Answer: Both cannonballs will hit the ground at the same time.
Explanation:
Suppose that a given object is on the air. The only force acting on the object (if we ignore air friction and such) will be the gravitational force.
then the acceleration equation is only on the vertical axis, and can be written as:
a(t) = -(9.8 m/s^2)
Now, to get the vertical velocity equation, we need to integrate over time.
v(t) = -(9.8 m/s^2)*t + v0
Where v0 is the initial velocity of the object in the vertical axis.
if the object is dropped (or it only has initial velocity on the horizontal axis) then v0 = 0m/s
and:
v(t) = -(9.8 m/s^2)*t
Now, if two objects are initially at the same height (both cannonballs start 1 m above the ground)
And both objects have the same vertical velocity, we can conclude that both objects will hit the ground at the same time.
You can notice that the fact that one ball is fired horizontally and the other is only dropped does not affect this, because we only analyze the vertical problem, not the horizontal one. (This is something useful to remember, we can separate the vertical and horizontal movement in these type of problems)
I think it’s either A or B
Switch because look A switch detects the speed that given device can handle and communicates with it at that speed
The answer is:
C. 361 m/s
The explanation:
To calculate the speed of sound at a given temperature (50°C) we are going to use this formula:
v = 331 + 0.6T
when V is the velocity
and T is the temperature = 50°C
by substitution:
v = 331 + 0.6(50)
v = 361 m/s
So, The correct answer is C.
because of the variation of the motion of the molecules of air with change of temperature so, the velocity (V) of the sound in the air is change with temperature.
Answer:
46.4 s
Explanation:
5 minutes = 60 * 5 = 300 seconds
Let g = 9.8 m/s2. And 
be the slope of the road, s be the distance of the road, a be the acceleration generated by Rob, 3a/4 is the acceleration generated by Jim . Both of their motions are subjected to parallel component of the gravitational acceleration 
Rob equation of motion can be modeled as s = a_Rt_R^2/2 = a300^2/2 = 45000a[/tex]
Jim equation of motion is 
As both of them cover the same distance
So Jim should start 346.4 – 300 = 46.4 seconds earlier than Rob in other to reach the end at the same time
