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Fynjy0 [20]
3 years ago
6

What is the definition of oxygen​

Physics
2 answers:
natima [27]3 years ago
7 0

Answer:

Something we convert into c02 to power our bodies

Explanation:

We use it to fuel ourselfs

DerKrebs [107]3 years ago
6 0
“ a colorless, odorless reactive gas, the chemical element of atomic number 8 and the life-supporting component of the air. “
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(b) 360 days into seconds.
Angelina_Jolie [31]

Explanation:

(b) We know that,

1 day = 24 hours

1 hour = 3600 s

So, we found that, 1 day = 86400 s

We need to find the 360 days into seconds. So,

1 day = 86400 s

360 days = 86400×360

360 days = 31104000 seconds

(d) Weight of a body, W = 600 N

Acceleration due to gravity on mars is 3.7 m/s²

Weight, W = mg

m is mass of body

m=\dfrac{W}{g}\\\\m=\dfrac{600}{3.7}\\\\m=162.16\ kg

(e) Mass of body, m = 100 kg

Acceleration due to gravity on the moon, 1.6 m/s²

Weight, W = 100 × 1.6

W = 160 N

8 0
3 years ago
A student compresses the spring in a pop up toy .020 meter if the sprinf has a spring constant of 340 newtons per meter how much
vodomira [7]
The answer is letter c-0.068 j
5 0
3 years ago
What are some charactersitics of pressure waves
True [87]
Soft target by impact and its contribution to indirect bone fractures.
7 0
3 years ago
If Scobie could drive a Jetson's flying car at a constant speed of 160.0 km/hr across oceans and space, approximately how long w
DochEvi [55]

Scobie will take 10 days to drive around Earth's equator.

To calculate the time that takes Scobie to drive around Earth's equator we need to find the distance, which is given by the equation of a circumference:

d = 2\pi r

<em>Where:</em>

r: is the Earth's radius = 6371 km

Then, the distance is:

d = 2\pi r = 2\pi*6371 km = 40030.2 km

Now, if we divide the above distance by the speed of the car we can find the time:

t = \frac{d}{v} = \frac{40030.2 km}{160.0 km/h} = 250.2 h*\frac{1 d}{24 h} = 10 d

Therefore, Scobie will take 10 days to drive around Earth's equator.

     

To learn more about distance and time here: brainly.com/question/14236800?referrer=searchResults

I hope it helps you!

6 0
3 years ago
A Carnot engine operates with an efficiency of 26.0% when the temperature of its cold reservoir is 281 K. Assuming that the temp
VLD [36.1K]

Answer:

The temperature of cold reservoir should be 246.818 K for efficiency of 35%

Explanation:

In first case we have given efficiency of Carnot engine = 26 % = 0.26

Temperature of cold reservoir T_L=281K

We know that efficiency of Carnot engine is given by

\eta =1-\frac{T_L}{T_H}

0.26 =1-\frac{281}{T_H}

T_H=379.72K

For second Carnot engine efficiency is given as 35% = 0.35

And temperature of hot reservoir is same so T_H=379.72K

So 0.35=1-\frac{T_L}{379.72}

T_L=246.818K

So the temperature of cold reservoir should be 246.818 K for efficiency of 35%

4 0
4 years ago
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