Answer:
The thrown rock will strike the ground
earlier than the dropped rock.
Explanation:
<u>Known Data</u>


, it is negative as is directed downward
<u>Time of the dropped Rock</u>
We can use
, to find the total time of fall, so
, then clearing for
.
![t_{D}=\sqrt[2]{\frac{300m}{4.9m/s^{2}}} =\sqrt[2]{61.22s^{2}} =7.82s](https://tex.z-dn.net/?f=t_%7BD%7D%3D%5Csqrt%5B2%5D%7B%5Cfrac%7B300m%7D%7B4.9m%2Fs%5E%7B2%7D%7D%7D%20%3D%5Csqrt%5B2%5D%7B61.22s%5E%7B2%7D%7D%20%3D7.82s)
<u>Time of the Thrown Rock</u>
We can use
, to find the total time of fall, so
, then,
, as it is a second-grade polynomial, we find that its positive root is
Finally, we can find how much earlier does the thrown rock strike the ground, so 
Answer:
F= 5.71 N
Explanation:
width of door= 0.91 m
door closer torque on door= 5.2 Nm
In order to hold the door in open position we need to exert an equal and opposite torque, to the door closer torque, on the door.
so wee need to exert 5.2 Nm torque on the door.
If we want to apply minimum force to exert the required torque we need to apply force perpendicularly on the door knob (end of door) so that to to greater moment arm.
T= r x F
T= r F sin∅
F= T/ (r * sin∅)
F= 5.2/ (0.91 * 1)
F= 5.71 N
Comets orbit the sun just like planets do. Except a comet usually has a very elongated orbit. Thanks to the laws of gravity comets obey the same laws. A comets orbit takes it very close to the sun and then far away again.
In my estimation I would say C, I was leaning towards A, but I believe that would merely be "incomplete combustion." I hope this was semi-helpful!