Answer:
3.50 m/s^2
Explanation:
the race is of 100 m.
time after which Marge crosses the finish line after t= 10.8 sec
time take by her to attain maximum speed t_o= 2.20 sec
a= ![\frac{v}{t_0}](https://tex.z-dn.net/?f=%5Cfrac%7Bv%7D%7Bt_0%7D)
we know that
s= ut +1/2at^2
100= v![100= v\times10.8+ 0.5a10.8^2](https://tex.z-dn.net/?f=100%3D%20v%5Ctimes10.8%2B%200.5a10.8%5E2)
put a=
here t_o= 2.20 sec
![100=v\times10.8+ 0.5 \frac{v}{2.20} 10.8^2](https://tex.z-dn.net/?f=100%3Dv%5Ctimes10.8%2B%200.5%20%5Cfrac%7Bv%7D%7B2.20%7D%2010.8%5E2)
v= 7.72 m/s
now Marge acceleration a=
= 3.50 m/s^2
It takes about 27.32 days
Answer: 7.578x10^-12
Explanation:
First, we find the power:
Power P = 140x4/100 =5,6W
Distance r = 14m
Then,
Intensity I = P/4πr2
= 5.6/(4π x 14 x 14)
=. 2.27 x10^3 W/m2
Radiation pressure:
P(rad) = I/c =0.00227÷{3 x 10^8)
=7.578x10^-12 N/m2
Answer:
b
Explanation:
again ?
parabola = quadratic equation.
so, only b can be right : y = kx²
Answer:
8159 W, 10.9 hp
Explanation:
The work done by the cheetah is equal to the kinetic energy gained, so we can write:
![W=\Delta K=\frac{1}{2}mv^2-\frac{1}{2}mu^2](https://tex.z-dn.net/?f=W%3D%5CDelta%20K%3D%5Cfrac%7B1%7D%7B2%7Dmv%5E2-%5Cfrac%7B1%7D%7B2%7Dmu%5E2)
where
m = 101 kg is the mass of the cheetah
u = 0 is the initial speed
v = 29.4 m/s is the final speed
Substituting,
![W=\frac{1}{2}(101)(29.4)^2-0=43650 J](https://tex.z-dn.net/?f=W%3D%5Cfrac%7B1%7D%7B2%7D%28101%29%2829.4%29%5E2-0%3D43650%20J)
Now we can calculate the power developed by the cheetah, given by:
![P=\frac{W}{t}](https://tex.z-dn.net/?f=P%3D%5Cfrac%7BW%7D%7Bt%7D)
where
t = 5.35 s
is the time taken. Substituting,
![P=\frac{43650}{5.35}=8159 W](https://tex.z-dn.net/?f=P%3D%5Cfrac%7B43650%7D%7B5.35%7D%3D8159%20W)
And keeping in mind that
1 hp = 746 W
We can also convert into horsepowers:
![P=\frac{8159}{746}=10.9 hp](https://tex.z-dn.net/?f=P%3D%5Cfrac%7B8159%7D%7B746%7D%3D10.9%20hp)