Possibilities . . .
-- nuclear reaction
-- nuclear fission
-- nuclear fusion
-- radioactive decay.
Any of these makes it a true statement.
Fine, lets do a retry of this.
Δd = -0.9m
v₁ = 0
v₂ = ?
a = -9.8 m/s²
Δt = ?
We can use the following kinematic equation and solve for Δt.
Δd = v₁Δt + 0.5(a)(Δt)²
Δd = 0.5(a)(Δt)²
2Δd = a(Δt)²
√2Δd/a = Δt
√2(-0.9m)/(-9.8 m/s²) = Δt
0.<u>4</u>28571428574048 = Δt
Therefore, it takes 0.4 seconds for the glass to hit the ground, or 0.43s as you said (even though I don't believe it follows significant digit rules)
A person on Earth would weigh a lot more on the sun due to increased gravity.
The sun is a massive object as compared to the sun. It is assumed that the gravity must be a lot greater as well. It is the strong gravitational pull of the sun that keeps our planets from drifting off into space. No matter where you are the mass of an object will remain constant, so will the energy and the number of atoms they hold.
The answer is gravity.
Answer:
Follows are the solution to this question:
Explanation:
Please find the correct question in the attachment file.
Let:
Calculating the value of 
![\to \left | \begin{array}{ccc}\hat{i}&\hat{j}&\hat{K}\\R_i&R_j&R_k\\S_i&S_j&S_k\end{array}\right | = \hat{i}[R_j S_k-S_jR_k]-\hat{j}[R_i S_k-S_iR_k]+\hat{k}[R_i S_j-S_iR_j]](https://tex.z-dn.net/?f=%5Cto%20%5Cleft%20%7C%20%5Cbegin%7Barray%7D%7Bccc%7D%5Chat%7Bi%7D%26%5Chat%7Bj%7D%26%5Chat%7BK%7D%5C%5CR_i%26R_j%26R_k%5C%5CS_i%26S_j%26S_k%5Cend%7Barray%7D%5Cright%20%7C%20%3D%20%5Chat%7Bi%7D%5BR_j%20S_k-S_jR_k%5D-%5Chat%7Bj%7D%5BR_i%20S_k-S_iR_k%5D%2B%5Chat%7Bk%7D%5BR_i%20S_j-S_iR_j%5D)
Calculating the value of 
![\to (R_i\hat{i}+R_j\hat{j}+R_k\hat{k}) \cdot ( \hat{i}[R_j S_k-S_jR_k]-\hat{j}[R_i S_k-S_iR_k]+\hat{k}[R_i S_j-S_iR_j])](https://tex.z-dn.net/?f=%5Cto%20%28R_i%5Chat%7Bi%7D%2BR_j%5Chat%7Bj%7D%2BR_k%5Chat%7Bk%7D%29%20%5Ccdot%20%28%20%5Chat%7Bi%7D%5BR_j%20S_k-S_jR_k%5D-%5Chat%7Bj%7D%5BR_i%20S_k-S_iR_k%5D%2B%5Chat%7Bk%7D%5BR_i%20S_j-S_iR_j%5D%29)
by solving this value it is equal to 0.
2.3 seconds
Ignoring air resistance, the flight time is merely a function of gravity and vertical velocity. The vertical velocity will be the initial velocity multiplied by the sine of the angle above the horizon. So:
V = sin(72)*12 m/s
V = 0.951056516 * 12 m/s
V = 11.4126782 m/s
Gravitational acceleration is 9.8 m/s, so divide the vertical velocity by gravitational acceleration to get how long it takes for the ball to reach its apex.
11.4126782 m/s / 9.8 m/s^2 = 1.164559 s
And the old saying "What goes up, must come down" really applies here. And conveniently, it's also symmetric, in that the time it takes to fall will match the time it takes to reach its apex. So multiply the time by 2.
1.164559 s * 2 = 2.329117999 s
Rounding the result to 2 significant figures gives 2.3 seconds.
