Answer:
It's a pretty simple suvat linear projectile motion question, using the following equation and plugging in your values it's a pretty trivial calculation.
V^2=U^2+2*a*x
V=0 (as it is at max height)
U=30ms^-1 (initial speed)
a=-g /-9.8ms^-2 (as it is moving against gravity)
x is the variable you want to calculate (height)
0=30^2+2*(-9.8)*x
x=-30^2/2*-9.8
x=45.92m
I do have a couple ideas and tips that may help you win. I don’t know how the guidelines are set up so if the ideas won’t be helpful I apologize.
First off put some ice cubes in the container then sprinkle salt on them, The reaction will create an effect and be super cold.
Another idea would be to get some dry ice if you able to, This will freeze it solid within seconds.
The last idea combines the the first. Take a bowl and fill it with with water and ice (Make sure the bowl is insulated) add a small handful of salt into the bowl, Put your drink into the cooler and before shutting stir then well then close and wait for the amount of time left, Your should have a cold water bottle.
I hoped this helped you out and I hope you also win the contest.
Answer:
emotions
Explanation:
emotions are how you feel and can happen any time
Hope it helps <333
Answer:
the final speed of object A changed by a factor of
= 0.58
the final speed of object B changed by a factor of
= 1.29
Explanation:
Given;
kinetic energy of object A, = 27 J
let the mass of object A = ![m_A](https://tex.z-dn.net/?f=m_A)
then, the mass of object B = ![m_B = \frac{m_A}{4}](https://tex.z-dn.net/?f=m_B%20%3D%20%5Cfrac%7Bm_A%7D%7B4%7D)
work done on object A = -18 J
work done on object B = -18 J
let
be the initial speed
let
be the final speed
For object A;
![K.E_A = 27\\\\\frac{1}{2} m_A v_i^2 = 27\\\\m_A v_i^2 = 54\\\\m_A = \frac{54}{v_i^2} ----Equation \ (1)\\\\Apply \ work-energy \ theorem;\\\\\delta K.E_A = -18\\\\\frac{1}{2} m_A v_f^2 - \frac{1}{2} m_A v_i^2 = -18\\\\\frac{1}{2} m_A ( v_f^2 \ - v_i^2 )\ =- 18\\\\v_f^2 \ - v_i^2 = -\frac{36}{m_A} ---Equation \ (2)\\\\v_f^2 \ - v_i^2 = -\frac{36v_i^2}{54}\\\\ v_f^2 \ =v_i^2 - \frac{36v_i^2}{54}\\\\ v_f^2 = \frac{54v_i^2 -36v_i^2 }{54} \\\\v_f^2 = \frac{18v_i^2}{54} \\\\v_f^2 = \frac{v_i^2}{3} \\\\](https://tex.z-dn.net/?f=K.E_A%20%3D%2027%5C%5C%5C%5C%5Cfrac%7B1%7D%7B2%7D%20m_A%20v_i%5E2%20%3D%2027%5C%5C%5C%5Cm_A%20v_i%5E2%20%20%3D%2054%5C%5C%5C%5Cm_A%20%3D%20%5Cfrac%7B54%7D%7Bv_i%5E2%7D%20----Equation%20%5C%20%281%29%5C%5C%5C%5CApply%20%5C%20work-energy%20%5C%20theorem%3B%5C%5C%5C%5C%5Cdelta%20K.E_A%20%3D%20-18%5C%5C%5C%5C%5Cfrac%7B1%7D%7B2%7D%20m_A%20v_f%5E2%20-%20%5Cfrac%7B1%7D%7B2%7D%20m_A%20v_i%5E2%20%3D%20-18%5C%5C%5C%5C%5Cfrac%7B1%7D%7B2%7D%20m_A%20%28%20v_f%5E2%20%5C%20-%20%20v_i%5E2%20%29%5C%20%3D-%2018%5C%5C%5C%5Cv_f%5E2%20%5C%20-%20%20v_i%5E2%20%20%3D%20-%5Cfrac%7B36%7D%7Bm_A%7D%20---Equation%20%5C%20%282%29%5C%5C%5C%5Cv_f%5E2%20%5C%20-%20%20v_i%5E2%20%20%3D%20-%5Cfrac%7B36v_i%5E2%7D%7B54%7D%5C%5C%5C%5C%20v_f%5E2%20%5C%20%3Dv_i%5E2%20-%20%5Cfrac%7B36v_i%5E2%7D%7B54%7D%5C%5C%5C%5C%20v_f%5E2%20%3D%20%5Cfrac%7B54v_i%5E2%20-36v_i%5E2%20%7D%7B54%7D%20%5C%5C%5C%5Cv_f%5E2%20%3D%20%5Cfrac%7B18v_i%5E2%7D%7B54%7D%20%5C%5C%5C%5Cv_f%5E2%20%3D%20%5Cfrac%7Bv_i%5E2%7D%7B3%7D%20%5C%5C%5C%5C)
![v_f = \sqrt{\frac{v_i^2}{3} }\\\\v_f = \frac{1}{\sqrt{3} } \ v_i\\\\](https://tex.z-dn.net/?f=v_f%20%3D%20%5Csqrt%7B%5Cfrac%7Bv_i%5E2%7D%7B3%7D%20%7D%5C%5C%5C%5Cv_f%20%3D%20%5Cfrac%7B1%7D%7B%5Csqrt%7B3%7D%20%7D%20%5C%20v_i%5C%5C%5C%5C)
Thus, the final speed of object A changed by a factor of
= 0.58
To obtain the change in the final speed of object B, apply the following equations.
![K.E_B_i = \frac{1}{2} m_Bv_i^2\\\\m_B = \frac{m_A}{4} \\\\K.E_B_i = \frac{1}{2}(\frac{m_A}{4} )v_i^2\\\\K.E_B_i = \frac{m_Av_i^2}{8} \\\\But, \ m_Av_i^2 = 54 \\\\K.E_B_i = \frac{54}{8} \\\\Apply \ work-energy \ theorem ;\\\\\delta K.E = -18\\\\K.E_f -K.E_i = -18\\\\\frac{1}{2}m_Bv_f^2 - \frac{1}{2} m_Bv_i^2 = -18\\\\Recall \ m_B = \frac{m_A}{4} \\\\\frac{1}{2}(\frac{m_A}{4} )v_f^2 - \frac{1}{2}(\frac{m_A}{4} )v_i^2 = -18\\\\\frac{1}{2}\times \frac{m_A}{4} (v_i^2 -v_f^2) = 18\\\\](https://tex.z-dn.net/?f=K.E_B_i%20%3D%20%5Cfrac%7B1%7D%7B2%7D%20m_Bv_i%5E2%5C%5C%5C%5Cm_B%20%3D%20%5Cfrac%7Bm_A%7D%7B4%7D%20%5C%5C%5C%5CK.E_B_i%20%3D%20%5Cfrac%7B1%7D%7B2%7D%28%5Cfrac%7Bm_A%7D%7B4%7D%20%29v_i%5E2%5C%5C%5C%5CK.E_B_i%20%3D%20%5Cfrac%7Bm_Av_i%5E2%7D%7B8%7D%20%5C%5C%5C%5CBut%2C%20%5C%20m_Av_i%5E2%20%3D%2054%20%5C%5C%5C%5CK.E_B_i%20%3D%20%5Cfrac%7B54%7D%7B8%7D%20%5C%5C%5C%5CApply%20%5C%20work-energy%20%5C%20theorem%20%3B%5C%5C%5C%5C%5Cdelta%20K.E%20%3D%20-18%5C%5C%5C%5CK.E_f%20-K.E_i%20%3D%20-18%5C%5C%5C%5C%5Cfrac%7B1%7D%7B2%7Dm_Bv_f%5E2%20-%20%5Cfrac%7B1%7D%7B2%7D%20m_Bv_i%5E2%20%3D%20-18%5C%5C%5C%5CRecall%20%5C%20m_B%20%3D%20%20%5Cfrac%7Bm_A%7D%7B4%7D%20%5C%5C%5C%5C%5Cfrac%7B1%7D%7B2%7D%28%5Cfrac%7Bm_A%7D%7B4%7D%20%29v_f%5E2%20-%20%5Cfrac%7B1%7D%7B2%7D%28%5Cfrac%7Bm_A%7D%7B4%7D%20%29v_i%5E2%20%3D%20-18%5C%5C%5C%5C%5Cfrac%7B1%7D%7B2%7D%5Ctimes%20%5Cfrac%7Bm_A%7D%7B4%7D%20%28v_i%5E2%20-v_f%5E2%29%20%3D%2018%5C%5C%5C%5C)
