I believe it would be weight. mass never changes.
Ignoring the air resistance it will take about 3 seconds for the object to reach the ground.We know that the acceleration due to gravity is 10m/s2.
We also know that the final velocity is 30 m/s while the initial velocity is 0 m/s
we can use the formulae for acceleration to calculate the time taken/
(final - initial velocity)/timetaken=10
(30-0)/timetaken=10
timetaken =30/10=3 seconds
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
Using the formula: E = kQ / d² where E is the electric field, Q is the test charge in coulomb, and d is the distance.
E = kQ / d²
k = 9 x 10^9 N-m²/C²
Q = 6.4 x 10^-5 C
d = 2.5 x 10^-2 m
Substituting the given values to the equation, we have:
E = (9 x 10^9)(6.4 x 10^-5) / (2.5 x 10^-2) ²
Electric field at the test charge is 921600000 N/C
