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2 years ago

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Dinitrogen oxide has?

1 answer:

Vesnalui [34]2 years ago

3 0

Two atoms of the element bind to form dinitrogen and it’s a colourless and odorless diatomic gas with the formula N2. Also Dinitrogen forms up about 78% of earths atmosphere

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What is your volume of the object?<br> 30 cm3<br> 35 cm3<br> 42 cm3<br> 54 cm3

Alinara [238K]

The answer is 54 cm3

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2 years ago

Any two application of gravity

Brrunno [24]

Answer:

Well the definition of an application is the act of putting to a special use or purpose so lam assuming that you want specific uses that scientists make of gravity in their work.

Well our first application has helped us to send satellites around the solar system with what Nasa calls gravity assist. Using a particular planets gravity to slingshot a satellite to another destination. Look it up.

The next application much simpler but here on Earth. There are many hydro-electric power stations in use all over the world. Water is stored at a high level and released falling 100s of metres to a turbine where it generates electricity.

Hope that helps.

Explanation:

5 0

2 years ago

Need help ASAP <br> Thankss + BRAINLIST only for correct answer

spayn [35]

Answer:

False

true

Hope this helps!

7 0

2 years ago

Read 2 more answers

Determine the critical crack length for a through crack contained within a thick plate of 7150-T651 aluminum alloy that is in un

Mila [183]

Explanation:

Formula to determine the critical crack is as follows.

$K_{IC} = \gamma \sigma_{f} \sqrt{\pi \times a}$

$\gamma$ = 1, $K_{IC}$ = 24.1

[/tex]\sigma_{y}[/tex] = 570

and, $\sigma_{f} = 570 \times \frac{3}{4}$

= 427.5

Hence, we will calculate the critical crack length as follows.

a = $\frac{1}{\pi} \times (\frac{K_{IC}}{\sigma_{f}})^{2}$

= $\frac{1}{3.14} \times (\frac{24.1}{427.5})^{2}$

= $10.13 \times 10^{-4}$

Therefore, largest size is as follows.

Largest size = 2a

= $2 \times 10.13 \times 10^{-4}$

= $20.26 \times 10^{-4}$

Thus, we can conclude that the critical crack length for a through crack contained within the given plate is $20.26 \times 10^{-4}$ .

4 0

2 years ago

Before starting this problem, review Conceptual Example 3 in your text. Suppose that the hail described there comes straight dow

bulgar [2K]

Answer:

0.9 N

Explanation:

The force exerted on an object is related to its change in momentum by:

$F=\frac{\Delta p}{\Delta t}$

where

F is the force exerted

$\Delta p$ is the change in momentum

$\Delta t$ is the time interval

The change in momentum can be rewritten as

$\Delta p = m(v-u)$

where

m is the mass

u is the initial velocity

v is the final velocity

So the formula can be rewritten as

$F=\frac{m(v-u)}{\Delta t}$

In this problem we have:

$\frac{m}{\Delta t}=0.030 kg/s$ is the mass rate

$u=-15 m/s$ is the initial velocity

$v=+15 m/s$ is the final velocity

Therefore, the force exerted by the hail on the roof is:

$F=(0.030)(+15-(-15))=0.9 N$

6 0

2 years ago