![\frac{1}{2}\times \frac{m_A}{4} (v_i^2 -v_f^2) = 18\\\\v_i^2 -v_f^2 = \frac{8}{m_A} \times 18\\\\v_i^2 -v_f^2 =\frac{144}{m_A} \\\\But , m_A = \frac{54}{v_i^2} \\\\v_i^2 -v_f^2 =\frac{144v_i^2}{54} \\\\v_f^2 = v_i^2 - \frac{144v_i^2}{54}\\\\v_f^2 = \frac{54v_i^2-144v_i^2}{54}\\\\ v_f^2 = \frac{-90v_i^2}{54} \\\\v_f^2 = \frac{-5v_i^2}{3} \\\\|v_f| = \sqrt{\frac{5v_i^2}{3}} \\\\|v_f| = \sqrt{\frac{5}{3}} \ v_i](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7D%5Ctimes%20%5Cfrac%7Bm_A%7D%7B4%7D%20%28v_i%5E2%20-v_f%5E2%29%20%3D%2018%5C%5C%5C%5Cv_i%5E2%20-v_f%5E2%20%3D%20%5Cfrac%7B8%7D%7Bm_A%7D%20%5Ctimes%2018%5C%5C%5C%5Cv_i%5E2%20-v_f%5E2%20%3D%5Cfrac%7B144%7D%7Bm_A%7D%20%5C%5C%5C%5CBut%20%2C%20m_A%20%3D%20%5Cfrac%7B54%7D%7Bv_i%5E2%7D%20%5C%5C%5C%5Cv_i%5E2%20-v_f%5E2%20%3D%5Cfrac%7B144v_i%5E2%7D%7B54%7D%20%5C%5C%5C%5Cv_f%5E2%20%3D%20v_i%5E2%20-%20%5Cfrac%7B144v_i%5E2%7D%7B54%7D%5C%5C%5C%5Cv_f%5E2%20%3D%20%5Cfrac%7B54v_i%5E2-144v_i%5E2%7D%7B54%7D%5C%5C%5C%5C%20v_f%5E2%20%3D%20%5Cfrac%7B-90v_i%5E2%7D%7B54%7D%20%5C%5C%5C%5Cv_f%5E2%20%3D%20%5Cfrac%7B-5v_i%5E2%7D%7B3%7D%20%5C%5C%5C%5C%7Cv_f%7C%20%3D%20%5Csqrt%7B%5Cfrac%7B5v_i%5E2%7D%7B3%7D%7D%20%5C%5C%5C%5C%7Cv_f%7C%20%3D%20%5Csqrt%7B%5Cfrac%7B5%7D%7B3%7D%7D%20%5C%20v_i)
Thus, the final speed of object B changed by a factor of
= 1.29
The answer is 0.245N.
<h3>What is kinetic energy?</h3>
- A particle or an item that is in motion has a sort of energy called kinetic energy. An item accumulates kinetic energy when work, which involves the transfer of energy, is done on it by exerting a net force.
- Kinetic energy comes in five forms: radiant, thermal, acoustic, electrical, and mechanical.
- The energy of a body in motion, or kinetic energy (KE), is essentially the energy of all moving objects. Along with potential energy, which is the stored energy present in objects at rest, it is one of the two primary types of energy.
- Explain that a moving object's mass and speed are two factors that impact the amount of kinetic energy it will possess.
(b) 0.100
For the block on the left, ![f_{k} =u_{k} n= 0.100(2.45N)=0.245N.](https://tex.z-dn.net/?f=f_%7Bk%7D%20%3Du_%7Bk%7D%20n%3D%200.100%282.45N%29%3D0.245N.)
∑
=![ma_{x}](https://tex.z-dn.net/?f=ma_%7Bx%7D)
–0.308N+0.245N=(0.250kg)a
a=−0.252
if the force of static friction is not too large.
For the block on the right,
=
=0.490N. The maximum force of static friction would be larger, so no motion would begin, and the acceleration is zero
To learn more about kinetic energy, refer to:
brainly.com/question/25959744
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