Answer:
If child weight is equal to rope force then child will move with uniform speed
or we can say that the child will remain at rest in his position
Explanation:
As we know that child is hanging by rope
so here there will be two forces on the child
1) Weight or gravitational force which act vertically downwards
2) Tension in the rope which act vertically upwards
Now if child will accelerate upwards then tension force must be more than the weight of the child
If tension force is less than the weight then child will decelerate and his speed will decrease
if tension force is equal to child weight then in that case the child will remain at rest or it will move with same speed
You got the formulas on the sheet on the top :) So just use those, exchanging v (as in velocity, expressed in m/s) and the d (in meters) and t (in seconds). Hope you will manage it.
Answer:
1.19 m/s²
Explanation:
The frequency of the wave generated in the string in the first experiment is f = n/2l√T/μ were T = tension in string = mg were m = 1.30 kg weight = 1300 g , μ = mass per unit length of string = 1.01 g/m. l = length of string to pulley = l₀/2 were l₀ = lent of string. Since f is the second harmonic, n = 2, so
f = 2/2(l₀/2)√mg/μ = 2(√mg/μ)/l₀ (1)
Also, for the second experiment, the period of the wave in the string is T = 2π√l₀/g. From (1) l₀ = 2(√mg/μ)/f and from (2) l₀ = T²g/4π²
Equating (1) and (2) we ave
2(√mg/μ)/f = T²g/4π²
Making g subject of the formula
g = 2π√(2√(m/μ)/f)/T
The period T = 316 s/100 = 3.16 s
Substituting the other values into , we have
g = 2π√(2√(1300 g/1.01 g/m)/200 Hz)/3.16
g = 2π√(2 × 35.877/200 Hz)/3.16
g = 2π√(71.753/200 Hz)/3.16
g = 2π√(0.358)/3.16
g = 2π × 0.599/3.16
g = 1.19 m/s²
a). The truck's distance covered for the trip is
(60) + (60 - 15) = 105 kilometers .
b). Its displacement for the whole trip is the distance
and direction from the start-point to the end-point.
15 kilometers east .