To solve this problem we will apply the concepts related to the equations of linear motion and angular motion in order to find the radius. Our values are given as,

Then

The relation between the acceleration and the angular velocity and the radius is,

The angular velocity and the lineal velocity can be related as,

The acceleration and the velocity was given, then


Replacing at the first equation we have,



Now using this expression to find the radius we have that



Therefore the correct answer is C.
Since the track is friction less, block’s kinetic energy at the bottom of the track is equal to its potential energy at the top of the track.
PE = 81 * 9.8 * 3.8 = 3016.44 J
Work = 1/2 * 1888 * d^2
PE = Kinetic energy at the base.
1/2 * 1888 * d^2 = 3016.44
d = 1.78 approx 1.8
F = Ke = 1888 * 1.8 = 3398.4N
Answer:
Explanation:
e. At the right edge of the beam
Check attachment for other solution
Answer:
4.5 x 10¹⁴ Hz
666.7 nm
1.8 x 10⁵ J
The color of the emitted light is red
Explanation:
E = energy of photons of light = 2.961 x 10⁻¹⁹ J
f = frequency of the photon
Energy of photons is given as
E = h f
2.961 x 10⁻¹⁹ = (6.63 x 10⁻³⁴) f
f = 4.5 x 10¹⁴ Hz
c = speed of light = 3 x 10⁸ m/s
λ = wavelength of photon
Using the equation
c = f λ
3 x 10⁸ = (4.5 x 10¹⁴) λ
λ = 0.6667 x 10⁻⁶ m
λ = 666.7 x 10⁻⁹ m
λ = 666.7 nm
n = number of photons in 1 mole = 6.023 x 10²³
U = energy of 1 mole of photons
Energy of 1 mole of photons is given as
U = n E
U = (6.023 x 10²³) (2.961 x 10⁻¹⁹)
U = 1.8 x 10⁵ J
The color of the emitted light is red
Answer:I’m pretty sure it’s spatial
Explanation: