Answer is B. ABAB. Hope it helped you, and have a great day.
-Charlie
Answer:
0.37sec
Explanation:
Period of oscillation of a simple pendulum of length L is:
T
=
2
π
×
√
(L
/g)
L=length of string 0.54m
g=acceleration due to gravity
T-period
T = 2 x 3.14 x √[0.54/9.8]
T = 1.47sec
An oscillating pendulum, or anything else in nature that involves "simple harmonic" (sinusoidal) motion, spends 1/4 of its period going from zero speed to maximum speed, and another 1/4 going from maximum speed to zero speed again, etc. After four quarter-periods it is back where it started.
The ball will first have V(max) at T/4,
=>V(max) = 1.47/4 = 0.37 sec
The force exerted on the box is 56 N
Explanation:
The work done by a force on an object is given by
where
F is the magnitude of the force
d is the displacement of the object
is the angle between the direction of the force and of the displacement
For the box in this problem, we have:
W = 2240 J is the work done
d = 40 m is the displacement of the box
Assuming that the force is parallel to the displacement, 
Solving the equation for F, we find the force exerted on the box:
Learn more about work:
brainly.com/question/6763771
brainly.com/question/6443626
#LearnwithBrainly
C. element only one substance
Answer:
#_photon = 5 10²⁰ photons / s
Explanation:
For this exercise let's calculate the energy of a single quantum of energy, use Planck's law
E = h f
c= λ f
E = h c / λ
λ= 1000 nm (1 m / 109 nm) = 1000 10⁻⁹ m
Let's calculate
E₀ = 6.6310⁻³⁴ 3 10⁸/1000 10⁻⁹
E₀ = 19.89 10⁻²⁰ J
This is the energy emitted by a photon let's use a proportions rule to find the number emitted in P = 100 w
#_photon = P / E₀
#_photon = 100 / 19.89 10⁻²⁰
#_photon = 5 10²⁰ photons / s
