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)
Answer:
All of the above
Explanation:
The correct answer is option E (All of the above)
All the options are alternative source of energy.
The option given are not the traditional way (energy production from coal) of extracting energy as the loss of natural resources does not take place in these source of energy.
energy extracted from wind, solar light , hydrogen ,etc are the source of energy the alternative source of production of energy because do not exploit the natural resources, reduce the carbon emission and energy produced by them is clean energy.
Answer:
varn=n1+1ehvkT–1
Explanation:
This is Einstein's equation.
The displacement vector (SI units) is
![\vec{r} =At\hat{i}+A[t^{3}-6t^{2}]\hat{j}](https://tex.z-dn.net/?f=%5Cvec%7Br%7D%20%3DAt%5Chat%7Bi%7D%2BA%5Bt%5E%7B3%7D-6t%5E%7B2%7D%5D%5Chat%7Bj%7D)
The speed is a scalar quantity. Its magnitude is
![v= \sqrt{A^{2}t^{2}+A^{2}(t^{3}-6t^{2})^{2}} \\ v=A \sqrt{t^{2}+t^{6}-12t^{5}+36t^{4}} \\ v=At \sqrt{t^{4}-12t^{3}+36t^{2}+1}](https://tex.z-dn.net/?f=v%3D%20%5Csqrt%7BA%5E%7B2%7Dt%5E%7B2%7D%2BA%5E%7B2%7D%28t%5E%7B3%7D-6t%5E%7B2%7D%29%5E%7B2%7D%7D%20%5C%5C%20v%3DA%20%5Csqrt%7Bt%5E%7B2%7D%2Bt%5E%7B6%7D-12t%5E%7B5%7D%2B36t%5E%7B4%7D%7D%20%5C%5C%20v%3DAt%20%5Csqrt%7Bt%5E%7B4%7D-12t%5E%7B3%7D%2B36t%5E%7B2%7D%2B1%7D%20%20)
Answer: At√(t⁴ - 12t³ + 36t² + 1)