Answer:
Options d and e
Explanation:
The pendulum which will be set in motion are those which their natural frequency is equal to the frequency of oscillation of the beam.
We can get the length of the pendulums likely to oscillate with the formula;
where g=9.8m/s
ω= 2rad/s to 4rad/sec
when ω= 2rad/sec
L = 2.45m
when ω= 4rad/sec
L = 9.8/16
L=0.6125m
L is between 0.6125m and 2.45m.
This means only pendulum lengths in this range will oscillate.Therefore pendulums with length 0.8m and 1.2m will be strongly set in motion.
Have a great day ahead
V = t^2 - 9t + 18
position, s
s = t^3 /3 - 4.5t^2 +18t + C
t = 0, s = 1 => 1=C => s = t^3/3 -4.5t^2 + 18t + 1
Average velocity: distance / time
distance: t = 8 => s = 8^3 / 3 - 4.5 (8)^2 + 18(8) + 1 = 27.67 m
Average velocity = 27.67 / 8 = 3.46 m/s
t = 5 s
v = t^2 - 9t + 18 = 5^2 - 9(5) + 18 = -2 m/s
speed = |-2| m/s = 2 m/s
Moving right
V > 0 => t^2 - 9t + 18 > 0
(t - 6)(t - 3) > 0
=> t > 6 and t > 3 => t > 6 s => Interval (6,8)
=> t < 6 and t <3 => t <3 s => interval (0,3)
Going faster and slowing dowm
acceleration, a = v' = 2t - 9
a > 0 => 2t - 9 > 0 => 2t > 9 => t > 4.5 s
Then, going faster in the interval (4.5 , 8) and slowing down in (0, 4.5)
Answer:
Im pretty sure 1 is gravity, 2 is force
Explanation:
The strength of the electric field is 5 N/C
Explanation:
The magnitude of the electric field produced by a single-point charge is given by:
where
is the Coulomb's constant
Q is the magnitude of the charge
r is the distance from the charge
In this problem, we have:
is the charge producing the field
r = 100 m is the distance from the charge at which we want to calculate the field
Substituting into the equation, we find the s trength of the electric field:
Learn more about electric field:
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