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prisoha [69]
3 years ago
9

The half-life of a radioactive isotope is the time it takes for a quantity of the isotope to be reduced to half its initial mass

. Starting with grams of a radioactive isotope, how much will be left after half-lives
Physics
1 answer:
lora16 [44]3 years ago
5 0

Answer:

Incomplete questions

This is the complete question

The half-life of a radioactive isotope is the time it takes for a quantity for the isotope to be reduced to half its initial mass. Starting with 150 grams of a radioactive isotope, how much will be left after 6 half-lives

Explanation:

Let analyse the question generally first,

The the mass of the radioactive element be M.

We want to know it mass after n half life

Then,

After first half life, it mass is

M1=M×½

After second half life, it mass is

M2= M×(½)²

After third half life, it mass is

M3= M×(½)³

But now we can see a pattern developing, because for each new half-life we are dividing the quantity by 2 to a power that increases as the number of half-lives.

Then we can take the original quantity and quickly compute for

nth half-lives:

So after nth half life will be

Mn= M × (½)ⁿ

Generally,

Now, let apply it to our questions

Give that the mass of the radioactive isotope is 150grams

It mass after 6th half life

Then, n=6

So applying the formula

Mn= M × (½)ⁿ

M6= 150 ×(½)^6

M6= 150×1/64

M6=2.34grams

The mass of the radioactive isotope after 6th half life is 2.34grams

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max2010maxim [7]

Answer:

The formula comes from Lorentz force law which includes both the electric and magnetic field. If the electric field is zero, the force law for just the magnetic field is <u>F=q(ν×B</u>) . Here, F  is force and is a vector because the force acts in a direction.  q  is the charge of the particle.  v  is velocity and is a vector because the particle is moving in some direction.  B is the magnetic flux density.

We can derive an expression for the magnetic force on a current by taking a sum of the magnetic forces on individual charges. (The forces add because they are in the same direction.) The force on an individual charge moving at the drift velocity vd.  Since the magnitude of B is constant at every line element of the loop (circle) and it dot product with the line element is B dl everywhere, therefore

                                                  ∮B dl=μ0 I

                                                  B ∮dl=μ0 I

                                                  B 2πr=μ0 I

                                                   B=μ02πr Id=μ0/4π I dl×rr3

Since, r can be written as r=(rcosθ,rsinθ,z) and dl as dl=(dl,0,0) And now, if we take the cross product we would get

                                               dl×r=−z dlj^+rsinθk^

and therefore the magnitude of dB is equal to

dB=μ0/4π I |dl×r|/r3=μ0/4π I z2+r2sin2θ−−−−−−−−−−√dl/r3

Thus, magnetic field is depending on r,θ,z.

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7 0
2 years ago
This should be correct
sweet-ann [11.9K]

Maybe it is, maybe it isn't.  We can't tell, until we see what "this" is. 

Show us a drawing, an equation, an expression, a statement ... something !

5 0
3 years ago
How does changes in distance affect the gravitational pull between two objects? Describe and give one example.
maxonik [38]
The formula is

F_grav = G * m1 * m2 / r^2

G m1 and m2 are going to stay the same once chosen no matter what the distance is. The only thing that will change is the distance.

As the distance increases, the Gravitational Force will decrease. It will decrease by quite a bit.

As the distance decreases, the gravitational force will Increase.

The relationship is inverse. The moon travelling around the earth is one example. The earth travelling around the sun is another.
8 0
4 years ago
Chromatic aberration comes from the fact that different wavelengths of light travel at different speeds through the material of
gtnhenbr [62]

Answer:

 y_red / y_blue = 1.11

Explanation:

Let's use the constructor equation to find the image for each wavelength

         1 /f = 1 /o + 1 /i

Where f is the focal length, or the distance to the object and i the distance to the image

Red light

           1 / i = 1 / f - 1 / o

           1 / i_red = 1 / f_red - 1 / o

           1 / i_red = 1 / 19.57 - 1/30

           1 / i_red = 1,776 10-2

           i_red = 56.29 cm

Blue light

            1 / i_blue = 1 / f_blue - 1 / o

            1 / i_blue = 1 / 18.87 - 1/30

            1 / i_blue = 1,966 10-2

            i_blue = 50.863 cm

Now let's use the magnification ratio

             m = y ’/ h = - i / o

             y ’= - h i / o

Red Light

            y_red ’= - 5 56.29 / 30

            y_red ’= - 9.3816 cm

Light blue

            y_blue ’= 5 50,863 / 30

            y_blue ’= - 8.47716 cm

The ratio of the height of the two images is

            y_red ’/ y_blue’ = 9.3816 / 8.47716

            y_red / y_blue = 1,107

            y_red / y_blue = 1.11

5 0
3 years ago
A wheel with a tire mounted on it rotates at the constant rate of 2.73 revolutions per second. Find the radial acceleration of a
Lostsunrise [7]

Answer:

110.9 m/s²

Explanation:

Given:

Distance of the tack from the rotational axis (r) = 37.7 cm

Constant rate of rotation (N) = 2.73 revolutions per second

Now, we know that,

1 revolution = 2\pi radians

So, 2.73 revolutions = 2.73\times 2\pi=17.153\ radians

Therefore, the angular velocity of the tack is, \omega=17.153\ rad/s

Now, radial acceleration of the tack is given as:

a_r=\omega^2 r

Plug in the given values and solve for a_r. This gives,

a_r=(17.153\ rad/s)^2\times 37.7\ cm\\a_r=294.225\times 37.7\ cm/s^2\\a_r=11092.28\ cm/s^2\\a_r=110.9\ m/s^2\ \ \ \ \ \ \ [1\ cm = 0.01\ m]

Therefore, the radial acceleration of the tack is 110.9 m/s².

4 0
3 years ago
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