Answer:
Coefficient of static friction will be equal to 0.642
Explanation:
We have given acceleration ![a=6.3m/sec^2](https://tex.z-dn.net/?f=a%3D6.3m%2Fsec%5E2)
Acceleration due to gravity ![g=9.8m/sec^2](https://tex.z-dn.net/?f=g%3D9.8m%2Fsec%5E2)
We have to find the coefficient of static friction between truck and a cabinet will
We know that acceleration is equal to
, here
is coefficient of static friction and g is acceleration due to gravity
So ![\mu =\frac{a}{g}=\frac{6.3}{9.8}=0.642](https://tex.z-dn.net/?f=%5Cmu%20%3D%5Cfrac%7Ba%7D%7Bg%7D%3D%5Cfrac%7B6.3%7D%7B9.8%7D%3D0.642)
So coefficient of static friction will be equal to 0.642
Answer:
![a=4.44\frac{m}{s^2}](https://tex.z-dn.net/?f=a%3D4.44%5Cfrac%7Bm%7D%7Bs%5E2%7D)
Explanation:
First we have to find the time required for train to travel 60 meters and impact the car, this is an uniform linear motion:
![t=\frac{d}{v}\\\\t=\frac{60m}{30\frac{m}{s}}=2s](https://tex.z-dn.net/?f=t%3D%5Cfrac%7Bd%7D%7Bv%7D%5C%5C%5C%5Ct%3D%5Cfrac%7B60m%7D%7B30%5Cfrac%7Bm%7D%7Bs%7D%7D%3D2s)
The reaction time of the driver before starting to accelerate was 0.50 seconds. So, remaining time for driver is 1.5 seconds.
Now, we have to calculate the distance traveled for the driver in this 0.5 seconds before he start to accelerate. Again, is an uniform linear motion:
![d=vt\\d=20\frac{m}{s}(0.5s)=10m](https://tex.z-dn.net/?f=d%3Dvt%5C%5Cd%3D20%5Cfrac%7Bm%7D%7Bs%7D%280.5s%29%3D10m)
The driver cover 10 meters in this 0.5 seconds. So, the remaining distance to be cover in 1.5 seconds by the driver are 35 meters. We calculate the minimum acceleration required by the car in order to cross the tracks before the train arrive, Since this is an uniformly accelerated motion, we use the following equation:
![d=v_0t+\frac{1}{2}at^2\\a=\frac{2(d-v_0t)}{t^2}\\a=\frac{2(35m-20\frac{m}{s}*1.5s}{(1.5s)^2}\\a=4.44\frac{m}{s^2}](https://tex.z-dn.net/?f=d%3Dv_0t%2B%5Cfrac%7B1%7D%7B2%7Dat%5E2%5C%5Ca%3D%5Cfrac%7B2%28d-v_0t%29%7D%7Bt%5E2%7D%5C%5Ca%3D%5Cfrac%7B2%2835m-20%5Cfrac%7Bm%7D%7Bs%7D%2A1.5s%7D%7B%281.5s%29%5E2%7D%5C%5Ca%3D4.44%5Cfrac%7Bm%7D%7Bs%5E2%7D)
Drop "moves" from the list for a moment.
You can also drop "stops moving", because that's included in "changes speed"
(from something to zero).
When an object changes speed or changes direction, that's called "acceleration".
I dropped the first one from the list, because an object can be moving,
and as long as it's speed is constant and it's moving in a straight line,
there's no acceleration.
I think you meant to say "starts moving". That's a change of speed (from zero
to something), so it's also acceleration.
The answer is Golgi apparatus.
HOPE THIS HELPS!
Answer:
Open- closed
Explanation:
It has Open-closed configuration of its end