-- reduce the length of a wire to 1/2 . . . cut the resistance in half
-- reduce the diameter to 1/4 . . . reduce the cross-section area by (1/4²) . . . increase the resistance by 16x .
-- R2 = (R1) · (1/2) · (16) = 8 · R1
<em>-- R2 / R1 = 8</em>
C: Litmus Paper. Red litmus paper turns blue in acids; blue litmus paper turns red in bases
Answer:
17.2 seconds
Explanation:
Given:
v₀ = 0 m/s
a₁ = 10.0 m/s²
t₁ = 3.0 s
a₂ = 16 m/s²
t₂ = 5.0 s
a₃ = -12 m/s²
v₃ = 0 m/s
Find: t
First, find v₁:
v₁ = a₁t₁ + v₀
v₁ = (10.0 m/s²) (3.0 s) + (0 m/s)
v₁ = 30 m/s
Next, find v₂:
v₂ = a₂t₂ + v₁
v₂ = (16 m/s²) (5.0 s) + (30 m/s)
v₂ = 110 m/s
Finally, find t₃:
v₃ = a₃t₃ + v₂
(0 m/s) = (-12 m/s²) t₃ + (110 m/s)
t₃ = 9.2 s
The total time is:
t = t₁ + t₂ + t₃
t = 3.0 s + 5.0 s + 9.2 s
t = 17.2 s
Round as needed.
Answer:
DONTKNOW NEED CAREER FILD FIRST TO ANWSER
Explanation:
Answer:
The answer is below
Explanation:
The length of the rope is equal to the radius of the circle formed by the complete rotation of the rope. Therefore the radius = 1.50 m.
a) The distance covered by the rope when completing one rotation is the same as the perimeter of the circle. Hence:
Distance covered in one rotation = 2π * radius = 2π * 1.5 = 3π meters
The velocity of the ball = Distance / time = 3π meters / 3.4 seconds = 2.77 m/s
b) The initial velocity (u) is 0 m/s, the final velocity is 2.77 m/s during time (t) = 3.4 s. Hence acceleration (a):
v = u + at
2.77 = 3.4a
a = 0.82 m/s²
c) Force on ball = mass * acceleration = 4 * 0.82 = 3.28 N
