Answer:
233.1 miles per hours
Explanation:
Speed: This is defined as the ratio of distance to time. The S.I unit of speed is m/s. speed is a vector quantity because it can only be represented by magnitude only. Mathematically, speed can be expressed as,
S = d/t ....................... Equation 1
Where S = speed of the runner, d = distance covered, t = time.
Given: d = 100 meter , t = 9.580 seconds
Conversion:
If, 1 meter = 0.00062 miles
Then, 100 meters = (0.00062×100) miles = 0.62 miles.
Also
If, 3600 s = 1 h
Then, 9.580 s = (1×9.580)/3600 = 0.00266 hours.
Substitute into equation 1
S = 0.62/0.00266
S = 233.1 miles per hours.
Hence the runner speed is 233.1 miles per hours
Answer:
The object is dropped, we know the initial velocity is zero. Once the object has left contact with whatever held or threw it, the object is in free-fall. Under these circumstances, the motion is one-dimensional and has constant acceleration of magnitude g.
No so sure
Explanation:
Hope it helps
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:
The equilibrium position will shift towards the left hand side or reactants side
Explanation:
Decreasing the volume (increasing the pressure) of the system will shift the equilibrium position towards the lefthand side or reactants side. This is because, decreasing the volume (increasing the pressure) implies shifting the equilibrium position towards the side having the least number of moles.
There are two moles of reactants and a total of three moles of products(total). Hence decreasing the volume and increasing the pressure of the gas phase reaction will shift the equilibrium position towards the lefthand side.