How is Forrest related to acceleration and gravity

Determine the amount of time for polonium-210 to decay to one fourth its original quantity. The half-life of polonium-210 is 138

ira [324]

Answer: 276 days

Explanation:

This problem can be solved using the Radioactive Half Life Formula:

$A=A_{o}.2^{\frac{-t}{h}}$ (1)

Where:

$A=\frac{1}{4}A_{o}$ is the final amount of the material

$A_{o}$ is the initial amount of the material

$t$ is the time elapsed

$h=138 days$ is the half life of polonium-210

Knowing this, let's substitute the values and find $t$ from (1):

$\frac{1}{4}A_{o}=A_{o}2^{\frac{-t}{138 days}}$ (2)

$\frac{A_{o}}{4A_{o}}=2^{\frac{-t}{138 days}}$ (3)

$\frac{1}{4}=2^{\frac{-t}{138 days}}$ (4)

Applying natural logarithm in both sides:

$ln(\frac{1}{4})=ln(2^{\frac{-t}{138 days}})$ (5)

$-1.386=-\frac{t}{138days}ln(2)$ (6)

Clearing $t$ :

$t=276days$ (7)

7 0

3 years ago

How do gases respond to changes in pressure and temperature?

kakasveta [241]

They compress or expand depending on amount of pressure or depending on the temperature

6 0

3 years ago

Read 2 more answers

A rigid tank initially contains 1.4 kg saturated liquid water at 200◦C. At this state, 25 percent of the volume is occupied by w

rjkz [21]

Answer:

(a) Volume of the tank is $6.47\times 10^{- 3}\ m^{3}$

(b) Temperature is $371^{\circ}C$

Pressure is 21.3 kPa

(c) The change in internal energy is 1373.54 kJ/kg

Solution:

As per the question:

Mass of liquid in the tank, m = 1.4 kg

Temperature, T = $200^{\circ}C$

Volume occupied by water, V = 25% $V_{t}$ = 0.25 $V_{t}$

Volume occupied by air, V' = 75% $V_{t}$

where

$V_{t}$ = Volume of tank

Now,

(a) In order to calculate the volume of the tank, we make use of the steam table for specific volume at as given temperature:

At $200^{\circ}C$ , the specific volume, v = 0.001157 $m^{3}/kg$

At $200^{\circ}C$ , the internal energy, u = 850.46 kJ/kg

Now, Volume of water, V = mv = $1.4\times 0.001157 = 1.62\times 10^{- 3} m^{3}$

Thus

$V = 0.25V_{t}$

$V_{t} = \frac{1.62\times 10^{- 3}}{0.25} = 6.47\times 10^{- 3}\ m^{3}$

(b) For the final temperature and pressure, we calculate the specific volume, v' and then find the corresponding temperature and pressure from the steam table:

v' = $\frac{V_{t}}{m} = \frac{6.47\times 10^{- 3}}{1.4} = 4.63\times 10^{- 3}\ m^{3}/kg$

The corresponding temperature to this specific volume, T' = $371^{\circ}C$

The corresponding pressure to this specific volume, P' = 21.3 kPa

The corresponding internal energy to this specific volume, u' = 2224 kJ/kg

(c) The change in the internal energy of water is given by:

$\Delta U = u' - u = 2224 - 850.46 = 1373.54 kJ/kg$

3 0

3 years ago

The unit of energy is a derived unit give reasons.<br>

liraira [26]

Answer:

ioinl

Explanation:

6 0

2 years ago

A bird watcher meanders through the woods, walking 1.62 km due east, 0.246 km due south, and 3.08 km in a direction 50.4 ° north

Liula [17]

Answer:

Displacement CO=2.15 Km

Average velocity = 1.99 Km/h

Explanation:

Given

Total time = 1.08 h

OA= 1.62 Km

AB=0.246 Km

BC=3.08 Km

θ=50.4°

CE=3.08 sin50.4° =2.37 Km

BE=3.08 cos50.4° =1.96 Km

CE=AB+OD

OD=CE-AB

OD=2.37-0.246

OD=2.124 Km

BE=CD+OA

CD=BE-OA

CD=1.96-1.62

CD=0.34 Km

Now CO

$CO=\sqrt{CD^2+OD^2}$

$CO=\sqrt{0.34^2+2.124^2}$

CO=2.15 Km

So the displacement CO=2.15 Km

Average velocity =Displacement/Time

Average velocity = 2.15/1.08 Km/h

Average velocity = 1.99 Km/h

8 0

4 years ago