The quantum mechanics theory
Answer:
whats the question...............
Explanation:
Answer:
a) 29.36 m
b) 2.44 s
c) 2.57 s
d) 25.117 m/s
Explanation:
t = Time taken
u = Initial velocity = 24 m/s
v = Final velocity
s = Displacement
a = Acceleration due to gravity = 9.81 m/s²
b)
![v=u+at\\\Rightarrow 0=24-9.81\times t\\\Rightarrow \frac{-24}{-9.81}=t\\\Rightarrow t=2.44 \s](https://tex.z-dn.net/?f=v%3Du%2Bat%5C%5C%5CRightarrow%200%3D24-9.81%5Ctimes%20t%5C%5C%5CRightarrow%20%5Cfrac%7B-24%7D%7B-9.81%7D%3Dt%5C%5C%5CRightarrow%20t%3D2.44%20%5Cs)
Time taken by the ball to reach the highest point is 2.44 seconds
a)
![s=ut+\frac{1}{2}at^2\\\Rightarrow s=24\times 2.44+\frac{1}{2}\times -9.81\times 2.44^2\\\Rightarrow s=29.35\ m](https://tex.z-dn.net/?f=s%3Dut%2B%5Cfrac%7B1%7D%7B2%7Dat%5E2%5C%5C%5CRightarrow%20s%3D24%5Ctimes%202.44%2B%5Cfrac%7B1%7D%7B2%7D%5Ctimes%20-9.81%5Ctimes%202.44%5E2%5C%5C%5CRightarrow%20s%3D29.35%5C%20m)
The highest point reached by the ball above its release point is 29.36 m
c) Total height is 3+29.35 = 32.35 m
![s=ut+\frac{1}{2}at^2\\\Rightarrow 32.35=0t+\frac{1}{2}\times 9.81\times t^2\\\Rightarrow t=\sqrt{\frac{32.35\times 2}{9.81}}\\\Rightarrow t=2.57\ s](https://tex.z-dn.net/?f=s%3Dut%2B%5Cfrac%7B1%7D%7B2%7Dat%5E2%5C%5C%5CRightarrow%2032.35%3D0t%2B%5Cfrac%7B1%7D%7B2%7D%5Ctimes%209.81%5Ctimes%20t%5E2%5C%5C%5CRightarrow%20t%3D%5Csqrt%7B%5Cfrac%7B32.35%5Ctimes%202%7D%7B9.81%7D%7D%5C%5C%5CRightarrow%20t%3D2.57%5C%20s)
The ball reaches the ground 2.57 seconds after reaching the highest point
d)
![v=u+at\\\Rightarrow v=0+9.81\times 2.57\\\Rightarrow v=25.2117\ m/s](https://tex.z-dn.net/?f=v%3Du%2Bat%5C%5C%5CRightarrow%20v%3D0%2B9.81%5Ctimes%202.57%5C%5C%5CRightarrow%20v%3D25.2117%5C%20m%2Fs)
The ball will hit the ground at 25.2117 m/s
Answer:
c. 100 N
Explanation:
= mass of each = 55 kg
= Time period of rotation = 7.0 sec
Angular frequency is given as
![w = \frac{2\pi }{T} \\w = \frac{2(3.14) }{7}\\w = 0.897 rads^{-1}](https://tex.z-dn.net/?f=w%20%3D%20%5Cfrac%7B2%5Cpi%20%7D%7BT%7D%20%5C%5Cw%20%3D%20%5Cfrac%7B2%283.14%29%20%7D%7B7%7D%5C%5Cw%20%3D%200.897%20rads%5E%7B-1%7D)
= Diameter of the merry-go-round = 4.5 m
Radius of the merry-go-round is given as
![r = \frac{d}{2} = \frac{4.5}{2} = 2.25 m](https://tex.z-dn.net/?f=r%20%3D%20%5Cfrac%7Bd%7D%7B2%7D%20%3D%20%5Cfrac%7B4.5%7D%7B2%7D%20%3D%202.25%20m)
Magnitude of outward force is same as the centripetal force and is given as
![F = m r w^{2} = (55) (2.25) (0.897)^{2} \\F = 100 N](https://tex.z-dn.net/?f=F%20%3D%20m%20r%20w%5E%7B2%7D%20%3D%20%2855%29%20%282.25%29%20%280.897%29%5E%7B2%7D%20%5C%5CF%20%3D%20100%20N)
The answer is:
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"sensors"
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"<span>Monitoring systems may also use ___<u>sensors</u>___, which are devices that respond to a stimulus (such as heat, light, or pressure) and generate an electrical signal that can be measured or interpreted."
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