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Sunny_sXe [5.5K]
4 years ago
12

Why do astronauts need to wear pressurized suits in space?

Physics
1 answer:
SashulF [63]4 years ago
6 0
The correct answer is C , because the space is vacuum and his body can explode and for this reason,  the astronaut need a special costum to be protected. It's the same on the moon, because there is no atmosphere 
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An overnight rainstorm has caused a major roadblock. Three massive rocks of mass m1=584 kg, m2=838 kg, and m3=322 kg have blocke
Elena-2011 [213]

Answer:

Force must be applied to m₁ to move the group of rocks from the road at 0.250 m/s² = 436 N

Explanation:

Total force required = Mass x Acceleration,

F = ma

Here we need to consider the system as combine, total mass need to be considered.

Total mass, a = m₁+m₂+m₃ = 584 + 838 + 322 = 1744 kg

We need to accelerate the group of rocks from the road at 0.250 m/s²

That is acceleration, a = 0.250 m/s²

Force required, F = ma = 1744 x 0.25 = 436 N

Force must be applied to m₁ to move the group of rocks from the road at 0.250 m/s² = 436 N

8 0
3 years ago
A blue liquid with a density of 1.2 g/cm3
lyudmila [28]
All you have to do is multiply that by the mass or volume.
3 0
3 years ago
Some hydrogen gas is enclosed within a chamber being held at 200^\ { C} with a volume of 0.025 \rm m^3. The chamber is fitted wi
vlada-n [284]

Answer:

The final volume is 0.039 m^3

Explanation:

<u>Data:</u>

Initial temperature: T1=200C

Final temperature: T2=200C

Initial pressure: P1=1.50 \times10^6 Pa

Final pressure: P2=0.950 \times10^6 Pa

Initial volume: V1=0.025m^{3}

Final volume: V2=?

Assuming hydrogen gas as a perfect gas it satisfies the perfect gas equation:

\frac{PV}{T}=nR (1)

With P the pressure, V the volume, T the temperature, R the perfect gas constant and n the number of moles. If no gas escapes the number of moles of the gas remain constant so the right side of equation (1) is a constant, that allows to equate:

\frac{P_{1}V_{1}}{T_{1}}=\frac{P_{2}V_{2}}{T_{2}}

Subscript 2 referring to final state and 1 to initial state.

solving for V2:

V_{2}=\frac{P_{1}V_{1}T_{2}}{T_{1}P_{2}}=\frac{(1.50 \times10^6)(0.025)(200)}{(200)(0.950 \times10^6)}

V_{2}=0.039 m^3

3 0
4 years ago
A wire loop of radius 0.40 m lies so that an external magnetic field of magnitude 0.30 T is perpendicular to the loop. The field
Marat540 [252]

Answer:

The magnitude of the average induced emf in the loop is 0.1 volts.

Explanation:

let Ф be the flux in the loop, B be the magnetic field and A be the area of the loop.

the induced emf in the loop is given by:

ε = - dФ/dt

  =  - d(B×A)/dt

  = - A×d(B)/dt

  = - π×r^2×d(B)/dt

  = -  π×(0.40)^2×(0.20 - (-0.30))/(2.5 - 0)

  = 0.1 volt

Therefore, the magnitude of the average induced emf in the loop is 0.1 volts.

8 0
4 years ago
The question is on he attachment
atroni [7]

Answer:

A. put water in a container under the pot water will disappear from the container .

3 0
3 years ago
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