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JulijaS [17]
3 years ago
13

A 60.0 kg astronaut is on a space walk away from the shuttle when her tether line breaks! She is able to throw her 10.0 kg oxyge

n tank away from the shuttle with a velocity of 12.0 m/s to propel herself back to the shuttle. Assuming that she starts from rest (relative to the shuttle), determine the maximum distance she can be from the craft when the line breaks and still return with 60.0 s ( the amount of time she can hold her breath).
Physics
1 answer:
IgorLugansk [536]3 years ago
3 0

Answer:

120 m

Explanation:

m_1 = Mass of Astronaut

m_1 = Mass of tank

v_1 = Velocity of Astronaut

v_2 = Velocity of tank

From conservation of linear momentum

m_1v_1=m_2v_2\\\Rightarrow v_1=\frac{m_2v_2}{m_1}\\\Rightarrow v_1=\frac{10\times 12}{60}\\\Rightarrow v_1=2\ m/s

Distance = Speed × Time

Distance=2\times 60=120\ m

So, the maximum distance she can travel is 120 m

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Two buses are driving along parallel freeways that are 5mi apart, one heading east and the other heading west. Assuming that eac
Oksanka [162]

Answer:

101.54m/h

Explanation:

Given that the buses are 5mi apart, and that they are both driving at the same speed of 55m/h, rate of change of distance can be determined using differentiation as;

Let l be the be the distance further away at which they will meet from the current points;

l=\sqrt{13^2-5^2}=12m\\\\\frac{dl}{dt}=-(55m/h+55m/h})\\\\=-110m/h#The speed toward each other.

\frac{dh}{dt}=0, \ \ \ \ h=constant\\\\h^2+l^2=b^2\\\\2h\frac{dh}{dt}+2l\frac{dl}{dt}=2b\frac{db}{dt}\\\\2\times5\times0+2\times12\times(-110)=2\times13\frac{db}{dt}\\\\\frac{db}{dt}=-101.54m/h

Hence, the rate at which the distance between the buses is changing when they are 13mi apart is 101.54m/h

4 0
3 years ago
Consider a venturi with a small hole drilled in the side of the throat. This hole is connected via a tube to a closed reservoir.
Marina CMI [18]

Answer:

(P_1-P_2)=1913.31 N/m^2

Explanation:

given:

\frac{A_t}{A_1}=0.85

V_1=90 m/s

γ∞=1.23 kg/m^3

solution:

since outside pressure is atm pressure vaccum can be defined by (P_1-P_2)

V_1=√2(P_1-P_2)/γ∞[\frac{A_t}{A_1}^2-1]

(P_1-P_2)=1913.31 N/m^2

6 0
3 years ago
A dumbell has a mass of 95 kg. What force must be applied to accelerate it upward at 2.2 m/s2?
Sveta_85 [38]
A :-) F = ma
Given - m = 95 kg
a = 2.2 m/s^2
Solution -
F = ma
F = 95 x 2.2
F = 209

.:. The force is 209 N
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3 years ago
Can someone help me with the exercise 3 only c,d,e please!!!
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I'm not sure...
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4 0
3 years ago
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Which refers to the sum of all the forces that act upon an object?
ElenaW [278]

Answer:

Net force refers to the sum of all the forces that act upon an object.

Explanation:

4 0
3 years ago
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