A(n) 71.1 kg astronaut becomes separated from the shuttle, while on a space walk. She finds herself 70.2 m away from the shuttle
1 answer:
Answer:
10.347 minutes.
Explanation:
According to F = ma, she exerts force on camera of the magnitude
F = 0.67Kg*12m/ = 8.04N, assuming it took her one second to accelerate camera to 12m/s, then by newtons third law, which says every action has equal and opposite reaction , the camera exerts the same amount of force on the astronaut which gives her acceleration of a =
.
and velocity of V = 0.1130801680m/s.
at this velocity , the astronaut has to cover the distance of 70.2 meters, it will take her 620.7985075s = 10.347 min to get to the shuttle (using S = vt).
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Answer:
4800N
Explanation:
Lets assume,
Mass of first object = m₁
Mass of second object = m₂
Distance between the two objects = r
Thus the force between the two objects will be
where, G = Universal gravitational constant
Given, F = 2400N
New mass of second object = 2m₂
Now, the force will be
Thus, F₂ = 4800N
Explanation:
First we will convert the given mass from lb to kg as follows.
157 lb =
= 71.215 kg
Now, mass of caffeine required for a person of that mass at the LD50 is as follows.
= 12818.7 mg
Convert the % of (w/w) into % (w/v) as follows.
0.65% (w/w) =
=
=
Therefore, calculate the volume which contains the amount of caffeine as follows.
12818.7 mg = 12.8187 g =
= 1972 ml
Thus, we can conclude that 1972 ml of the drink would be required to reach an LD50 of 180 mg/kg body mass if the person weighed 157 lb.