Answer:
4.44s
Explanation:
A 34-kg child on an 18-kg swing set swings back and forth through small angles. If the length of the very light supporting cables for the swing is 4.9 m, how long does it take for each complete back-and-forth swing? Assume that the child and swing set are very small compared to the length of the cables
since the mass of the child and that of the swing is negligible, the masses wont be involved in the calculation
T=2π√L/g
g=acceleration due to gravity which is 9.81m/s2
the length of the supporting cable is 4.9m
T the period
period is the time required to make a complete oscillation
T=2*π√4.9/9.81
T=2*π*0.706
T=4.44s
4.44s
Acceleration is constant in free fall. So it would again be 9.8 m/s².
Answer:
Explanation:
A track 100m
The sprinter passes 12m mark in 1.8seconds
Let this be the initial point
d1 = 12m
t1 = 1.8seconds
And Passes the 56m mark in 6.7 seconds
Then, the second point is
d2 = 56m
t2 = 6.7 seconds
The sprinter velocity between this two position
Velocity is the rate of change of displacement
V = displacement / time taken
V = ∆x / ∆t
V = (x2 - x1) / (t2 - t1)
V = (56 - 12) / (6.7 - 1.8)
V = 44 / 4.9
V = 8.98 m/s
Then, the sprinter velocity between this two position is 8.98 m/s
Answer:
I = 0.058 [A]
Explanation:
Ohm's law is defined as voltage equal to the product of current by resistance.
where:
V = voltage = 4.2 [V]
I = current [amp]
R = resistivity = 72 [ohm]
Now clearing I, we have:
R = D/2 = 4.0 / 2 = 2 m
a = V² / R = 2,8² / 2 = 7.84 / 2 = 3.92 m/s²