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MAXImum [283]
4 years ago
11

How many atoms are present in 3 molecules of chromium

Physics
1 answer:
iren2701 [21]4 years ago
5 0
Answer:

1,8066 x 1024 atoms

Explanation:

1 mole of chromium contains 6,022 x 1023 atoms

so

3 moles of chromium contains 3 times as many atoms

3 x 6,022 x 1023 = 1,8 066 x 1024

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Need an answer, please need an answer, please​
Karolina [17]

9514 1404 393

Answer:

  • moderate low: 82
  • moderate high: 112.75
  • vigorous low: 123
  • vigorous high: 174.25

Explanation:

When calculations are repetitive, I find it convenient to use a calculator that can work with tables.

The PMHR is (220 -15) = 205.

Each of the other heart rates is computed as the formula shows. For example, the low moderate heart rate is 205×0.40 = <u> 82 </u> bpm

The other rates are shown in the attached table. They are computed the same way.

6 0
3 years ago
What would the circuit resistance be if a 7.5 Amp draw was present with the engine running and the charging system producing 15
AlladinOne [14]

Answer:

2 ohms

Explanation:

Hi there!

Ohm's law states that V=IR where V is the voltage, I is the current and R is the resistance.

Plug in the given information (I=7.5, V=15) and solve for R

V=IR\\15=(7.5)R

Divide both sides by 7.5 to isolate R

\frac{15}{7.5}= \frac{7.5R}{7.5}  \\2=R

Therefore, the circuit resistance would be 2 ohms.

I hope this helps!

4 0
3 years ago
based on the information in the graph why is energy released during the fission of a uranium (U) nucleus?
frutty [35]

Answer:

B

Explanation:

Nuclear fission is the process of splitting apart nuclei. When nuclei fission energy is released. So much energy is released that there is a measurable decrease in mass, from the mass-energy equivalence. This means that some of the mass is converted to energy.

4 0
3 years ago
Read 2 more answers
A 18 kg sled starts up a 28 degree incline with a speed of 2.3 m/s. The coefficient of kinetic friction between the sled and the
sergey [27]

Answer:

d = 0.391 m

Explanation:

given,

mass of sled = 18 kg

inclined at an angle of  = 28°

kinetic friction between the sled and inclined = 0.25

using energy equation

\dfrac{1}{2}mv^2- Fd - mgh = 0

\dfrac{1}{2}mv^2- (\mu mg cos\theta )d - mgdsin\theta = 0

\dfrac{1}{2}mv^2 = (\mu mg cos\theta )d + mgdsin\theta

\dfrac{1}{2}mv^2 = d mg (\mu cos\theta+sin\theta)

d = \dfrac{v^2}{2g (\mu cos\theta+sin\theta)}

d = \dfrac{2.3^2}{2\times 9.8 (0.25\times cos28^0+sin28^0)}

d = 0.391 m

hence, the distance traveled by the sled is equal to 0.391 m

7 0
3 years ago
Consider the four free body diagrams. In which case does the net force equal 15N up
Mila [183]

The answer is D.

40-25=15.

Hope this helps.

3 0
3 years ago
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