Answer:
a.) a = 0 ms⁻²
b.) a = 9.58 ms⁻²
c.) a = 7.67 ms⁻²
Explanation:
a.)
Acceleration (a) is defined as the time rate of change of velocity
Given data
Final velocity = v₂ = 0 m/s
Initial velocity = v ₁ = 0 m/s
As the space shuttle remain at rest for the first 2 minutes i.e there is no change in velocity so,
a = 0 ms⁻²
b.)
Given data
As the space shuttle start from rest, So initial velocity is zero
Initial velocity = v₁ = 0 ms⁻¹
Final velocity = v₂ = 4600 ms⁻¹
Time = t = 8 min = 480 s
By the definition of Acceleration (a)

a = 9.58 ms⁻²
c.)
Given data
As the space shuttle is at rest for first 2 min then start moving, So initial velocity is zero
Initial velocity = v₁ = 0 ms⁻¹
Final velocity = v₂ = 4600 ms⁻¹
Time = t = 10 min = 600 s
By the definition of Acceleration (a)

a = 7.67 ms⁻²
Photon energy is directly proportional to the frequency of electromagnetic radiation.
(That would also mean that it's inversely proportional to the wavelength.)
So the photon energy increases as you scan the chart of visible colors
moving from the red end of the rainbow to the blue end.
first off lemme just say this is really easy man, just look at the directions
Blank #1: -23
Blank #2: 23
A wave is a result of the disturbance in the equilibrium state. There are two types of wave, transverse and longitudinal. Transverse wave affects amplitude while longitudinal wave affects the frequency of the wave. As for the transverse wave, the magnitude of the perpendicular disturbance of the wave is directly proportional to the amplitude of the wave. The higher the transverse disturbance the higher the amplitude.
<h2>distance = 523 cm</h2>
Explanation:
( a ) The rotational speed of the ladybug = 25 r.p.m = 25/60 r.p.s
= 5/12 rev/sec
( b ) The definition of frequency is the number of rotations per second .
Here the number of rotations per second is 5/12 . Thus frequency = 5/12 Hz
( c ) The tangential speed is v = angular velocity x radius of rotation
The angular velocity ω = 2π x n , where n is the number of rotations per second
Thus angular velocity = 2π x 5/12 = 5π/6 rad/sec
The linear velocity = angular velocity x distance from center of record
Thus tangential speed = 5π/6 x 10 = 25π/3 cm/sec
Angular displacement in 20 sec = ω x t = 5π/6 x 20 = 50π/3 rad
Linear displacement = angular displacement x distance from center of record
= 50π/3 x 10 = 500π/3 = 523 cm