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3 years ago

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Half-life is the time a. For half of a material to decay into energy b. For half of a material to decay into free protons, neutr

ons, and electrons c. For half of a material to decay into a new material d. It takes to get out of school

1 answer:

aniked [119]3 years ago

6 0

Answer:

Explanation:

The half life of a radioactive substance is the time in which the activity of the radioactive substance becomes half of its initial activity.

Or

The time in which the amount of radioactive substance becomes half of its initial amount in a sample.

It is denoted by T .

The relation between the decay constant and the half life is given by

$\lambda =\frac{0.693}{T}$

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A contestants spin a wheel when it is their turn in a game show. One contestant gives the wheel an initial angular speed of 3.40

guajiro [1.7K]

Answer:

4.62 s

Explanation:

We are given that

Initial angular speed, $\omega=3.4 rad/s$

$\theta=1\frac{1}{4} rev=\frac{5}{4}\times 2\pi=2.5\pi rad$

$\omega'=0$

$\omega'^2-\omega^2=2\alpha \theta$

Substitute the values

$0-(3.4)^2=2\times 2.5\pi \alpha$

$\alpha=\frac{-(3.4)^2}{2\times 2.5\pi}=-0.736 rad/s^2$

$\omega'=\omega+\alpha t$

$0=3.4-0.736 t$

$-0.736t=-3.4$

$t=\frac{-3.4}{-0.736}=4.62 s$

Hence, the wheel takes 4.62 s to come to rest.

3 0

3 years ago

Muscles that are fatigued ____. A. Are injured , B. Should always be rested , C. Can sometimes be exercised further D. Should be

Lorico [155]

Answer:

<u>B. Should be rested</u>

Explanation:

That is the only reasonable answer.

5 0

3 years ago

Read 2 more answers

Which of the following statements is true of chords? Select all that apply.

tankabanditka [31]

All of those are true, except the one about the radius.

Both ends of a chord have to be on the circle, but one end
of a radius is at the center, so a radius can't be a chord.

5 0

3 years ago

A tiny object carrying a charge of +44 μC and a second tiny charged object are initially very far apart. If it takes 21 J of wor

STatiana [176]

Answer:

The magnitude of the second charge is $\rm 1.062\times 10^{-7}\ C$ or $\rm 0.1062\ \mu C.$

Explanation:

The work done in bringing a charged particle from one point to another in the presence of some electric field is equal to the change in the electric potential energy of the charge in moving from one point to another.

The electric potential energy of some charge $q_o$ at a point in the electric field of another charge $q$ is given by the product of the amount of charge $q_o$ and electric potential at that point due to the charge $q$ .

$U = q_o\ V.$

The electric potential at that point is given by

$V = \dfrac{kq}{r}.$

where $k$ is the Coulomb's constant.

Therefore,

$U=q_o\ \dfrac{kq}{r}.$

Now, We have given two charges $q_1 = +44\ \mu C = +44\times 10^{-6}\ C$ and $q_2$ , whose value is to be found.

When the two charges are infinitely dar apart, the electric potential energy of the system is given by

$U_i = \dfrac{kq_1q_2}{\infty}=0.$

When the coordinates of position of the two charges are

$(x_1,\ y_1) = (1.00\ mm,\ 1.00\ mm).\\(x_2,\ y_2) = (1.00\ mm,\ 3.00\ mm).$

The distance between the two charges is given by

$r=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}=\sqrt{(1.00-1.00)^2+(3.00-1.00)^2}=2.00\ mm = 2.00\times 10^{-3}\ m.$

The electric potential energy of the charges in this configuration is given by

$U_f = \dfrac{kq_1q_2}{r}\\=\dfrac{(8.99\times 10^9)\times (+44\times 10^{-6})\times q_2}{2.00\times 10^{-3}}\\=1.9778\times 10^8\times q_2.$

The change in the electric potential energy of the system is equal to the work done to bring the system from inifinitely far apart position to given configuration.

Therefore,

$W = U_f-U_i\\21=(1.9778\times 10^8\times q_2)-0\\\Rightarrow q_2 = \dfrac{21}{1.9778\times 10^8}\\=1.062\times 10^{-7}\ C\\=0.1062\times 10^{-6}\ C\\=0.1062\ \mu C.$

6 0

3 years ago

un avión se mueve en línea recta a una velocidad constante de 400 km/h durante 1,5h de su recorrido ¿que distancia recorrió en e

Amanda [17]

The distance covered by the plane is 600 km

Explanation:

The motion of the plane is a uniform motion, so at constant velocity, therefore we can use the following equation

$d=vt$

where

d is the distance covered

v is the velocity

t is the time interval considered

For the plane in this problem,

v = 400 km/h

t = 1.5 h

Substituting, we find the distance covered:

$d=(400)(1.5)=600 km$

Learn more about distance:

brainly.com/question/3969582

#LearnwithBrainly

6 0

3 years ago