In the first case, the force acting on the spring is the weight of the mass:
This force causes a stretching of
on the spring, so we can use these data to find the spring constant:
In the second case, the first mass is replaced with a second mass, whose weight is
And since we know the spring constant, we can calculate the new elongation of the spring:
<h3>
Answer:</h3><h3>we can say that:-</h3>
- A reading with more no of significant figures is considered to be more precise.
- Kyra recorded a reading of 24.3 sec. Since all non 0 digits are considered to be significant this reading has 3 significant figures.
- Pari recorded a reading of 24 sec. Since all non 0 digits are considered to be significant this reading has 3 significant figures.
<h3>hence we can say that kyra's reading has more significant figures nd so it is more precise.</h3>
The answer for this question is 5 m
===> Distance fallen from rest in free fall =
(1/2) (acceleration) (time²)
(122.5 m) = (1/2) (9.8 m/s²) (time²)
Divide each side by (4.9 m/s²): (122.5 m / 4.9 m/s²) = time²
(122.5/4.9) s² = time²
Take the square root of each side: 5.0 seconds
===> (Accelerating at 9.8 m/s², he will be dropping at
(9.8 m/s²) x (5.0 s) = 49 m/s
when he goes 'splat'. We'll need this number for the last part.)
===> With no air resistance, the horizontal component of velocity
doesn't change.
Horizontal distance = (10 m/s) x (5.0 s) = 50 meters .
===> Impact velocity = (10 m/s horizontally) + (49 m/s vertically)
= √(10² + 49²) = 50.01 m/s arctan(10/49)
= 50.01 m/s at 11.5° from straight down,
away from the base of the cliff.
Answer:
Convection? I'm pretty sure that's it
