Answer:
I don't get it?
like yhu want us to rate it or?
Explanation:
Answer:
The acceleration of Trout is largest.
Explanation:
Speed of Thomson's gazelle, v = 13 m/s in 3 s. Let us assumed that initial speed is 0. Acceleration is given by :
![a_T=\dfrac{v-u}{t}\\\\a_T=\dfrac{v}{t}\\\\a_T=\dfrac{13\ m/s}{3\ s}\\\\a_T=4.34\ m/s^2](https://tex.z-dn.net/?f=a_T%3D%5Cdfrac%7Bv-u%7D%7Bt%7D%5C%5C%5C%5Ca_T%3D%5Cdfrac%7Bv%7D%7Bt%7D%5C%5C%5C%5Ca_T%3D%5Cdfrac%7B13%5C%20m%2Fs%7D%7B3%5C%20s%7D%5C%5C%5C%5Ca_T%3D4.34%5C%20m%2Fs%5E2)
Speed of lion is 9.5 m/s in 10 s. Let us assumed that initial speed is 0. Acceleration is given by :
![a=\dfrac{v-u}{t}\\\\a_l=\dfrac{v}{t}\\\\a_l=\dfrac{9.5\ m/s}{1\ s}\\\\a_l=9.5\ m/s^2](https://tex.z-dn.net/?f=a%3D%5Cdfrac%7Bv-u%7D%7Bt%7D%5C%5C%5C%5Ca_l%3D%5Cdfrac%7Bv%7D%7Bt%7D%5C%5C%5C%5Ca_l%3D%5Cdfrac%7B9.5%5C%20m%2Fs%7D%7B1%5C%20s%7D%5C%5C%5C%5Ca_l%3D9.5%5C%20m%2Fs%5E2)
A trout can reach a speed of 2.8 m/s in 0.12 s. Let us assumed that initial speed is 0. Acceleration is given by :
![a=\dfrac{v-u}{t}\\\\a_t=\dfrac{v}{t}\\\\a_t=\dfrac{2.8\ m/s}{0.12\ s}\\\\a_t=23.34\ m/s^2](https://tex.z-dn.net/?f=a%3D%5Cdfrac%7Bv-u%7D%7Bt%7D%5C%5C%5C%5Ca_t%3D%5Cdfrac%7Bv%7D%7Bt%7D%5C%5C%5C%5Ca_t%3D%5Cdfrac%7B2.8%5C%20m%2Fs%7D%7B0.12%5C%20s%7D%5C%5C%5C%5Ca_t%3D23.34%5C%20m%2Fs%5E2)
It is clear that the acceleration of Thomson's gazelle, lion and trout is
respectively. So, the acceleration of Trout is largest.
Answer:
n=142 N
Explanation:
solution:
pic 1 is attached
<em>There is no vertical acceleration (ay = 0), so net sum of vertical forces is 0. n acts up, weight and the vertical component of F act down, so </em>
∑F_y =m*a_y
n-mg-Fsin43°=0
n=142 N
KE=(1/2)mv² ⇒ v=√(2KE/m)
Data:
KE=kenetic energy=810 J
m=mass=80 Kg
v=velocity (speed)
v=√(2KE/m)
v=√(2*810 J/80 Kg)=4.5 m/s.
answer: his speed is 4.5 m/s.