Ask question

Ask question

All categories

3 years ago

13

A drop of liquid tends to have a spherical shape due to the property of 1. viscosity. 2. capillary action. 3. surface tension. 4

. vapor pressure. 5. close packing.

1 answer:

posledela3 years ago

3 0

Answer:

surface tension.

Explanation:

You might be interested in

The —— of melting is freezing

natima [27]

Should be reverse or opposite

4 0

3 years ago

Read 2 more answers

I WILL MARK Brainest

wolverine [178]

It should be sketch,identify, dig, then set water for the watering times, hope this helps, hope you feel better soon

5 0

3 years ago

5 points<br> What do you think photosynthesis means ?

Helga [31]

The process of turning photons (sunlight) and carbon dioxide Co^2 and water H^2O into glucose (sugar)

7 0

3 years ago

The initial activity of 37Ar is 8540 disintegrations per minute. After 10.0 days, the activity is 6990 disintegrations per minut

mylen [45]

Answer:

Approximately 3318 disintegrations per minute.

Explanation:

The activity $A$ of a radioactive decay at time $t$ can be found with the following equation:

$\displaystyle A(t) = A_0 \cdot \mathrm{e}^{-\lambda\cdot t}$ .

In this equation,

$\mathrm{e}$ is the natural base. $\mathrm{e}\approx 2.71828$ .
$A_0$ is the initial activity of the decay. For this question, $A_0 = \rm 8540\; min^{-1}$ .
The decay constant $\lambda$ of this sample needs to be found.

The decay constant here can be found using the activity after 10 days. As long as both times are in the same unit (days in this case,) conversion will not be necessary.

$A(10) = A_0\cdot \mathrm{e}^{\rm -\lambda \times 10\;day}= (\mathrm{8540\; min^{-1}})\cdot \mathrm{e}^{\rm -\lambda \times 10\;day}$ .

$A(10) = \rm 6690\; min^{-1}$ .

$\displaystyle \frac{\rm 6690\; min^{-1}}{\mathrm{8540\; min^{-1}}} = \mathrm{e}^{\rm -\lambda \times 10\;day}$

Apply the natural logarithm to both sides of this equation.

$\displaystyle \ln{\left(\frac{\rm 6690\; min^{-1}}{\mathrm{8540\; min^{-1}}}\right)} = \ln{\left(\mathrm{e}^{\rm -\lambda \times 10\;day}\right)}$ .

$\displaystyle \rm -\lambda \cdot (10\;day) = \ln{\left(\frac{\rm 6690\; min^{-1}}{\mathrm{8540\; min^{-1}}}\right)}$ .

$\displaystyle \rm \lambda= \rm \frac{\displaystyle \ln{\left(\frac{\rm 6690\; min^{-1}}{\mathrm{8540\; min^{-1}}}\right)}}{-10 \; day} \approx 0.0200280\; day^{-1}$ .

Note that the unit of the decay constant $\lambda$ is $\rm day^{-1}$ (the reciprocal of days.) The exponent $-\lambda \cdot t$ should be dimensionless. In other words, the unit of $t$ should also be days. This observation confirms that there's no need for unit conversion as long as the two times are in the same unit.

Apply the equation for decay activity at time $t$ to find the decay activity after 47.2 days.

$\displaystyle \begin{aligned}A(t)& = A_0 \cdot \mathrm{e}^{-\lambda\cdot t}\\&\approx \rm \left(8540\; min^{-1}\right)\cdot \mathrm{e}^{-0.0200280\; day^{-1}\times 47.2\;day}\\&\approx \rm 3318\; min^{-1}\end{aligned}$ .

By dimensional analysis, the unit of activity here should also be disintegrations per minute. The activity after 47.2 days will be approximately 3318 disintegrations per minute.

8 0

3 years ago

Identify the reaction<br> In photo

MrRa [10]

Answer:

single displacement

Explanation:

4 0

2 years ago