Answer:
0.8m per second
Explanation:
we measure speed as distance over time
distance=8m
time=10s
therefore speed = 8/10
=0.8m per second
Answer:
The height of the water slide is 0.878 m
Explanation:
Given that,
Distance = 2.52 m
Suppose Children slide down a friction less water slide that ends at a height of 1.80 m above the pool.
We need to calculate the time
Using equation of motion
![s=ut+\dfrac{1}{2}gt^2](https://tex.z-dn.net/?f=s%3Dut%2B%5Cdfrac%7B1%7D%7B2%7Dgt%5E2)
Put the value in the equation
![1.80=0+\dfrac{1}{2}\times9.8\times t^2](https://tex.z-dn.net/?f=1.80%3D0%2B%5Cdfrac%7B1%7D%7B2%7D%5Ctimes9.8%5Ctimes%20t%5E2)
![t^2=\dfrac{1.80\times2}{9.8}](https://tex.z-dn.net/?f=t%5E2%3D%5Cdfrac%7B1.80%5Ctimes2%7D%7B9.8%7D)
![t=\sqrt{\dfrac{1.80\times2}{9.8}}](https://tex.z-dn.net/?f=t%3D%5Csqrt%7B%5Cdfrac%7B1.80%5Ctimes2%7D%7B9.8%7D%7D)
![t=0.606\ sec](https://tex.z-dn.net/?f=t%3D0.606%5C%20sec)
We need to calculate the velocity
Using formula of velocity
![v = \dfrac{d}{t}](https://tex.z-dn.net/?f=v%20%3D%20%5Cdfrac%7Bd%7D%7Bt%7D)
Put the value into the formula
![v=\dfrac{2.52}{0.606}](https://tex.z-dn.net/?f=v%3D%5Cdfrac%7B2.52%7D%7B0.606%7D)
![v=4.15\ m/s](https://tex.z-dn.net/?f=v%3D4.15%5C%20m%2Fs)
We need to calculate height
Using conservation of energy
![\dfrac{1}{2}mv^2=mgh](https://tex.z-dn.net/?f=%5Cdfrac%7B1%7D%7B2%7Dmv%5E2%3Dmgh)
![h=\dfrac{v^2}{2g}](https://tex.z-dn.net/?f=h%3D%5Cdfrac%7Bv%5E2%7D%7B2g%7D)
Put the value into the formula
![h=\dfrac{4.15^2}{2\times9.8}](https://tex.z-dn.net/?f=h%3D%5Cdfrac%7B4.15%5E2%7D%7B2%5Ctimes9.8%7D)
![h=0.878\ m](https://tex.z-dn.net/?f=h%3D0.878%5C%20m)
Hence, The height of the water slide is 0.878 m.
All of the above. They all right
B) 7.87 m/s
The gravitational pull is the rate of change of velocity which is the acceleration. Formula for acceleration is;
![a = \frac{final \: velocity - initial \: velocity}{time \: taken}](https://tex.z-dn.net/?f=a%20%3D%20%20%5Cfrac%7Bfinal%20%5C%3A%20velocity%20-%20initial%20%5C%3A%20velocity%7D%7Btime%20%5C%3A%20taken%7D%20)
Given:
• Initial velocity = 0m/s; I dropped the ball, and didn't throw it, so it was at rest firstly
• Time taken = 2.40s
• Acceleration = 3.28m/s^2
We're require to find the final velocity, at which the ball hit the ground with. Ignoring air resistance, keep in mind that the velocity of an object increases as it comes closer to the ground.
![3.28 = \frac{final \: velocity - 0}{2.40}](https://tex.z-dn.net/?f=3.28%20%20%3D%20%20%5Cfrac%7Bfinal%20%5C%3A%20velocity%20-%200%7D%7B2.40%7D%20)
![3.28 \times 2.40 = final \: velocity](https://tex.z-dn.net/?f=3.28%20%5Ctimes%202.40%20%3D%20final%20%5C%3A%20velocity)
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