Answer:
Time, t = 0.23 seconds
Explanation:
It is given that,
Initial speed of the ranger, u = 52 km/h = 14.44 m/s
Final speed of the ranger, v = 0 (as brakes are applied)
Acceleration of the ranger, ![a=-4\ m/s^2](https://tex.z-dn.net/?f=a%3D-4%5C%20m%2Fs%5E2)
Distance between deer and the vehicle, d = 87 m
Let d' is the distance covered by the deer so that it comes top rest. So,
![d'=\dfrac{v^2-u^2}{2a}](https://tex.z-dn.net/?f=d%27%3D%5Cdfrac%7Bv%5E2-u%5E2%7D%7B2a%7D)
![d'=\dfrac{-(14.44)^2}{2\times -4}](https://tex.z-dn.net/?f=d%27%3D%5Cdfrac%7B-%2814.44%29%5E2%7D%7B2%5Ctimes%20-4%7D)
d' = 26.06 m
Distance between the point where the deer stops and the vehicle is :
D=d-d'
D=87 - 26.06 = 60.94 m
Let t is the maximum reaction time allowed if the ranger is to avoid hitting the deer. It can be calculated as :
![t=\dfrac{v}{D}](https://tex.z-dn.net/?f=t%3D%5Cdfrac%7Bv%7D%7BD%7D)
![t=\dfrac{14.44}{60.94}](https://tex.z-dn.net/?f=t%3D%5Cdfrac%7B14.44%7D%7B60.94%7D)
t = 0.23 seconds
Hence, this is the required solution.
Answer:
Velocity (magnitude) is 98.37 m/s
Explanation:
We use the vertical component of the initial velocity, which is:
![v_{0y}=v_0*sin(45)=\frac{\sqrt{2} }{2}v_0](https://tex.z-dn.net/?f=v_%7B0y%7D%3Dv_0%2Asin%2845%29%3D%5Cfrac%7B%5Csqrt%7B2%7D%20%7D%7B2%7Dv_0)
Using kinematics expression of vertical velocity (in y direction) for an accelerated motion (constant acceleration, which is gravity):
![v_{y}=v_{0y}+a*t=\frac{\sqrt{2} }{2}v_0-9.8t](https://tex.z-dn.net/?f=v_%7By%7D%3Dv_%7B0y%7D%2Ba%2At%3D%5Cfrac%7B%5Csqrt%7B2%7D%20%7D%7B2%7Dv_0-9.8t)
Now we need to find
as a function of
. We use the horizontal velocity, which is always the same as follow:
![v_x=v_0cos(45\º)=\frac{\sqrt{2} }{2}v_0=v_{t=3}*cos(30\º) \\](https://tex.z-dn.net/?f=v_x%3Dv_0cos%2845%5C%C2%BA%29%3D%5Cfrac%7B%5Csqrt%7B2%7D%20%7D%7B2%7Dv_0%3Dv_%7Bt%3D3%7D%2Acos%2830%5C%C2%BA%29%20%5C%5C)
We know the angle at 3 seconds:
![v_y(t=3)=v_{t=3}*sin(30\º)\\v_{t=3}=\frac{v_y}{sin(30\º)}](https://tex.z-dn.net/?f=v_y%28t%3D3%29%3Dv_%7Bt%3D3%7D%2Asin%2830%5C%C2%BA%29%5C%5Cv_%7Bt%3D3%7D%3D%5Cfrac%7Bv_y%7D%7Bsin%2830%5C%C2%BA%29%7D)
Substitute
in
and then solve for ![v_y](https://tex.z-dn.net/?f=v_y)
![\frac{\sqrt{2} }{2}v_0=\frac{v_y*cos(30\º) }{sin(30\º)} \\v_y=\frac{\sqrt{6} }{6}v_0](https://tex.z-dn.net/?f=%5Cfrac%7B%5Csqrt%7B2%7D%20%7D%7B2%7Dv_0%3D%5Cfrac%7Bv_y%2Acos%2830%5C%C2%BA%29%20%7D%7Bsin%2830%5C%C2%BA%29%7D%20%5C%5Cv_y%3D%5Cfrac%7B%5Csqrt%7B6%7D%20%7D%7B6%7Dv_0)
With this expression we go back to the kinematic equation and solve it for initial speed
