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geniusboy [140]
3 years ago
6

If an astronaut can throw a certain wrench 10.0 m vertically upward on earth, how high could he throw it on our moon if he gives

it the same starting speed in both places?
Physics
1 answer:
Jobisdone [24]3 years ago
7 0

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>
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A projectile is fired directly upward from the ground with an initial velocity of 112 f/sec its distance s above the ground afte
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Answer:

3.48 seconds

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⇒g*t= Vi

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7 0
3 years ago
Please answer the number 5 6 and 7. I don't know what to do its our hw in physics. (new lesson as well)
Aleks04 [339]

From the picture, I see that you had no trouble at all with #4.
Well, #5, 6, and 7 are easily handled in exactly the same way.

Just as you did with #4, please sketch these on paper
as I walk you through the solutions.  That'll help you see
immediately what's going on.

#5.b).
Traveling east at 3 m/s for 4 seconds,
he covers  (3 m/s) x (4 sec) = 12 meters.

Traveling south at 5 m/s for 2 seconds,
he covers (5 m/s) x (2 sec) = 10 meters.

The total distance he covers is  (12m + 10m) = 22 meters.

#5.c).
Average speed (scalar)

                           = (distance covered)/(time to cover the distance)

                           = (22 meters)/(6 sec) = 3-2/3 m/s .

#5.d).
Displacement (vector)

                       = distance between the start-point and the end-point,
                          regardless of the route traveled,
                      
  in the direction from the start-point to the end-point.

Distance from the start-point to the end-point =

               √(12² + 10²) = √(144 + 100) = √(244) = 15.62 meters

in the direction of  arctan(10/12) south of east

                             =  39.8° south of east.
 
#5.e).
Average velocity (vector) =

             (displacement vector) / (time)

         =  15.62 meters directed 39.8° south of east / 6 seconds

         =  2.603 m/s directed 39.8° south of east.

 #6).
Magnitude = √(5.2² + 2.1²) = √(27.04 + 4.41) = √31.45 = 5.608 km.

Direction = arctan(5.2/2.1) south of east

                =   68° south of east  =  158° bearing . 

#7).
Magnitude = √(39² + 57²) = √(1521 + 3249) = √( 4770)

                                                                         =  69.07 m/s .

Direction = arctan (57/39) south of west

               =   55.6° south of west

                    Bearing = 214.4°

                    Compass: 0.65° past "southwest by south".  


I'm grateful for the privilege and opportunity to practice my math,
and I shall cherish the bounty of 5 points that came with it.

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