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4 years ago

One atomic mass unit is 0.0000000000000000000000166 what is the weight in scientific notation?

Mathematics

2 answers:

jeka57 [31]4 years ago

4 0

1.7 x 10 ^-23

Or maybe 1.6 x 10^-23

(The -23 is how many 0’s there are).

Ulleksa [173]4 years ago

3 0

Answer:

One atomic mass unit (AMU) in scientific notation is 1.66 x 10^26

Step-by-step explanation:

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Oscar needs to purchase soft drinks for a staff party. He is

PilotLPTM [1.2K]

Answer:

sorry getting points

Step-by-step explanation:

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3 years ago

Which equation is represented by the graph below?

garik1379 [7]

Answer:c

Step-by-step explanation:

5 0

3 years ago

Read 2 more answers

A given field mouse population satisfies the differential equation dp dt = 0.5p − 410 where p is the number of mice and t is the

ohaa [14]

Answer:

a) $t = 2 *ln(\frac{82}{5}) =5.595$

b) $t = 2 *ln(-\frac{820}{p_0 -820})$

c) $p_0 = 820-\frac{820}{e^6}$

Step-by-step explanation:

For this case we have the following differential equation:

$\frac{dp}{dt}=\frac{1}{2} (p-820)$

And if we rewrite the expression we got:

$\frac{dp}{p-820}= \frac{1}{2} dt$

If we integrate both sides we have:

$ln|P-820|= \frac{1}{2}t +c$

Using exponential on both sides we got:

$P= 820 + P_o e^{1/2t}$

Part a

For this case we know that p(0) = 770 so we have this:

$770 = 820 + P_o e^0$

$P_o = -50$

So then our model would be given by:

$P(t) = -50e^{1/2t} +820$

And if we want to find at which time the population would be extinct we have:

$0=-50 e^{1/2 t} +820$

$\frac{820}{50} = e^{1/2 t}$

Using natural log on both sides we got:

$ln(\frac{82}{5}) = \frac{1}{2}t$

And solving for t we got:

$t = 2 *ln(\frac{82}{5}) =5.595$

Part b

For this case we know that p(0) = p0 so we have this:

$p_0 = 820 + P_o e^0$

$P_o = p_0 -820$

So then our model would be given by:

$P(t) = (p_o -820)e^{1/2t} +820$

And if we want to find at which time the population would be extinct we have:

$0=(p_o -820)e^{1/2 t} +820$

$-\frac{820}{p_0 -820} = e^{1/2 t}$

Using natural log on both sides we got:

$ln(-\frac{820}{p_0 -820}) = \frac{1}{2}t$

And solving for t we got:

$t = 2 *ln(-\frac{820}{p_0 -820})$

Part c

For this case we want to find the initial population if we know that the population become extinct in 1 year = 12 months. Using the equation founded on part b we got:

$12 = 2 *ln(\frac{820}{820-p_0})$

$6 = ln (\frac{820}{820-p_0})$

Using exponentials we got:

$e^6 = \frac{820}{820-p_0}$

$(820-p_0) e^6 = 820$

$820-p_0 = \frac{820}{e^6}$

$p_0 = 820-\frac{820}{e^6}$

8 0

4 years ago

Dafna11 [192]

The approximate length of rail that needs to be replaced is 7.1 ft

<h3>Length of arc</h3>

Since the pool is circular, the approximate length that needs to be replaced is an arc of length, L = Ф/360° × πD where

Ф = central angle of rail section = 27° and
D = diameter of circular pool = 30 ft

<h3>Approximate length of rail</h3>

So, substituting the values of the variables into the equation, we have

L = Ф/360° × πD

L = 27°/360° × π × 30 ft

L = 3/40 × π × 30 ft

L = 3/4 × π × 3 ft

L = 9/4 × π ft

L = 2.25 × π ft

L = 7.07 ft

L ≅ 7.1 ft

So, the approximate length of rail that needs to be replaced is 7.1 ft

Learn more about length of an arc here:

brainly.com/question/8402454

5 0

2 years ago

Find the sum of the arithmetic series given ai = 45, an = 85, and n = 5.

seraphim [82]

Answer:

C. 325.

Step-by-step explanation:

The last term a5 = 85

a1 = 45

Sum of n terms = n/2 (a1 + l)

So here we have n = 5, a1 = 45 and the last term l = 85

= (5/2)(45 + 85)

= 5/2 * 130

= 325.

5 0

3 years ago