All categories

2 years ago

When does the radioactive decay of a radioisotope stop? Give one example.

Physics

2 answers:

Romashka-Z-Leto [24]2 years ago

5 0

Answer:

When does the radioactive decay of a radioisotope stop? Give one example. An unstable isotope continues the decay process until it reaches a stable form. One example is the decay of carbon-14 to nitrogen-14.

Explanation:

Rudik [331]2 years ago

4 0

Answer:

Sample response: An unstable isotope continues the decay process until it reaches a stable form. One example is the decay of carbon-14 to nitrogen-14.

Explanation:

You might be interested in

How is frequency related to the sound we hear?

Leviafan [203]

Frequency is the vibration of noise and the vibration determines the pitch, which we depend on to be a pitch or frequency we can hear. If it's too high or too low our ears can't hear it

8 0

3 years ago

PSYCHOLOGY! A _________ is a graphical representation of association between variables.

pshichka [43]

Answer:

A. Scatterplot

Explanation:

because it is

5 0

2 years ago

Read 2 more answers

Concept Simulation 20.4 provides background for this problem and gives you the opportunity to verify your answer graphically. Ho

77julia77 [94]

Answer:

The time constant is 1.049.

Explanation:

Given that,

Charge $q{t}= 0.65 q_{0}$

We need to calculate the time constant

Using expression for charging in a RC circuit

$q(t)=q_{0}[1-e^{-(\dfrac{t}{RC})}]$

Where, $\dfrac{t}{RC}$ = time constant

Put the value into the formula

$0.65q_{0}=q_{0}[1-e^{-(\dfrac{t}{RC})}]$

$1-e^{-(\dfrac{t}{RC})}=0.65$

$e^{-(\dfrac{t}{RC})}=0.35$

$-\dfrac{t}{RC}=ln (0.35)$

$-\dfrac{t}{RC}=-1.049$

$\dfrac{t}{RC}=1.049$

Hence, The time constant is 1.049.

6 0

2 years ago

A person is trying to lift a crate that has a mass of 30 kg. The normal force of the floor is currently supplying 150N of force.

alexdok [17]

Here when an object is placed on the level floor then in that case there are two forces on the object

1). Weight of object downwards (mg)

2). Normal force due to floor which will counterbalance the weight (N)

so when no force is applied on the box at that time normal force is counter balanced by weight.

Now here it is given that A person tried to lift the box upwards

So now there are two forces on the box

1). Applied force of person

2). Normal force due to ground

So now these two forces will counter balance the weight of the crate

So we can write an equation for force balance like

$F_g = F_n + F_a$

given that

$F_g = mg$

here

m = 30 kg and

g = acceleration due to gravity = 10 m/s^2

$F_n = 150 N$

now from above equation

$30*10 = 150 + F_a$

$F_a = 300 - 150 = 150 N$

So force applied by the person must be 150 N

7 0

3 years ago

If you touch the two terminals of a power supply with your two fingertips on opposite hands, the potential difference will produ

LiRa [457]

Answer:

Yes the body will receive a dangerous shock in both cases.

Explanation:

Different parts of the body has different resistance. skin has the high resistance as compared to other organs of the body.

Dry skin has high resistance than wet skin this is because water is relatively good conductor of electricity, it adds parallel path to the current flow and hence reduces skin resistance.

Dry hands body has approximately 500 kΩ resistance and if 120 V electricity supply current received will be:

I = V/R= 120/ 500*10^3

I= 0.24 mA

Even the current seems is much lower than the safe zone but this is the case in case of DC voltage in case of AC voltage the body will receive a shock this is because the skin pass more current when the voltage is changing i.e. AC.

Similarly for wet hands body resistance is 1 kΩ. so the current through the body seems to be:

I = 120 / 1000

I = 12 mA

The current is higher than safe zone so the body will receive a dangerous shock.

7 0

3 years ago