Answer:
83,900 J
Explanation:
First, find the acceleration:
F = ma
1150 N = (1600 kg) a
a = 0.719 m/s²
Now find the final velocity.
Given:
Δx = 45.8 m
v₀ = 6.25 m/s
a = 0.719 m/s²
Find: v
v² = v₀² + 2aΔx
v² = (6.25 m/s)² + 2 (0.719 m/s²) (45.8 m)
v = 10.2 m/s
Now find the final KE:
KE = ½ mv²
KE = ½ (1600 kg) (10.2 m/s)²
KE = 83,920 J
Rounded to three significant figures, the final kinetic energy is 83,900 J.
Answer:
1 x 10¹⁷
Explanation:
Given data:
Radius of the earth = 6000km
Radius of an atom = 60pm
Now, how many orders is the radius of the earth larger than an atom
Solution:
To solve this problem, let us express both quantity as the same unit;
1000m = 1km
6000km = 6000 x 10³m = 6 x 10⁶m
60pm;
1 x 10⁻¹²m = 1pm
60pm = 60 x 1 x 10⁻¹²m = 6 x 10⁻¹¹m
Now;
The order:
= 1 x 10¹⁷
The atoms furthest from the nucleus
Instead of asking this question go and study
===> 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.