Answer:
9241.6 W or 12.39318 hp
Explanation:
u = Initial velocity = 0
v = Final velocity
m = Mass
t = Time taken
Energy
![KE=\frac{1}{2}m(v^2-u^2)\\\Rightarrow KE=\frac{1}{2}108(30.4^2-0^2)\\\Rightarrow KE=49904.64\ Joules](https://tex.z-dn.net/?f=KE%3D%5Cfrac%7B1%7D%7B2%7Dm%28v%5E2-u%5E2%29%5C%5C%5CRightarrow%20KE%3D%5Cfrac%7B1%7D%7B2%7D108%2830.4%5E2-0%5E2%29%5C%5C%5CRightarrow%20KE%3D49904.64%5C%20Joules)
Power
![P=\frac{KE}{t}\\\Rightarrow P=\frac{49904.64}{5.4}\\\Rightarrow P=9241.6\ W](https://tex.z-dn.net/?f=P%3D%5Cfrac%7BKE%7D%7Bt%7D%5C%5C%5CRightarrow%20P%3D%5Cfrac%7B49904.64%7D%7B5.4%7D%5C%5C%5CRightarrow%20P%3D9241.6%5C%20W)
Converting to hp
![1\ W=\frac{1}{745.7}\ hp](https://tex.z-dn.net/?f=1%5C%20W%3D%5Cfrac%7B1%7D%7B745.7%7D%5C%20hp)
![\\\Rightarrow 9241.6\ W=\frac{9241.6}{745.7}\ hp=12.39318\ hp](https://tex.z-dn.net/?f=%5C%5C%5CRightarrow%209241.6%5C%20W%3D%5Cfrac%7B9241.6%7D%7B745.7%7D%5C%20hp%3D12.39318%5C%20hp)
The power developed by the cheetah is 9241.6 W or 12.39318 hp
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:
Acceleration (b) not sure tho
Explanation:
Answer:
9800 m
Explanation:
During acceleration, given:
v₀ = 0 m/s
a = 39.2 m/s²
t = 10.0 s
Find: v and Δy
v = at + v₀
v = (39.2 m/s²) (10.0 s) + 0 m/s
v = 392 m/s
Δy = v₀ t + ½ at²
Δy = (0 m/s) (10.0 s) + ½ (39.2 m/s²) (10.0 s)²
Δy = 1960 m
During free fall, given:
v₀ = 392 m/s
v = 0 m/s
a = -9.8 m/s²
Find: Δy
v² = v₀² + 2aΔy
(0 m/s)² = (392 m/s)² + 2 (-9.8 m/s²) Δy
Δy = 7840 m
Therefore, h = 1960 m + 7840 m = 9800 m.
Answer:
T' = 0.9677T
Explanation:
The period of a pendulum is given by the following formula:
![T=2\pi \sqrt{\frac{l}{g}}](https://tex.z-dn.net/?f=T%3D2%5Cpi%20%5Csqrt%7B%5Cfrac%7Bl%7D%7Bg%7D%7D)
l: length of the pendulum
g: gravitational acceleration
If the length of the pendulum is decreased in 6.35% the length of the pendulum becomes:
![l'=l-0.0635l=0.9365l](https://tex.z-dn.net/?f=l%27%3Dl-0.0635l%3D0.9365l)
The new period for a length of l' is:
![T'=2\pi \sqrt{\frac{l'}{g}}=2\pi \sqrt{\frac{0.9365l}{g}}=\sqrt{0.9365}(2\pi \sqrt{\frac{l}{g}})=0.9677(2\pi \sqrt{\frac{l}{g}})\\\\T'=0.9677T](https://tex.z-dn.net/?f=T%27%3D2%5Cpi%20%5Csqrt%7B%5Cfrac%7Bl%27%7D%7Bg%7D%7D%3D2%5Cpi%20%5Csqrt%7B%5Cfrac%7B0.9365l%7D%7Bg%7D%7D%3D%5Csqrt%7B0.9365%7D%282%5Cpi%20%5Csqrt%7B%5Cfrac%7Bl%7D%7Bg%7D%7D%29%3D0.9677%282%5Cpi%20%5Csqrt%7B%5Cfrac%7Bl%7D%7Bg%7D%7D%29%5C%5C%5C%5CT%27%3D0.9677T)
hence, the new period is 0.9677T