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Vikentia [17]
3 years ago
8

What does the half life for a radioactive material mean?

Physics
1 answer:
marissa [1.9K]3 years ago
6 0
Half-life, in radioactivity, the interval of time required for one-half of the atomic nuclei of a radioactive sample to decay (change spontaneously into other nuclear species by emitting particles and energy), or, equivalently, the time interval required for the number of disintegrations per second of a radioactive
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Name of compounds<br>a. H2Co3<br>b. NH4OH<br>c.AlPO4<br>d.NaOH<br>e.AuCI3<br>​
seraphim [82]

Answer:

<em><u>a) </u></em><em><u> </u></em><em><u>Carbonic acid</u></em>

<em><u>b</u></em><em><u>)</u></em><em><u> </u></em><em><u>ammonium hydroxide</u></em>

<em><u>c</u></em><em><u>)</u></em><em><u> </u></em><em><u>Aluminum phosphate</u></em>

<em><u>d</u></em><em><u>)</u></em><em><u> </u></em><em><u>Sodium hydroxide</u></em>

<em><u>e</u></em><em><u>)</u></em><em><u> </u></em><em><u>Gold trichloride</u></em>

Explanation:

<em>I</em><em> </em><em>hope this</em><em> </em><em>will help</em><em> </em><em>you</em><em> </em><em>buddy</em><em> </em>

3 0
3 years ago
When the atom's electrons step down to lower energy levels in a thin cloud of hot gas, what is produced?
Reika [66]

Answer: The correct answer is an emission line spectrum.

Explanation:

When the electrons are excited to the higher energy level, the energy is absorbed in this case.

When the electrons in the atom step down to lower energy levels in a thin cloud of hot gas then the radiation will emit.

The electron will lose energy in this case in the form of radiation. There will be an emission line spectrum.

8 0
3 years ago
Read 2 more answers
In severe head-on automobile accidents, a deceleration of 60 g’s or more (1 g 5 32.2 ft/s2) often results in a fatality. What fo
Musya8 [376]

Answer:

The resulting force on the child is 3000 lbf

Explanation:

To find the force that acts on a child of 50 lb with a deceleration of 60 g's, we can use the formula:

Force = mass * acceleration

To find the force in lbf, we need to use the mass in lb and the acceleration in g (standard unit of gravity).

So we have that:

Force = 50 * 60

Force = 3000 lbf

So the resulting force on the child is 3000 lbf.

3 0
3 years ago
A construction worker drags a box across the floor. If the frictional force between the floor and the box is 12.2 newtons, how m
victus00 [196]

The work done is 140.3 J

Explanation:

  • The work done is defined as the amount of force required or used to move an object over a particular distance. Hence, work done is the product of force in Newton and the distance in which the object moved in meters.
  • Here, in the question, it is given that there is a frictional force acting on the box by the floor and the distance it moves. Hence, to get the answer to multiply frictional force and the distance. So, will get the work done as 140.3 J.
8 0
3 years ago
Joey, whose mass is m = 36 kg, stands at rest at the outer edge of a frictionless merry-go-round with the mass M = 300 kg and th
Evgen [1.6K]

Answer:

\omega=0.24\ rad.s^{-1}

Explanation:

Given:

mass of person, m=36\ kg

mass of merry go-round, M=300\ kg

radius of merry go-round, R=2\ m

velocity of the person running, v=4\ m.s^{-1}

<u>We consider merry go-round as a ring:</u>

Now the moment of inertial of the ring is given as,

I=M.R^2

I=300\times 2^2

I=1200\ kg.m^{-2}

<u>Moment of inertia of the person considering as a point mass:</u>

I_p=m.R^2

I_p=36\times 2^2

I_p=144\ kg.m^2

<u>Now according to the conservation of angular momentum:</u>

I.\omega=I_p.\omega_p

where:

\omega = angular velocity of the merry-go-round

\omega_p= angular velocity of the person running

1200\times \omega=144\times \frac{v}{R}

\omega=\frac{144}{1200} \times \frac{4}{2}

\omega=0.24\ rad.s^{-1}

4 0
3 years ago
Read 2 more answers
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