Answer:
630.75 j
Explanation:
from the question we have the following
total mass (m) = 54.5 kg
initial speed (Vi) = 1.4 m/s
final speed (Vf) = 6.6 m/s
frictional force (FF) = 41 N
height of slope (h) = 2.1 m
length of slope (d) = 12.4 m
acceleration due to gravity (g) = 9.8 m/s^2
work done (wd) = ?
- we can calculate the work done by the boy in pushing the chair using the law of law of conservation of energy
wd + mgh = (0.5 mVf^2) - (0.5 mVi^2) + (FF x d)
wd = (0.5 mVf^2) - (0.5 mVi^2) + (FF x d) - (mgh)
where wd = work done
m = mass
h = height
g = acceleration due to gravity
FF = frictional force
d = distance
Vf and Vi = final and initial velocity
wd = (0.5 x 54.5 x 6.9^2) - (0.5 x 54.5 x 1.4^2) + (41 x 12.4) - (54.5 X 9.8 X 2.1)
wd = 630.75 j
Answer:42.43m/s
Explanation:According to vf=vi+at, we can calculate it since v0 equals to 0. vf=0+9.8m/s^2*4.33s= 42.434m/s
Total distance = 36500 m
The average velocity = 19.73 m/s
<h3>Further explanation</h3>
Given
vo=initial velocity=0(from rest)
a=acceleration= 1 m/s²
t₁ = 20 s
t₂ = 0.5 hr = 1800 s
t₃= 30 s
Required
Total distance
Solution
State 1 : acceleration


State 2 : constant speed

State 3 : deceleration


Total distance : state 1+ state 2+state 3

the average velocity = total distance : total time

The one fact that needs to be mentioned but isn't given anywhere on or around the graph is: The distance, on the vertical axis, is the distance FROM home. So any point on the graph where the distance is zero ... the point is in the x-axis ... is a point AT home.
Segment D ...
Walking AWAY from home; distance increases as time increases.
Segment B ...
Not walking; distance doesn't change as time increases.
Segment C ...
Walking away from home, but slower than before; distance increases as time increases, but not as fast. Slope is less than segment-D.
Segment A ...
Going home; distance is DEcreasing as time increases. Walking pretty fast ... the slope of the line is steep.
A magnetic domain is a group of atoms aligns with magnetic poles. Domains are usually <span>light and dark stripes visible within each grain.</span>