\\Rightarrow \sqrt{\dfrac{g}{l}}=\dfrac{1}{2}\sqrt{\dfrac{k}{m}}\\\Rightarrow \dfrac{g}{l}=\dfrac{1}{4}\dfrac{k}{m}\\\Rightarrow l=\dfrac{4gm}{k}\\\Rightarrow l=\dfrac{4\times 9.81\times \dfrac{1}{9.81}}{1.5}\\\Rightarrow l=2.67\ m$