Answer:
12.0 meters
Explanation:
Given:
v₀ = 0 m/s
a₁ = 0.281 m/s²
t₁ = 5.44 s
a₂ = 1.43 m/s²
t₂ = 2.42 s
Find: x
First, find the velocity reached at the end of the first acceleration.
v = at + v₀
v = (0.281 m/s²) (5.44 s) + 0 m/s
v = 1.53 m/s
Next, find the position reached at the end of the first acceleration.
x = x₀ + v₀ t + ½ at²
x = 0 m + (0 m/s) (5.44 s) + ½ (0.281 m/s²) (5.44 s)²
x = 4.16 m
Finally, find the position reached at the end of the second acceleration.
x = x₀ + v₀ t + ½ at²
x = 4.16 m + (1.53 m/s) (2.42 s) + ½ (1.43 m/s²) (2.42 s)²
x = 12.0 m
A because the girl in that instant is not moving up or down so
( up forces)=(down forces )
The up forces is the tension of the rope and down forces us mg -the gravitational force on the girl by the earth
(3 m) / (2 mm/yr) = (3,000mm)/(2mm/yr) = 1,500 yrs.
This is the time required to age approx 75 generations of the best wine.
Answer:
Explanation:
a ) After the attainment of terminal speed , object takes 4.5 s to cover a distance of 2 m
So terminal speed V = 2 / 4.5
= .444 m /s
When it attains terminal speed , acceleration becomes zero
0 = g - B x .444
B = 22.25 s⁻¹
b ) At t = 0 , v = 0
a = g - B v
a = g at t = 0
c ) When v = .15
a = g - 22.25 x .15
= 9.8 - 3.31
= 6.5 m /s²
Answer:
Electric field E = kQ/r^2
Distance between charges = 6.30 - (-4.40) = 10.70m
Say the neutral point, P, is a distance d from q1. This means it is a distance (10.70 - d) from q2.
Field from q1 at P = k(-9.50x^10^-6) / d^2
Field from q2 at P = k(-8.40x^10^-6) / (10.70-d)^2
These fields are in opposite directions and are equal magnitudes if the resultant field = 0
k(-9.50x^10^-6) / d^2 = k(-8.40x^10^-6) / (10.70-d)^2
9.50 / d^2 =8.40 / (10.70-d)^2
d^2 / (10.70-d)^2 = 9.50/8.40 = 1.131
d/(10.70-d) = sqrt(1.1331) = 1.063
d = 1.063 ((10.70-d)
= 10.63 - 1.063d
2.063d = 10.63
d = 5.15m
The y coordinate where field is zero is 6.30 - 5.15 = 1.15m
Explanation:
