Answer:
d = 265 ft
Therefore, an object fall 265 ft in the first ten seconds after being dropped
Explanation:
This scenario can be represented by an arithmetic progression AP.
nth term = a + nd
Where a is the first term given as 2.63 ft.
d is the common difference and is given as 5.3ft.
n is the particular second/time.
To calculate how far the object would fall in the first 10 seconds, we can derive it using the sum of an AP.
d = nth sum = (n/2)(2a+(n-1)d)
Where n = 10 seconds
a = 2.65 ft
d = 5.3 ft
Substituting the values we have;
d = (10/2)(2×2.65 + (10-1)5.3)
d = 265 ft
Therefore, an object fall 265 ft in the first ten seconds after being dropped
1 N = 1 kgm/s²
1 N / (1m/s²) = 1 kg
2540 N / (9,8 m/s²) = 259,18 kg
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)
The answer is going to be A
