anonymous
  • anonymous
Cleo and Clare are looking from their balcony to a swimming pool below that is located 15 m horizontally from the bottom of their building. They estimate the balcony is 45 m high and wonder how fast they would have to jump horizontally to succeed in reaching the pool. What calculations would you show to help them determine the answer? Evaluate the practicality of their being able to succeed at jumping into the pool.
Physics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
@iGreen
anonymous
  • anonymous
@Michele_Laino
anonymous
  • anonymous
@radar

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
In three scenarios a truck is towing a boat on a trailer. Using 3 – 4 complete sentences indicate which scenario(s) (if any) is the force of the truck on the trailer greater than the force of the trailer on the truck: Scenario A: masstruck = 1000 kg, massboat and trailer = 500 kg; velocity = 20 m/s Scenario B: masstruck = 800 kg, massboat and trailer = 800 kg; velocity = 20 m/s Scenario C: masstruck = 500 kg, massboat and trailer = 1000 kg; velocity = 15 m/s
Michele_Laino
  • Michele_Laino
hello!
Michele_Laino
  • Michele_Laino
have you got any ideas?
anonymous
  • anonymous
Not at all, i got some some earlier but theres a few that I'm lost on.
Michele_Laino
  • Michele_Laino
I think that in all of your three cases, both forces have the same magnitude, since we have to apply the third law of Newton
Michele_Laino
  • Michele_Laino
furthermore, since the velocity is constant, then we can consider the system truck+boat, an inertial system
Michele_Laino
  • Michele_Laino
and, as you know, the laws of Physics are the same into all inertial systems
anonymous
  • anonymous
so none of them are greater than the other? correct?
Michele_Laino
  • Michele_Laino
yes! correct, since the velocity of the system truck+boat, is constant in all of your three cases
anonymous
  • anonymous
how could i try and explain that in a few sentences?
Michele_Laino
  • Michele_Laino
here is my sentence: "since the velocity of the system truck+boat" is constant, as vector, then the same system "truck+boat" is an inertial system. Now within an inertial system, all law of Physics, have the same shape, as formula, so we can apply the third law of Newton, and we can conclude that the force applied by truck on boat have the same direction and magnitude of the force applied by the boat on truck. Furthermore those forces have opposite orientation each other"
anonymous
  • anonymous
what about the first question?
Michele_Laino
  • Michele_Laino
we have to consider these two equations: \[\left\{ \begin{gathered} h = \frac{1}{2}g{t^2} \hfill \\ \hfill \\ d = vt \hfill \\ \end{gathered} \right.\] where d=15 meters and h= 45 meters
Michele_Laino
  • Michele_Laino
now, I solve the second equation for t, and I get: \[t = \frac{d}{v}\]
Michele_Laino
  • Michele_Laino
Now I substitute into the first equation, and I get: \[45 = \frac{1}{2}g{\left( {\frac{{15}}{v}} \right)^2}\] please solve that equation for velocity v, here g=9.81 m/sec^2
anonymous
  • anonymous
i dont understand you math I'm confused
anonymous
  • anonymous
@Michele_Laino
Michele_Laino
  • Michele_Laino
the subsequent drawing represents the situation of your exercise: |dw:1432832271358:dw|
Michele_Laino
  • Michele_Laino
this time: \[t = \sqrt {\frac{{2 \times 45}}{{9.81}}} \] is the time needed to Cleo and Clare to reach the base of their hotel. During that time they have to travel, horizonatally by a distance L=15 meters, in order to reach the swimming pool, and the time needed to travel that distance L, is: \[{t_1} = \frac{{15}}{v}\] where v is the horizonatl velocity of Cleo and Clear. Now we have to equate those two times, so we can write: \[\sqrt {\frac{{2 \times 45}}{{9.81}}} = \frac{{15}}{v}\] |dw:1432832762849:dw|
Michele_Laino
  • Michele_Laino
so, what is v?
anonymous
  • anonymous
