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4 years ago

A sample contains 10 g of a radioactive isotope. How much radioactive isotope will remain in the sample after 2 half-lives?

Physics

2 answers:

xxMikexx [17]4 years ago

5 0

The answer c - Half life how to figure, you first divide it by half then divide than number by half

k0ka [10]4 years ago

3 0

Answer:

c) 2.5 g

Explanation:

half-life is the time it takes the radioactive isotope to decay to half of the intial sample.

so if the isotope decays to 2 half-lives, it will decay half the intial sample twice.

in the first half-life, the sample will decay from 10 g to 5 g. during the second half-life, the isotope will decay from 5 g to 2.5 g.

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WILL GIVE BRAINLIEST AND 30 POINTS!

mariarad [96]

Answer:

Gallium

Explanation:

Gallium is one such element used as a do/pant in a p-type semiconductor.

A do/pant is an impurity added to a semi-conductor used to alter its properties. Semi-conductors have a wide range of applications. They will conduct heat and electricity only under certain conditions. This property is highly desirable and find a wide application in electronics.

For p-type conductors, they are best do/ped with elements with 3 valence electrons. These are group 3 elements. From the choices, only gallium belongs to this group.

Other elements given are good do/pants for n-type semiconductors. They have 5 valence electrons.

8 0

4 years ago

A uniform magnetic field of magnitude 0.23 T is directed perpendicular to the plane of a rectangular loop having dimensions 7.8

Arlecino [84]

Answer:

0.0025116weber/m²

Explanation:

Magnetic field density (B) is the ratio of the magnetic flux (¶) through the loop to its cross sectional area (A).

Mathematically;

B = ¶/A

¶ = BA

Given B = 0.23Tesla which is the magnitude of the magnetic field

Dimension of the rectangular loop = 7.8 cm by 14 cm

Area of the rectangular loop perpendicular to the field B = 7.8cm×14cm

= 109.2cm²

Converting this value to m²

Area of the loop = 109.2 × 10^-4

Area of the loop = 0.01092m²

Magneto flux = 0.23×0.01092

Magnetic flux = 0.0025116weber/m²

3 0

3 years ago

Read 2 more answers

Init.

nikklg [1K]

Answer:

The kilogram, symbol kg, is the SI unit of mass. It is defined by taking the fixed numerical value of the Planck constant h to be 6.62607015×10−34 when expressed in the unit J⋅s, which is equal to kg⋅m2⋅s−1, where the metre and the second are defined in terms of c and ΔνCs.

8 0

3 years ago

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You know the power for a machine and the time it takes to produce that power, what value can you calculate?

Sati [7]

We can calculate the work done by the machine by the mathematical formula,

P = W/t

where, P = Power

W = work done

t = time

6 0

3 years ago

g A thin-walled hollow cylinder and a solid cylinder, both have same mass 2.0 kg and radius 20 cm, start rolling down from rest

ArbitrLikvidat [17]

Answer:

a. i. 3.43 m/s ii. 2.8 m/s

b. The thin-walled cylinder

Explanation:

a. Find translational speed of each cylinder upon reaching the bottom

The potential energy change of each mass = total kinetic energy gain = translational kinetic energy + rotational kinetic energy

So, mgh = 1/2mv² + 1/2Iω² where m = mass of object = 2.0 kg, g =acceleration due to gravity = 9.8 m/s², h = height of incline = 1.2 m, v = translational velocity of object, I = moment of inertia of object and ω = angular speed = v/r where r = radius of object.

i. translational speed of thin-walled cylinder upon reaching the bottom

So, For the thin-walled cylinder, I = mr², we find its translational velocity, v

So, mgh = 1/2mv² + 1/2Iω²

mgh = 1/2mv² + 1/2(mr²)(v/r)²

mgh = 1/2mv² + 1/2mv²

mgh = mv²

v² = gh

v = √gh

v = √(9.8 m/s² × 1.2 m)

v = √(11.76 m²/s²)

v = 3.43 m/s

ii. translational speed of solid cylinder upon reaching the bottom

So, For the solid cylinder, I = mr²/2, we find its translational velocity, v'

So, mgh = 1/2mv'² + 1/2Iω²

mgh = 1/2mv² + 1/2(mr²/2)(v'/r)²

mgh = 1/2mv'² + mv'²

mgh = 3mv'²/2

v'² = 2gh/3

v' = √(2gh/3)

v' = √(2 × 9.8 m/s² × 1.2 m/3)

v' = √(23.52 m²/s²/3)

v' = √(7.84 m²/s²)

v' = 2.8 m/s

b. Determine which cylinder has the greatest translational speed upon reaching the bottom.

Since v = 3.43 m/s > v'= 2.8 m/s,

the thin-walled cylinder has the greatest translational speed upon reaching the bottom.

3 0

3 years ago