Answer:
The answer is "
"
Explanation:
The formula for velocity:


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.
Answer:
Distance of the point where electric filed is 2.45 N/C is 1.06 m
Explanation:
We have given charge per unit length, that is liner charge density 
Electric field E = 2.45 N/C
We have to find the distance at which electric field is 2.45 N/C
We know that electric field due to linear charge is equal to
, here
is linear charge density and r is distance of the point where we have to find the electric field
So 
r = 1.06 m
So distance of the point where electric filed is 2.45 N/C is 1.06 m
Answer:
16.2 s
Explanation:
Given:
Δx = 525 m
v₀ = 0 m/s
a = 4.00 m/s²
Find: t
Δx = v₀ t + ½ at²
525 m = (0 m/s) t + ½ (4.00 m/s²) t²
t = 16.2 s
Answer:
32000 N
Explanation:
From the question given above, the following data were obtained:
Initial velocity (u) = 40 m/s
Distance (s) = 10 m
Final velocity (v) = 0 m/s
Mass (m) of car = 400 Kg
Force (F) =?
Next, we shall determine the acceleration of the the car. This can be obtained as follow:
Initial velocity (u) = 40 m/s
Distance (s) = 10 m
Final velocity (v) = 0 m/s
Acceleration (a) =?
v² = u² + 2as
0² = 40² + (2 × a × 10)
0 = 1600 + 20a
Collect like terms
0 – 1600 = 20a
–1600 = 20a
Divide both side by –1600
a = –1600 / 20
a = –80 m/s²
The negative sign indicate that the car is decelerating i.e coming to rest.
Finally, we shall determine the force needed to stop the car. This can be obtained as follow:
Mass (m) of car = 400 Kg
Acceleration (a) = –80 m/s²
Force (F) =?
F = ma
F = 400 × –80
F = – 32000 N
NOTE: The negative sign indicate that the force is in opposite direction to the motion of the car.