All categories

3 years ago

Can someone help me with this Physics question please?

Physics

2 answers:

Ghella [55]3 years ago

7 0

How many half-lifes is 49.2 years ?

(49.2 years) / (12.3 years per half-life) = 4 half-lifes.

In 4 half-lifes, (1/2) · (1/2) · (1/2) · (1/2) of the original sample remains.

That's (1/2⁴) or (1/16) of the original.

(1/16) of 48.0 mg = <em>3 mg</em> .

=======================

Step-by-step:

== Start with 48 mg .

== After one half-life, 24 mg remains.

== Then, after the second half-life, 12 mg remains.

== Then, after the third half-life, 6 mg remains.

== Then, after the fourth half-life, <em>3 mg</em> remains.

strojnjashka [21]3 years ago

6 0

Answer:

Explanation:

The formula for this, the easy one, is

$N=N_0(\frac{1}{2})^{\frac{t}{H}$ where No is the initial amount of the element, t is the time in years, and H is the half life. Filling in:

$N=48.0(\frac{1}{2})^{\frac{49.2}{12.3}$ and simplifying a bit:

$N=48.0(.5)^4$ and

N = 48.0(.0625) so

N = 3 mg left after 12.3 years

You might be interested in

Andre the Giant and Thumbelina are standing in the middle of an ice-skating rink facing each other. Andre the Giant pushes Thumb

IceJOKER [234]

I don't know for sure but i think that it is because the force him pushing her pushed him back as well

Hope i helped

3 0

3 years ago

Read 2 more answers

An Olympic track runner starts from rest and has an acceleration of 2.4 m/s2 for 3.6 s, then has zero acceleration for the remai

rjkz [21]

Answer:

The runner's speed at the following times would remain 8.64 m/s.

Explanation:

Acceleration definition: Acceleration is rate of change in velocity of an object with respect to time.

In this case, after 3.6 seconds the acceleration is zero, it means that the velocity of the runner after 3.6 seconds is not changing and it will remain constant for the remainder of the race. Now, we have to find the velocity of the runner that he had after 3.6 seconds and that would be the runner's speed for the remainder of the race. For this we use first equation of motion.

First equation of motion: Vf = Vi + a×t

Vf stands for final velocity

Vi stands for initial velocity

a stands for acceleration

t stands for time

In the question, it is mentioned that the runner starts from rest so its initial velocity (Vi) will be 0 m/s.

The acceleration (a) is given as 2.4 m/s²

The time (t) is given as 3.6 s

Now put the values of Vi, a and t in first equation of motion

Vf = Vi + a×t

Vf = 0 + 2.4×3.6

Vf = 2.4×3.6

Vf = 8.64 m/s

So,the runner's speed at the following times would remain 8.64 m/s.

5 0

3 years ago

Terry usually rides his bike at 15mph. If his speed is reduced by 3 mph , how far can he ride in 1.7 hours ?

kakasveta [241]

105 miles because you have to use the gif arable

7 0

3 years ago

Consider a model of a hydrogen atom in which an electron is in a circular orbit of radius r = 5.92×10−11 m around a stationary p

DaniilM [7]

Answer:

2.068 x 10^6 m / s

Explanation:

radius, r = 5.92 x 10^-11 m

mass of electron, m = 9.1 x 10^-31 kg

charge of electron, q = 1.6 x 10^-19 C

As the electron is revolving in a circular path, it experiences a centripetal force which is balanced by the electrostatic force between the electron and the nucleus.

centripetal force = $\frac{mv^{2}}{r}$

Electrostatic force = $\frac{kq^{2}}{r^{2}}$

where, k be the Coulombic constant, k = 9 x 10^9 Nm^2 / C^2

So, balancing both the forces we get

$\frac{kq^{2}}{r^{2}}=\frac{mv^{2}}{r}$

$v=\sqrt{\frac{kq^{2}}{mr}}$

$v=\sqrt{\frac{9\times 10^{9}\times1.6\times 10^{-19}\times 1.6\times 10^{-19}}{9.1\times 10^{-31}\times 5.92\times10^{-11}}}$

v = 2.068 x 10^6 m / s

Thus, the speed of the electron is give by 2.068 x 10^6 m / s.

6 0

3 years ago

Compare the maximum rate of heat transfer to the basal metabolic rate by converting a bmr of 88 kcal/hr into watts. what is the

elena-14-01-66 [18.8K]

Explanation :

It is given that,

BMR i.e basal metabolic rate is 88 kcal/hr. So, BMR in watts is converted by the following :

We know that, 1 kilocalorie = 4184 joules

So, $1\ kcal/h=\dfrac{1\times 4184\ J}{3600\ sec}$

$1\ kcal/h=1.16\ J/sec$

J/sec is nothing but watts.

So, $1\ kcal/h=1.16\ watts$

and $88\ kcal/h=88\times 1.16\ watts = 102.08\ watts$

So, it can be seen that the body can accommodate a modes amount of activity in hot weather but strenuous activity would increase the metabolic rate above the body's ability to remove heat.

8 0

3 years ago