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Ede4ka [16]
3 years ago
9

An astronaut in her space suit has a total mass of 87.0kg including suit and oxygen tank. Her tether line loses its attachment t

o her spacecraft while she's on a spacewalk. Initially at rest with respect to her spacecraft, she throws her 12.0kg oxygen tank away from her spacecraft with a speed of 8.00 m/s to propel herself back toward it. (a) determine the maximum distance she can be from the craft and still return withn 2.00 min( the amount of time the air in her helmet remains breathable). (b) Explain in terms of Newton's laws of motion why this strategy works.
Physics
1 answer:
Shtirlitz [24]3 years ago
6 0

Answer:

Explanation:

a )

In space due to weightlessness both astronaut and her oxygen tank will float .

when she throws the tank away from spacecraft , she will have a velocity in opposite direction ie towards the spacecraft . This happens due to conservation of momentum . She creates a momentum away so that she can get a momentum towards the spaceship.

So

m₁ v₁ = m₂v₂

12 x 8 = ( 87 - 12 ) x v₂

v₂ = 1.28 m /s

Time allowed = 2 x 60

= 120 s

So maximum distance upto which she can remain away from spacecraft

= 120 x 1.28

= 153 m .

b )

The Newton's law which explains the theory behind it is "third law of motion"  . This law gives law of conservation of momentum .  

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3 years ago
State Newton's three laws of motion.​
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Newton's second law states that the acceleration of an object is directly related to the net force and inversely related to its mass. Acceleration of an object depends on two things, force and mass.

Newton's third law states that if an object A exerts a force on object B, then object B must exert a force of equal magnitude and opposite direction back on object A. This law represents a certain symmetry in nature: forces always occur in pairs, and one body cannot exert a force on another without experiencing a force itself.

Explanation:

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3 years ago
Se lanza verticalmente una esfera con una rapidez de 30m/se. Determinar la rapidez de la esfera a una altura de 40m (g=10m/s2)
sergeinik [125]

v^2-{v_0}^2=2a(x-x_0)

dónde v es la velocidad de la esfera, v_0 es suya velocidad inicial, a=-g la aceleración debida a la gravedad, x la posición, y x_0 la posición inicial. Tomamos x_0=0\,\mathrm m a referirse a la posición de la esfera en el momento que la esfera fue lanzada.

Entonces

v^2-\left(30\,\dfrac{\mathrm m}{\mathrm s}\right)^2=2\left(-10\,\dfrac{\mathrm m}{\mathrm s^2}\right)(40\,\mathrm m)

\implies v^2=100\,\dfrac{\mathrm m^2}{\mathrm s^2}\implies v=\pm10\,\dfrac{\mathrm m}{\mathrm s}

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4 0
3 years ago
A thermistor is placed in a 100 °C environment and its resistance measured as 20,000 Ω. The material constant, β, for this therm
Karo-lina-s [1.5K]

Answer:

the thermistor temperature = 325.68 \ ^0 \ C

Explanation:

Given that:

A thermistor is placed in a 100 °C environment and its resistance measured as 20,000 Ω.

i.e Temperature

T_1 = 100^0C\\T_1 = (100+273)K\\\\T_1 = 373\ K

Resistance of the thermistor R_1 = 20,000 ohms

Material constant \beta = 3650

Resistance of the thermistor R_2 = 500 ohms

Using the equation :

R_1 = R_2  \ e^{\beta} (\frac{1}{T_1}- \frac{1}{T_2})

\frac{R_1}{ R_2} =   \ e^{\beta} (\frac{1}{T_1}- \frac{1}{T_2})

Taking log of both sides

In \ \frac{R_1}{ R_2} = In \  \ e^{\beta} (\frac{1}{T_1}- \frac{1}{T_2})

In \ \frac{R_1}{ R_2} = {\beta} (\frac{1}{T_1}- \frac{1}{T_2})

\frac{ In \ \frac{R_1}{ R_2}}{ {\beta}} = (\frac{1}{T_1}- \frac{1}{T_2})

\frac{1}{T_2} =   \frac{1}{T_1}  -          \frac{ In \ \frac{R_1}{ R_2}}{ {\beta}}

{T_2} =  \frac{\beta T_1}{\beta - In (\frac{R_1}{R_2})T}

Replacing our values into the above equation :

{T_2} =  \frac{3650*373}{3650 - In (\frac{20000}{500})373}

{T_2} =  \frac{1361450}{3650 - 3.6888*373}

{T_2} =  \frac{1361450}{3650 - 1375.92}

{T_2} =  \frac{1361450}{2274.08}

{T_2} = 598.68 \ K

{T_2} = 325.68 \ ^0 \ C

Thus, the thermistor temperature = 325.68 \ ^0 \ C

4 0
4 years ago
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