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Darina [25.2K]
3 years ago
8

PLZZ HELP ASAP!!

Physics
1 answer:
Nimfa-mama [501]3 years ago
8 0
Modern space suits augment the basic pressure garment with a complex system of equipment and environmental systems designed to keep the wearer comfortable, and to minimize the effort required to bend the limbs, resisting a soft pressure garment's natural tendency to stiffen against the vacuum. A self-contained oxygen supply and environmental control system is frequently employed to allow complete freedom of movement, independent of the spacecraft.
Three types of spacesuits exist for different purposes: IVA (intravehicular activity), EVA (extravehicular activity), and IEVA (intra/extravehicular activity). IVA suits are meant to be worn inside a pressurized spacecraft, and are therefore lighter and more comfortable. IEVA suits are meant for use inside and outside the spacecraft, such as the Gemini G4C suit. They include more protection from the harsh conditions of space, such as protection from micrometeorites and extreme temperature change. EVA suits, such as the EMU, are used outside spacecraft, for either planetary exploration or spacewalks. They must protect the wearer against all conditions of space, as well as provide mobility and functionality.
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Which of the following criteria can help Linda classify gases as greenhouse gases? gas molecules having three or more atoms gas
ratelena [41]
I think the correct answer would be the third  option. The criteria that could help Linda in classifying whether the gases are greenhouse gases would be gas molecules having at least one oxygen atom. Most of the greenhouse gases has an oxygen atom in their structures especially those that naturally occurs. These gases are CO2, H2O vapor and nitrous oxide. 
7 0
3 years ago
Read 2 more answers
The Moon requires about 1 month (0.08 year) to orbit Earth. Its distance from us is about 400,000 km (0.0027 AU). Use Kepler’s t
dem82 [27]

Answer:

\frac{M_e}{M_s} = 3.07 \times 10^{-6}

Explanation:

As per Kepler's III law we know that time period of revolution of satellite or planet is given by the formula

T = 2\pi \sqrt{\frac{r^3}{GM}}

now for the time period of moon around the earth we can say

T_1 = 2\pi\sqrt{\frac{r_1^3}{GM_e}}

here we know that

T_1 = 0.08 year

r_1 = 0.0027 AU

M_e = mass of earth

Now if the same formula is used for revolution of Earth around the sun

T_2 = 2\pi\sqrt{\frac{r_2^3}{GM_s}}

here we know that

r_2 = 1 AU

T_2 = 1 year

M_s = mass of Sun

now we have

\frac{T_2}{T_1} = \sqrt{\frac{r_2^3 M_e}{r_1^3 M_s}}

\frac{1}{0.08} = \sqrt{\frac{1 M_e}{(0.0027)^3M_s}}

12.5 = \sqrt{(5.08 \times 10^7)\frac{M_e}{M_s}}

\frac{M_e}{M_s} = 3.07 \times 10^{-6}

4 0
3 years ago
A 25 kg rock resting on the bottom of a lake must be moved from the paths of boats. The rock has a density of 2350 kg/m^3. What
algol13

Answer:

The force needed is the weight of the rock minus the buoyant force.

Explanation:

7 0
3 years ago
If you wanted to soundproof a room what insulator would you use
olasank [31]

Answer:

foam but thats what i heard from other people

Explanation:

4 0
3 years ago
Read 2 more answers
Two separate masses on two separate springs undergo simple harmonic motion indefinitely (the surface is frictionless). In CASE 1
Artemon [7]

Answer:

\frac{F_1}{F_2} =\frac{1}{2}

Explanation:

Assuming the we have to find ratio maximum forces on the mass in each case

we know that in a spring mass system

F= Kx

K= spring constant

x= spring displacement

Case 1:

F_1=2k\times d

case 2:

F_2=2k\times 2d

therefore, \frac{F_1}{F_2} = \frac{2K\times d}{2K\times 2d}

\frac{F_1}{F_2} =\frac{1}{2}

4 0
4 years ago
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