Force applied by the machine to over come resistance
Answer:
Physics is the answer to all the forces in nature and it is important to know all of that in a day to day life to succeed in life
Explanation:
Answer:
Explanation:
The volume of a sphere is:
V = 4/3 * π * a^3
The volume charge density would then be:
p = Q/V
p = 3*Q/(4 * π * a^3)
If the charge density depends on the radius:
p = f(r) = k * r
I integrate the charge density in spherical coordinates. The charge density integrated in the whole volume is equal to total charge.
Since p = k*r
Q = p*π^2*r^3 / 2
Then:
p(r) = 2*Q / (π^2*r^3)
Answer:
0-4 acceleration comes at 12 m/s where (B) stagnates at 12 m/s and remains for 4 seconds (C) is breaks being activated slowing the car to 6 m/s in 2 seconds and (D) over the course of 4 seconds brings the car to 10 m/s.
Explanation:
Answer:
269 m
45 m/s
-58.6 m/s
Explanation:
Part 1
First, find the time it takes for the package to land. Take the upward direction to be positive.
Given (in the y direction):
Δy = -175 m
v₀ = 0 m/s
a = -9.8 m/s²
Find: t
Δy = v₀ t + ½ at²
(-175 m) = (0 m/s) t + ½ (-9.8 m/s²) t²
t = 5.98 s
Next, find the horizontal distance traveled in that time:
Given (in the x direction):
v₀ = 45 m/s
a = 0 m/s²
t = 5.98 s
Find: Δx
Δx = v₀ t + ½ at²
Δx = (45 m/s) (5.98 s) + ½ (0 m/s²) (5.98 s)²
Δx = 269 m
Part 2
Given (in the x direction):
v₀ = 45 m/s
a = 0 m/s²
t = 5.98 s
Find: v
v = at + v₀
v = (0 m/s²) (5.98 s) + (45 m/s
v = 45 m/s
Part 3
Given (in the y direction):
Δy = -175 m
v₀ = 0 m/s
a = -9.8 m/s²
Find: v
v² = v₀² + 2aΔy
v² = (0 m/s)² + 2 (-9.8 m/s²) (-175 m)
v = -58.6 m/s
