When the diver reaches maximum height, the upward velocity will be zero.
We shall use the formula
v^2 = u^2 - 2gh
where
v = 0 (velocity at maximum height)
u = 1.2 m/s, intial upward velocity
g = -9.8 m/s^2, gravitational acceleration (downward)
h = maximum height attained above the diving board.
Therefore
0 = 1.2^2 - 2*9.8*h
h = 1.2^2/(2*9.8) = 0.0735 m
Answer: 0.074 m (nearest thousandth)
Answer:
(a) Magnitude of Vector = 207.73 m
(b) Direction = 65.48°
Explanation:
(a)
The formula to find out the magnitude of a resultant vector with the help of its x and y components is given as follows:
![Magnitude\ of\ Vector = \sqrt{d_{x}^{2} + d_{y}^{2}} = \sqrt{(86.2\ m)^{2} + (189\ m)^{2}}\\\\](https://tex.z-dn.net/?f=Magnitude%5C%20of%5C%20Vector%20%3D%20%5Csqrt%7Bd_%7Bx%7D%5E%7B2%7D%20%2B%20d_%7By%7D%5E%7B2%7D%7D%20%3D%20%5Csqrt%7B%2886.2%5C%20m%29%5E%7B2%7D%20%2B%20%28189%5C%20m%29%5E%7B2%7D%7D%5C%5C%5C%5C)
<u>Magnitude of Vector = 207.73 m</u>
(b)
For the direction of the vector we have the formula:
![Direction = tan^{-1}(\frac{y}{x})\\\\Direction = tan^{-1}(\frac{189\ m}{86.2\ m})\\\\](https://tex.z-dn.net/?f=Direction%20%3D%20tan%5E%7B-1%7D%28%5Cfrac%7By%7D%7Bx%7D%29%5C%5C%5C%5CDirection%20%3D%20tan%5E%7B-1%7D%28%5Cfrac%7B189%5C%20m%7D%7B86.2%5C%20m%7D%29%5C%5C%5C%5C)
<u>Direction = 65.48°</u>
No it shouldn't, a hypothesis doesn't need to be correct but must have an idea for why x variable effects y variable and have good reasoning. In the conclusion you should state if it's correct or not and explain why it's correct/incorrect and what answer you've determined from data.
Answer:
Accelrtation:a vehicle's capacity to gain speed within a short time.EX:An object was moving north at 10 meters per second.
Velociy:the speed of something in a given direction EX:Velocity is the rate of motion, speed or action. An example of velocity is a car driving at 75 miles per hour.
Explanation:
Answer:
Emitted power will be equal to ![7.85\times 10^{-5}watt](https://tex.z-dn.net/?f=7.85%5Ctimes%2010%5E%7B-5%7Dwatt)
Explanation:
It is given factory whistle can be heard up to a distance of R=2.5 km = 2500 m
Threshold of human hearing ![I=10^{-12}W/m^2](https://tex.z-dn.net/?f=I%3D10%5E%7B-12%7DW%2Fm%5E2)
We have to find the emitted power
Emitted power is equal to ![P=I\times A](https://tex.z-dn.net/?f=P%3DI%5Ctimes%20A)
![P=I\times 4\pi R^2](https://tex.z-dn.net/?f=P%3DI%5Ctimes%204%5Cpi%20R%5E2)
![P=10^{-12}\times 4\times 3.14\times 2500^2=7.85\times 10^{-5}watt](https://tex.z-dn.net/?f=P%3D10%5E%7B-12%7D%5Ctimes%204%5Ctimes%203.14%5Ctimes%20%202500%5E2%3D7.85%5Ctimes%2010%5E%7B-5%7Dwatt)
So emitted power will be equal to ![7.85\times 10^{-5}watt](https://tex.z-dn.net/?f=7.85%5Ctimes%2010%5E%7B-5%7Dwatt)