I think you forgot to include the acceleration due to
gravity of astronauts. I assume that it is = 0.170 g. To get the answer we have
to use the formula s = v0t – (1/2) At². Where s is the altitude, A is the
acceleration of gravity, t is the time after throwing.
v = v0 –At
v = 0 at max altitude so v0 – At = 0
t = v0/A at max altitude
Using the formula above for the altitude:
s = v0t – (1/2) At²
s = v0(v0/A) – (1/2) A (v0/A)²
s = v0²/A – (1/2) v0²/A
s = (1/2) v0²/A
The earth: E = (1/2) v0²/g
The moon: M = (1/2)v0²(0.17g)
So, take the ratio of M/E = g/0.17g = 1/0.17 = 588
M = 5.88 E
He can throw the wrench 5.88 times higher on the moon
<span>M =5.88 (10 m) = 58.8 meters that the can throw
the wrench a little over on the moon.</span>
Answer:
I believe its C: Secretary of War. I hope this helped :)
Explanation:
Answer:
3 Objects that carry instruments into the stratosphere to measure atmospheric conditions
Explanation:
Weather balloons are objects that carries instruments into the stratosphere to measure atmospheric conditions.
They are very important weather measuring device that helps meteorologists in their fore casts.
These balloons are specially designed for high altitudes. They carry very sensitive devices that are very useful in collecting information about temperature, precipitation, pressure e.t.c
Answer:
e. the number of active motor units increases.
Explanation:
There is a direct relationship between the number of active motor units and the grip strength in a given scenario. <em>For example, increase in the grip strength leads to increase in the number of active motor units. In the other-hand, the decrease in grip strength leads to the decrease in the number of active motor units.</em>
Answer:
The width of the slit is 0.167 mm
Explanation:
Wavelength of light, 
Distance from screen to slit, D = 88.5 cm = 0.885 m
The distance on the screen between the fifth order minimum and the central maximum is 1.61 cm, y = 1.61 cm = 0.0161 m
We need to find the width of the slit. The formula for the distance on the screen between the fifth order minimum and the central maximum is :
where
a = width of the slit
a = 0.000167 m
a = 0.167 mm
So, the width of the slit is 0.167 mm. Hence, this is the required solution.
