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
Answer:
hope this helps you're welcome
To solve the problem it is necessary to apply the concepts related to Kepler's third law as well as the calculation of distances in orbits with eccentricities.
Kepler's third law tells us that
Where
T= Period
G= Gravitational constant
M = Mass of the sun
a= The semimajor axis of the comet's orbit
The period in years would be given by
PART A) Replacing the values to find a, we have
Therefore the semimajor axis is 
PART B) If the semi-major axis a and the eccentricity e of an orbit are known, then the periapsis and apoapsis distances can be calculated by
Answer:
∑F = 10.2 N
Explanation:
We have:
Initial velocity: 0.5 m/s
Final velocity: 3 m/s
Time: 1.5 s
We have all of the components needed to calculate acceleration. Let's do that, shall we?
a = vf-vo/t
a = 2.5/1.5
a = 1.7 
/
Now, look at the Net Force equation:
∑F = ma
Plug in the variables, to get:
∑F = (6)(1.7)
∑F = 10.2 N (You can round this according to significant digits)
