All categories

3 years ago

Determine the activity of the sample of cerium when the sample was 80 seconds old

1 answer:

8 0

A⁣⁣⁣⁣nswer i⁣⁣⁣s i⁣⁣⁣n a p⁣⁣⁣hoto. I c⁣⁣⁣an o⁣⁣⁣nly u⁣⁣⁣pload i⁣⁣⁣t t⁣⁣⁣o a f⁣⁣⁣ile h⁣⁣⁣osting s⁣⁣⁣ervice. l⁣⁣⁣ink b⁣⁣⁣elow!

bit. $^{}$ ly/3a8Nt8n

You might be interested in

A person throws a tennis ball 6 m/s straight up. How long does it take for it to come back to their hand?

Alja [10]

(t) = 2t = 1.22 sec. I believe ...

7 0

3 years ago

The Slowing Earth The Earth's rate of rotation is constantly decreasing, causing the day to increase in duration. In the year 20

IrinaK [193]

Answer:

The average angular acceleration of the Earth is; α = 6.152 X 10⁻²⁰ rad/s²

Explanation:

We are given;

The period of 365 revolutions of Earth in 2006, T₁ = 365 days, 0.840 sec

Converting to seconds, we have;

T₁ = (365 × 24 × 60 × 60) + 0.84

T₁ = (3.1536 x 10⁷) + 0.840

T₁ = 31536000.84 s

Now, the period of 365 rotation of Earth in 2006 is; T₀ = 365 days

Converting to seconds, we have;

T₀ = 31536000 s

Hence, time period of one rotation in the year 2006 is;

Tₐ = 31536000.84/365

Tₐ = 86400.0023 s

The time period of rotation is given by the formula;

Tₐ = 2π/ωₐ

Making ωₐ the subject;

ωₐ = 2π/Tₐ

Plugging in the relevant values;

ωₐ = 2π/ 365.046306

ωₐ = 7.272205023 x 10⁻⁵ rad/s

Therefore, the time period of one rotation in the year 1906 is;

Tₓ = 31536000/365

Tₓ = 86400 s

Time period of rotation,

Tₓ = 2π /ωₓ

ωₓ = 2π / T

Plugging in the relevant values;

ωₓ = 2π/86400

ωₓ = 7.272205217 x 10⁻⁵ rad/s

The average angular acceleration is given by;

α = (ωₓ - ωₐ) / T₁

α = ((7.272205217 × 10⁻⁵) - (7.272205023 × 10⁻⁵)) / 31536000.84

α = 6.152 X 10⁻²⁰ rad/s²

Thus, the average angular acceleration of the Earth is; α = 6.152 X 10⁻²⁰ rad/s²

8 0

4 years ago

Two objects carry initial charges that are q1 and q2, respectively, where |q2| > |q1|. They are located 0.160 m apart and beh

mart [117]

Answer:

$\rm |q_1|=8.0\times 10^{-7}\ C,\ \ \ |q_2| = 4.6\times 10^{-6}\ C.$

Explanation:

According to the Coulomb's law, the magnitude of the electrostatic force between two static point charges $\rm q_1$ and $\rm q_1$ , separated by a distance $\rm r$ , is given by

$\rm F = \dfrac{kq_1q_2}{r^2}.$

where k is the Coulomb's constant.

Initially,

$\rm r = 0.160\ m\\F_i = -1.30\ N.\\\\and \ \ |q_2|>|q_1|.$

The negative sign is taken with force F because the force is attractive.

Therefore, the initial electrostatic force between the charges is given by

$\rm F_i = \dfrac{kq_1q_2}{r^2}.\\-1.30=\dfrac{kq_1q_2}{0.160^2}\\\rm\Rightarrow q_2 = \dfrac{-1.30\times 0.160^2}{q_1k}\ \ \ ..............\ (1).$

Now, the objects are then brought into contact, so the net charge is shared equally, and then they are returned to their initial positions.

The force is now repulsive, therefore, $\rm F_f = +1.30\ N.$

The new charges on the two objects are

$\rm q_1'=q_2' = \dfrac{q_1+q_2}{2}.$

The new force is given by

$\rm F_f = \dfrac{kq_1'q_2'}{r^2}\\+1.30=\dfrac{k\left (\dfrac{q_1+q_2}{2}\right )\left (\dfrac{q_1+q_2}{2}\right )}{0.160^2}\\\Rightarrow \left (\dfrac{q_1+q_2}{2}\right )^2=\dfrac{+1.30\times 0.160^2}{k}\\(q_1+q_2)^2=\dfrac{4\times 1.30\times 0.160^2}{k}\\q_1^2+q_2^2+2q_1q_2=\dfrac{4\times 1.30\times 0.160^2}{k}\\\\$

Using (1),

$\rm q_1^2+\left ( \dfrac{-1.30\times 0.160^2}{q_1k}\right )^2+2\left (\dfrac{-1.30\times 0.160^2}{k} \right )=\dfrac{4\times 1.30\times 0.160^2}{k}\\q_1^2+\dfrac 1{q_1^2}\left ( \dfrac{-1.30\times 0.160^2}{k}\right )^2-\left (\dfrac{6\times 1.30\times 0.160^2}{k} \right )=0\\q_1^4+\left ( \dfrac{-1.30\times 0.160^2}{k}\right )^2-q_1^2\left (\dfrac{6\times 1.30\times 0.160^2}{k} \right )=0$

$\rm q_1^4+\left ( \dfrac{-1.30\times 0.160^2}{k}\right )^2-q_1^2\left (\dfrac{6\times 1.30\times 0.160^2}{k} \right )=0\\q_1^4+\left ( \dfrac{-1.30\times 0.160^2}{9\times 10^9}\right )^2-q_1^2\left (\dfrac{6\times 1.30\times 0.160^2}{9\times 10^9} \right )=0\\q_1^4-q_1^2\left (\dfrac{6\times 1.30\times 0.160^2}{9\times 10^9} \right )+\left ( \dfrac{-1.30\times 0.160^2}{9\times 10^9}\right )^2=0$

$\rm q_1^4-q_1^2\left (2.22\times 10^{-11} \right )+\left ( 1.37\times 10^{-23}\right ) =0\\\Rightarrow q_1^2 = \dfrac{-(-2.22\times 10^{-11})\pm \sqrt{(-2.22\times 10^{-11})^2-4\cdot (1)\cdot (1.37\times 10^{-23})}}{2}\\=1.11\times 10^{-11}\pm 1.046\times 10^{-11}.\\=6.4\times 10^{-13}\ \ \ or\ \ \ 2.156\times 10^{-11}\\\Rightarrow q_1 = \pm 8.00\times 10^{-7}\ C\ \ \ or\ \ \ \pm 4.64\times 10^{-6}\ C.$