Answer:
The spring force constant is 
.
Explanation:
We are told the mass of the ball is 
, the height above the spring where the ball is dropped is 
, the length the ball compresses the spring is 
and the acceleration of gravity is 
.
We will consider the initial moment to be when the ball is dropped and the final moment to be when the ball stops, compressing the spring. We supose that there is no friction so the initial mechanical energy 
is equal to the final mechanical energy 
:
Initially there is only gravitational potential energy because the force of the spring isn't present and the speed is zero. In the final moment there is only elastic potential energy because the height is zero and the ball has stopped. So we have that:
If we manipulate the equation we have that:
Explanation:
Fgravity = G*(mass1*mass2)/D²
so, if you double one of the masses, what does that do to our equation ?
Fgravitynew = G*(2*mass1*mass2)/D²
due to the commutative property of multiplication
Fgravitynew = 2* G*(mass1*mass2)/D² = 2* Fgravity
so, the correct answer will be 2×45 = 90 units.
It would take roughly 3 seconds.
Answer:
The torque about the origin is 
Explanation:
Torque 
is the cross product between force 
and vector position 
respect a fixed point (in our case the origin):
There are multiple ways to calculate a cross product but we're going to use most common method, finding the determinant of the matrix:
![\overrightarrow{r}\times\overrightarrow{F} =-\left[\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k}\\ F1_{x} & F1_{y} & F1_{z}\\ r_{x} & r_{y} & r_{z}\end{array}\right]](https://tex.z-dn.net/?f=%20%5Coverrightarrow%7Br%7D%5Ctimes%5Coverrightarrow%7BF%7D%20%3D-%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D%20%5Chat%7Bi%7D%20%26%20%5Chat%7Bj%7D%20%26%20%5Chat%7Bk%7D%5C%5C%20F1_%7Bx%7D%20%26%20F1_%7By%7D%20%26%20F1_%7Bz%7D%5C%5C%20r_%7Bx%7D%20%26%20r_%7By%7D%20%26%20r_%7Bz%7D%5Cend%7Barray%7D%5Cright%5D%20)
Answer:
0.280 s
Explanation:
I set it up as 5.22=(55)(0.0266)/x and then solved for x to be 2.80.
