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maria [59]
3 years ago
13

What are the various types of technologies used for exploring space and understanding space

Physics
1 answer:
fgiga [73]3 years ago
7 0

The answer is flight technology, satelite technology, and  exploration technology.

The flight technology include advances in jet propulsion and navigation. Satelite have become more and more powerful and hecne communication with distant objects is increasing in efficiency. The emergence of AI robots in space exploration are reducing risks to humans in these deep ventures.  

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What do lungs do in your body ?
kolbaska11 [484]
We have millions of alveoli under our lungs, they diffuse oxygen molecules in our blood, then they get transported to every organ of our body through our Heart

Hope this helps!
5 0
4 years ago
Read 2 more answers
The lons entering the mass spectrometer have the same charges. After being accelerated through a potential difference of 8.20 kV
Ratling [72]

The calculated magnitude is  6.73 x 10³ V/m.

AMU is described as being one-twelfth the mass of a carbon-12 atom (12C). C makes up more than 98% of the carbon that can be found in nature, making it the most prevalent isotope. The magnitude of the field is the change in potential across a small distance in the indicated direction divided by that distance.

Potential difference = 8.20 kV= 8.20 x 10³ V

radius= 19.4/100=0.194 m

total distance that is circumference of the circle= 2πr =2 x 3.14 x 0.194

                                                                               = 1.218 m

therefore Magnitude= 8.20 x 10³ / 1.218

                                  =6.73 x 10³ V/m

Learn more about Magnitude here-

brainly.com/question/15681399

#SPJ9

4 0
1 year ago
A wheel accelerates from rest to 34.7 rad/s at a rate of 47.0 rad/s^2. Through what angle (in radians) did the wheel turn while
dem82 [27]

12.8 rad

Explanation:

The angular displacement \theta through which the wheel turned can be determined from the equation below:

\omega^2 = \omega_0^2 + 2\alpha\theta (1)

where

\omega_0 = 0

\omega = 34.7\:\text{rad/s}

\alpha = 47.0\:\text{rad/s}^2

Using these values, we can solve for \theta from Eqn(1) as follows:

2\alpha\theta = \omega^2 - \omega_0^2

or

\theta = \dfrac{\omega^2 - \omega_0^2}{2\alpha}

\:\:\:\:= \dfrac{(34.7\:\text{rad/s})^2 - 0}{2(47.0\:\text{rad/s}^2)}

\:\:\:\:= 12.8\:\text{rad}

7 0
2 years ago
At 5.0 minutes, the temperature of the water reaches 100 °C. The volume of the water in the urn
serious [3.7K]

Answer:

fdvevddvevkejokef0jeovdlvkjeuiyv

Explanation:

ddf4edscd

4 0
3 years ago
Which of the following statements apply to electric charges?
Gre4nikov [31]

Answer:

The statement "If a positively charged rod is brought close to a positively charged object, the two objects will repel " applies to electric charges.

Explanation:

There are only two types of electric charges. Both having own magnitude but different charge.

1. Positive charge

2. Negative charge

Like charges repel each other and opposite charges always attract each other.

When a positively charged rod is brought close to a positively charged object, the rod and the object will repel.

6 0
3 years ago
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