i dont understand the equation like how its written so i dont know
Michele_Laino
  • Michele_Laino
do you know the law of falling bodies?
anonymous
  • anonymous
no
anonymous
  • anonymous
would the answer be 3.03 seconds ?
anonymous
  • anonymous
A bug and the windshield of a fast-moving car collide. Indicate whether each of the following statements is True or False (1 point each) • The forces of impact on the bug and on the car are the same magnitude. • The impulses on the bug and non the car are the same magnitude • The changes in speed of the bug and of the car are the same magnitude. • The changes in momentum of the bug and of the car are the same magnitude. • The accelerations of the bug and the car are the same magnitude.
anonymous
  • anonymous
@Michele_Laino
Michele_Laino
  • Michele_Laino
please wait a moment
anonymous
  • anonymous
okay :)
Michele_Laino
  • Michele_Laino
the answer to the first question is: \[v = \sqrt {\frac{{9.81}}{{2 \times 45}}} \times 15\]
anonymous
  • anonymous
45=4.9t^2 9.184=t^2 t=3.03sec That is what i got for the first
Michele_Laino
  • Michele_Laino
yes I know, nevertheless, you have to compute the velocity v, since your problem asks for a velocity
anonymous
  • anonymous
so it that right? and the 2nd one i got true, true, false, true, false?
Michele_Laino
  • Michele_Laino
we have: \[v = \sqrt {\frac{{9.81}}{{2 \times 45}}} \times 15 = 4.95\;m/\sec \]
anonymous
  • anonymous
could you right the equation using the thing on here? thats the part that is getting me a little mixed up
Michele_Laino
  • Michele_Laino
I'm pondering....
Michele_Laino
  • Michele_Laino
the requested equation is: \[v = \sqrt {\frac{g}{{2 \times h}}} \times d = 4.95\;m/\sec \]
Michele_Laino
  • Michele_Laino
|dw:1432834613891:dw|
anonymous
  • anonymous
ohhhhhh okay i think i get it, what about the second?
Michele_Laino
  • Michele_Laino
please, I'm pondering...
Michele_Laino
  • Michele_Laino
I think the first option, since we have to apply the third law of Newton, and the fourth option since for an isolate system the total momentum is conserved
anonymous
  • anonymous
its asking which ones are true and which are false, what are you saying those are?
Michele_Laino
  • Michele_Laino
first option is true, second and third options are false, fourth option is true, fifth option is false
anonymous
  • anonymous
Using 6400 km as the radius of Earth, calculate how high above Earth’s surface you would have to be in order to weigh 1/16th of your current weight. Show all work leading to your answer OR describe your solution using 3 -4 complete sentences.
Michele_Laino
  • Michele_Laino
here we have to apply the formula of Newton, about the interaction force between two object of masses M, and m respectively, which are separated by a distance d \[F = G\frac{{Mm}}{{{d^2}}}\]
anonymous
  • anonymous
okay, so what goes where in the equation?
Michele_Laino
  • Michele_Laino
the weight W of an object whose mass is m, on the earth surface is given by the subsequent formula: \[\Large W = G\frac{{{M_T}m}}{{{R^2}}}\] where R is the earth radius
Michele_Laino
  • Michele_Laino
the weight of the same object, whose mass is m, at an altitude h with respect to the earth surface is given by the subsequent formula: \[\Large {W_1} = G\frac{{{M_T}m}}{{{{\left( {R + h} \right)}^2}}}\]
Michele_Laino
  • Michele_Laino
in both formulas M_T is the mass of the earth
Michele_Laino
  • Michele_Laino
now, we have to use this condition: \[\Large \frac{{{W_1}}}{W} = \frac{1}{{16}}\]
Michele_Laino
  • Michele_Laino
after a substitution, we get this condition: \[\Large {\left( {\frac{R}{{R + h}}} \right)^2} = \frac{1}{{16}}\] what is h?
anonymous
  • anonymous
the height?
Michele_Laino
  • Michele_Laino
yes!
Michele_Laino
  • Michele_Laino
|dw:1432838068670:dw|

Looking for something else?

Not the answer you are looking for? Search for more explanations.