![\frac{\sqrt{6} }{6} v_0 =\frac{\sqrt{2} }{2}v_0-29.4\\v_0(\frac{\sqrt{6}-3\sqrt{2}}{6} )=-29.4\\v_0=98.37 m/s](https://tex.z-dn.net/?f=%5Cfrac%7B%5Csqrt%7B6%7D%20%7D%7B6%7D%20v_0%20%3D%5Cfrac%7B%5Csqrt%7B2%7D%20%7D%7B2%7Dv_0-29.4%5C%5Cv_0%28%5Cfrac%7B%5Csqrt%7B6%7D-3%5Csqrt%7B2%7D%7D%7B6%7D%20%29%3D-29.4%5C%5Cv_0%3D98.37%20m%2Fs)
Answer:
Part a)
![v = \sqrt{xg + \frac{MLg}{m}}](https://tex.z-dn.net/?f=v%20%3D%20%5Csqrt%7Bxg%20%2B%20%5Cfrac%7BMLg%7D%7Bm%7D%7D)
Part b)
t = 12 s
Explanation:
Part a)
Tension in the rope at a distance x from the lower end is given as
![T = \frac{m}{L}xg + Mg](https://tex.z-dn.net/?f=T%20%3D%20%5Cfrac%7Bm%7D%7BL%7Dxg%20%2B%20Mg)
so the speed of the wave at that position is given as
![v = \sqrt{\frac{T}{\mu}}](https://tex.z-dn.net/?f=v%20%3D%20%5Csqrt%7B%5Cfrac%7BT%7D%7B%5Cmu%7D%7D)
here we know that
![\mu = \frac{m}{L}](https://tex.z-dn.net/?f=%5Cmu%20%3D%20%5Cfrac%7Bm%7D%7BL%7D)
now we have
![v = \sqrt{\frac{ \frac{m}{L}xg + Mg}{m/L}](https://tex.z-dn.net/?f=v%20%3D%20%5Csqrt%7B%5Cfrac%7B%20%5Cfrac%7Bm%7D%7BL%7Dxg%20%2B%20Mg%7D%7Bm%2FL%7D)
![v = \sqrt{xg + \frac{MLg}{m}}](https://tex.z-dn.net/?f=v%20%3D%20%5Csqrt%7Bxg%20%2B%20%5Cfrac%7BMLg%7D%7Bm%7D%7D)
Part b)
time taken by the wave to reach the top is given as
![t = \int \frac{dx}{\sqrt{xg + \frac{MLg}{m}}}](https://tex.z-dn.net/?f=t%20%3D%20%5Cint%20%5Cfrac%7Bdx%7D%7B%5Csqrt%7Bxg%20%2B%20%5Cfrac%7BMLg%7D%7Bm%7D%7D%7D)
![t = \frac{1}{g}(2\sqrt{xg + \frac{MLg}{m}})](https://tex.z-dn.net/?f=t%20%3D%20%5Cfrac%7B1%7D%7Bg%7D%282%5Csqrt%7Bxg%20%2B%20%5Cfrac%7BMLg%7D%7Bm%7D%7D%29)
![t = \frac{2}{9.8}(\sqrt{(39.2\times 9.8) + \frac{8(39.2)(9.8)}{1}})](https://tex.z-dn.net/?f=t%20%3D%20%5Cfrac%7B2%7D%7B9.8%7D%28%5Csqrt%7B%2839.2%5Ctimes%209.8%29%20%2B%20%5Cfrac%7B8%2839.2%29%289.8%29%7D%7B1%7D%7D%29)
![t = 12 s](https://tex.z-dn.net/?f=t%20%3D%2012%20s)
Answer:
4. It is the force of the road on the tires (an external force) that stops the car.
Explanation:
If there is no friction between the road and the tires, the car won't stop.
You can see this, for example, when there is ice on the road. You can still apply the brakes (internal force), but since there is no friction (external force) the car won't stop.
The force of the brakes on the wheels is not what makes the car stop, it is the friction of the road against still tires that makes it stop.
How might a suit of armor be a good analogy for a function of the skeletal system?
It's a frame for your body and protects organs and armor protects your body from injury