The y-component of the velocity of the carrion is equal to zero. That being said, the time it takes for the carrion to reach the ground (as close as possible to the fox) can be calculated through the equation,
d = Vot + 0.5gt²
where d is the distance, Vo is initial velocity (in this case, zero), g is the acceleration due to gravity (9.8 m/s²). Substituting the known values,
14 = 0.5(9.8)(t²)
t = 1.69 seconds
Since the horizontal component of the velocity is 1.5 m/s, the distance from the base of the tree to the point where the carrion will fall is equal to,
(1.5 m/s)(1.69 s) = 2.535 m
We add this to the given distance of the fox from the base of the tree to determine the distance of the fox from the carrion.
total distance = 2.535 m + 7 m = 9.535 m
Given that the time it takes for it to travel would only be 1.69 seconds, the speed would then be,
speed = (9.535 m) / (1.69 s) = 5.64 m/s
<em>ANSWER: speed = 5.64 m/s</em>
Answer:
![(a)F_{max}=588N\\(b)acceleration=1.96m/s^2](https://tex.z-dn.net/?f=%28a%29F_%7Bmax%7D%3D588N%5C%5C%28b%29acceleration%3D1.96m%2Fs%5E2)
Explanation:
Given data
Mass m=120.0 kg
Coefficient of static friction ![u_{k}=0.500](https://tex.z-dn.net/?f=u_%7Bk%7D%3D0.500)
The coefficient of sliding friction ![u_{d}=0.300](https://tex.z-dn.net/?f=u_%7Bd%7D%3D0.300)
For Part (a) Maximum force
According to Newtons second law the net force aced on the body is given by
![F_{net}=ma=F_{max}-f_{stat}=0\\so\\F_{max}=f_{stat}](https://tex.z-dn.net/?f=F_%7Bnet%7D%3Dma%3DF_%7Bmax%7D-f_%7Bstat%7D%3D0%5C%5Cso%5C%5CF_%7Bmax%7D%3Df_%7Bstat%7D)
The friction force is given by
![f_{stat}=u_{stat}*N\\f_{stat}=0.5*(9.8*120)\\f_{stat}=588N](https://tex.z-dn.net/?f=f_%7Bstat%7D%3Du_%7Bstat%7D%2AN%5C%5Cf_%7Bstat%7D%3D0.5%2A%289.8%2A120%29%5C%5Cf_%7Bstat%7D%3D588N)
Conclude that
![F_{max}=f_{stat}=588N](https://tex.z-dn.net/?f=F_%7Bmax%7D%3Df_%7Bstat%7D%3D588N)
For Part (b) Acceleration
The acceleration due to dynamic friction is given by:
![ma=F_{max}-f_{dyn}](https://tex.z-dn.net/?f=ma%3DF_%7Bmax%7D-f_%7Bdyn%7D)
The dynamic friction is given by:
![f_{dyn}=u_{dyn}*N\\f_{dyn}=0.300*(9.8*120)\\f_{dyn}=353N](https://tex.z-dn.net/?f=f_%7Bdyn%7D%3Du_%7Bdyn%7D%2AN%5C%5Cf_%7Bdyn%7D%3D0.300%2A%289.8%2A120%29%5C%5Cf_%7Bdyn%7D%3D353N)
So the acceleration given by
![a=\frac{F_{max}-f_{dyn}}{m}\\ a=\frac{588N-353N}{120kg}\\ a=1.96m/s^2](https://tex.z-dn.net/?f=a%3D%5Cfrac%7BF_%7Bmax%7D-f_%7Bdyn%7D%7D%7Bm%7D%5C%5C%20a%3D%5Cfrac%7B588N-353N%7D%7B120kg%7D%5C%5C%20a%3D1.96m%2Fs%5E2)
Answer:
Potential gravitational energy is the energy that the body has due to the Earth's gravitational attraction. In this way, the potential gravitational energy depends on the position of the body in relation to a reference level.
Explanation:
Answer:
Explanation:
Given
Free fall acceleration on mars ![g_{m}=3.7\ m/s^2](https://tex.z-dn.net/?f=g_%7Bm%7D%3D3.7%5C%20m%2Fs%5E2)
Time Period of pendulum on earth ![T=1\ s](https://tex.z-dn.net/?f=T%3D1%5C%20s)
Time period of Pendulum is given by
![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)
for earth
![1=2\pi\cdot \sqrt{\frac{L}{9.8}}](https://tex.z-dn.net/?f=1%3D2%5Cpi%5Ccdot%20%5Csqrt%7B%5Cfrac%7BL%7D%7B9.8%7D%7D)
![L=\frac{9.8}{(2\pi )^2}](https://tex.z-dn.net/?f=L%3D%5Cfrac%7B9.8%7D%7B%282%5Cpi%20%29%5E2%7D)
![L=0.498\ m](https://tex.z-dn.net/?f=L%3D0.498%5C%20m)
(b)For same time period on mars length is given by
![L'=\frac{g_m}{(2\pi )^2}](https://tex.z-dn.net/?f=L%27%3D%5Cfrac%7Bg_m%7D%7B%282%5Cpi%20%29%5E2%7D)
![L'=\frac{3.7}{39.48}](https://tex.z-dn.net/?f=L%27%3D%5Cfrac%7B3.7%7D%7B39.48%7D)
![L'=0.0936\ m](https://tex.z-dn.net/?f=L%27%3D0.0936%5C%20m)