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2 years ago

a sample contains 100 g of radioactive isotope. How much radioactive isotope will remain in the sample after 1 half-life?

Physics

1 answer:

kap26 [50]2 years ago

8 0

Answer:

$\huge\boxed{50g}$

Definition:

Half-life- The time taken for half of the radioactive isotopes to decay.

Explanation:

How does radioactive decay work? Radioactive decay is a process by which unstable nuclei become more stable through the emission of alpha or beta particles or gamma rays.

Since a half-life is the time taken for half of the isotopes to decay, we can simply divide the initial mass of 100 grams by 2; this gives us 50 grams.

1) Divide 100g by 2.

$\frac{100g}{2}=50g$

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The oscillating electric field in a plane electromagnetic wave is given by <img src="https://tex.z-dn.net/?f=%7B50%5Csin%20%28%5

Scorpion4ik [409]

Here, we are given with:

${:\implies \quad \longrightarrow \begin{cases}\sf E_{0}= 50\: V/m\\ \sf \nu = 2\times 10^{7}Hz\end{cases}}$

(a) So now, we can thus obtain

${:\implies \quad \sf \omega =2\pi \nu}$

${:\implies \quad \sf \omega =2\pi (2\times 10^{7})}$

${:\implies \quad \boxed{\bf{\omega = 4\pi \times 10^{7}\: s^{-1}}}}$

Now, finding $\lambda$

${:\implies \quad \sf \lambda =\dfrac{c}{\omega}=\dfrac{3\times 10^{8}}{4\pi \times 10^{7}}}$

${:\implies \quad \boxed{\bf{\lambda \approx 2.39\:m}}}$

Now, finding k

${:\implies \quad \sf k=\dfrac{2\pi}{\lambda}=\dfrac{200\pi}{239}}$

${:\implies \quad \boxed{\bf{k\approx 2.63\:m^{-1}}}}$

Thus, expression for the electric field is:

${:\implies \quad \boxed{\bf{E=50\sin \bigg(4\pi \times 10^{7}t-\dfrac{263x}{100}\bigg)}}}$

(b) Now, here

${:\implies \quad \sf B_{0}=\dfrac{E_{0}}{c}=\dfrac{50}{3\times 10^{8}}}$

${:\implies \quad \boxed{\bf{B_{0}\approx 16.67×10^{-7}\:T}}}$

Thus, expression for the magnetic field:

${:\implies \quad \boxed{\bf{B_{y}=16.67\times 10^{-7}\sin \bigg(4\pi \times 10^{7}t-\dfrac{263x}{100}\bigg)\:T}}}$

4 0

2 years ago

A long wire carrying a 5.8 A current perpendicular to the xy-plane intersects the x-axis at x=−2.3cm. A second, parallel wire ca

adoni [48]

Answer:

a) v r = 0.7318 cm , b) r = 7.23 cm

Explanation:

The magnetic field generated by a wire carrying a current can be found with Ampere's law

∫ B. ds = μ₀ I

the length of a surface circulates around the wire is

s = 2π r

where r is the point of interest of the calculation of the magnetic field

B = μ₀ I / 2π r

In this exercise we have two wires, write the equation of the magnetic field of each one

wire 1 I = 5.8 A

B₁ = μ₀ 5,8 / 2π r₁

wire 2 I = 3.0 A

B₂ = μ₀ 3/2π r₂

the direction of the field is given by the rule of the right hand, the thumb indicates the direction of the current and the other fingers the direction of the magnetic field

Let's apply these expressions to our case

a) the two streams go in the same direction

using the right hand rule for each wire we see that between the two wires the magnetic fields have opposite directions so there is some point where the total value is zero

B₁ - B₂ = 0

B₁ = B₂

μ₀ 5,8 / 2π r₁ = μ₀ 3 / 2π r₂

5.8 / r₁ = 3 / r₂

5.8 r₂ = 3r₁

the value of r is measured from each wire, therefore

r₁ = 2.3 + r

r₂ = 2.3 -r

we substitute

5.8 (2.3 - r) = 3 (2.3 + r)

r (3 + 5.8) = 2.3 (5.8 - 3)

r = 2.3 2.8 / 8.8

r = 0.7318 cm

b) the two currents have directional opposite

with the right hand rule in the field you have opposite directions outside the wires

suppose it is zero on the right side where the wire with the lowest current is

B₁ = B₂

5.8 / r₁ = 3 / r₂

5.8 r₂ = 3 r₁

r₁ = 2.3 + r

r₂ = r - 2.3

5.8 (r - 2.3) = 3 (2.3 + r)

r (5.8 -3) = 2.3 (3 + 5.8)

r = 2.3 8.8 / 2.8

r = 7.23 cm

8 0

3 years ago

A cannon of mass 6.43 x 103 kg is rigidly bolted to the earth so it can recoil only by a negligible amount. The cannon fires a 7

Nata [24]

Answer:

The velocity of the shell when the cannon is unbolted is 500.14 m/s

Explanation:

Given;

mass of cannon, m₁ = 6430 kg

mass of shell, m₂ = 73.8-kg

initial velocity of the shell, u₂ = 503 m/s

Initial kinetic energy of the shell; when the cannon is rigidly bolted to the earth.

K.E = ¹/₂mv²

K.E = ¹/₂ (73.8)(503)²

K.E = 9336032.1 J

When the cannon is unbolted from the earth, we apply the principle of conservation of linear momentum and kinetic energy

change in initial momentum = change in momentum after

0 = m₁u₁ - m₂u₂

m₁v₁ = m₂v₂

where;

v₁ is the final velocity of cannon

v₂ is the final velocity of shell

$v_1 = \frac{m_2v_2}{m_1}$

Apply the principle of conservation kinetic energy

$K = \frac{1}{2}m_1v_1^2 + \frac{1}{2}m_2v_2^2\\\\K = \frac{1}{2}m_1(\frac{m_2v_2}{m_1})^2 + \frac{1}{2}m_2v_2^2\\\\K = \frac{1}{2}m_2v_2^2(\frac{m_2}{m_1}) + \frac{1}{2}m_2v_2^2 \\\\K = \frac{1}{2}m_2v_2^2 (\frac{m_2}{m_1} + 1)\\\\2K = m_2v_2^2 (\frac{m_2}{m_1} + 1)\\\\v_2^2 = \frac{2K}{M_2(\frac{m_2}{m_1} + 1)} \\\\v_2^2 = \frac{2*9336032.1}{73.8(\frac{73.8}{6430} + 1)}\\\\$

$v_2^2 = 250138.173\\\\v_2 = \sqrt{250138.173} \\\\v_2 = 500.14 \ m/s$

Therefore, the velocity of the shell when the cannon is unbolted is 500.14 m/s

3 0

3 years ago

Find the electric force acting on an alpha particle in a horizontal electric field of 600N/C

Sergeeva-Olga [200]

Since an alpha particle has 2 protons and no negative particles (electrons) to balance the net charge, its charge is
Q=2(1.6e-19)=3.2e-19C.
The force on a charged particle is F=QE so
(3.2e-19C)(600N/C)=1.92e-16N

6 0

3 years ago

Carole is surprised when she makes an emergency stop and the tissue box that had been in the back window of her car hits the bac

krek1111 [17]

B An object in motion tends to stay in motion .

4 0

3 years ago

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