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PIT_PIT [208]
3 years ago
12

(please help asap)

Physics
2 answers:
OLEGan [10]3 years ago
6 0

Answer:

A. Yard

Explanation:

Should be correct lol :)

krok68 [10]3 years ago
5 0
I believe answer is A
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A 91-kg astronaut and a 1300-kg satellite are at rest relative to the space shuttle. The astronaut pushes on the satellite, givi
WARRIOR [948]

Answer:

18.2145 meters

Explanation:

Using the conservation of momentum, we have that:

m1v1 + m2v2 = m1'v1' + m2'v2'

m1 = m1' is the mass of the astronaut, m2=m2' is the mass of the satellite, v1 and v2 are the inicial speed of the astronaut and the satellite (v1 = v2 = 0), and v1' and v2' are the final speed of the astronaut and the satellite. Then we have that:

0 + 0 = 91*v1' + 1300*0.17

v1' = -1300*0.17/91 = -2.4286\ m/s

The negative sign of this speed just indicates the direction the astronaut goes, which is the opposite direction of the satellite.

If the astronaut takes 7.5 seconds to come into contact with the shuttle, their initial distance is:

distance = 2.4286 * 7.5 = 18.2145\ meters

8 0
3 years ago
The doppler effect is used to determine?
Korolek [52]
Uhh it is used to detirmine heat

7 0
3 years ago
Read 2 more answers
You can comfortably hold your fingers close beside a candle flame, but not very close above the flame. why? challenge (optional)
Inessa05 [86]
The candle flame releases hot gases, which directly go in upwards directions. Due to which the air near the flame of the candle is very hot and dense. The particles along with vapour move up. And since the sideways, the air is not very dense and hot, we are able to hold the candle. In anti-gravity region, there will be no density differences and also, the convection process wont occur. So, the candle quickly snuffs off.
6 0
3 years ago
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A sample of carbon dioxide (C°p,m = 37.11 J K−1 mol−1) of mass 2.80 g at 27°C is allowed to expand reversibly and adiabatically
My name is Ann [436]

Answer:

182 to 3 s.f

Explanation:

Workdone for an adiabatic process is given as

W = K(V₂¹⁻ʸ - V₁¹⁻ʸ)/(1 - γ)

where γ = ratio of specific heats. For carbon dioxide, γ = 1.28

For an adiabatic process

P₁V₁ʸ = P₂V₂ʸ = K

K = P₁V₁ʸ

We need to calculate the P₁ using ideal gas equation

P₁V₁ = mRT₁

P₁ = (mRT₁/V₁)

m = 2.80 g = 0.0028 kg

R = 188.92 J/kg.K

T₁ = 27°C = 300 K

V₁ = 500 cm³ = 0.0005 m³

P₁ = (0.0028)(188.92)(300)/0.0005

P₁ = 317385.6 Pa

K = P₁V₁¹•²⁸ = (317385.6)(0.0005¹•²⁸) = 18.89

W = K(V₂¹⁻ʸ - V₁¹⁻ʸ)/(1 - γ)

V₁ = 0.0005 m³

V₂ = 2.10 dm³ = 0.002 m³

1 - γ = 1 - 1.28 = - 0.28

W =

18.89 [(0.002)⁻⁰•²⁸ - (0.0005)⁻⁰•²⁸]/(-0.28)

W = -67.47 (5.698 - 8.4)

W = 182.3 = 182 to 3 s.f

7 0
3 years ago
A particle is traveling in the positive direction along an x axis, at a constant 5 m/s. Which of the following
kvasek [131]
The particle will accelerate 5m/s every second until it reaches a maximum of whatever your graph/diagram goes to, I'm in physical science and this is somewhat similar to what I am doing now but I'm not sure if that was what your looking for.
8 0
3 years ago
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