Answer:
A. 
B. 
C. 
Explanation:
Given:
- spring constant,

- mass attached,

A)
for a spring-mass system the frequency is given as:



B)
frequency is given as:



C)
Time period of a simple harmonic motion is given as:


Answer: See the explanation below.
Explanation: For this assignment, I chose to display how eclipses are created.
My model was made utilizing a 3D displaying device program for all intents and purposes. The items utilized are three models I made for this presentation, Earth, the moon, and the sun. These three models will be utilized for the showcase.
The light that shines from the sun would create a shadow on the moon. The moon would then catch the light that should've arrived on Earth, making the shadow we call an eclipse. Earth gets a shadow of the moon and the remainder of Earth is lit up from the rest of the light, making an eclipse.
The individual I demonstrated my project to was [<em>Someone you know</em>], [<em>Pronoun</em>] said it precisely took after the occasion of an eclipse. The light from the sun being shined on to the moon rather than the Earth, creating the shadow we call an eclipse.
Answer:
a)
b)
Explanation:
a)
The width of the central bright in this diffraction pattern is given by:
when m is a natural number.
here:
- m is 1 (to find the central bright fringe)
- D is the distance from the slit to the screen
- a is the slit wide
- λ is the wavelength
So we have:
b)
Now, if we do m=2 we can find the distance to the second minima.

Now we need to subtract these distance, to get the width of the first bright fringe :
I hope it heps you!
Answer:
The final velocity of the object is 330 m/s.
Explanation:
To solve this problem, we first must find the acceleration of the object. We can do this using Newton's Second Law, given by the following equation:
F = ma
If we plug in the values that we are given in the problem, we get:
42 = 7 (a)
To solve for a, we simply divide both sides of the equation by 7.
42/7 = 7a/7
a = 6 m/s^2
Next, we should write out all of the information we have and what we are looking for.
a = 6 m/s^2
v1 = 0 m/s
t = 55 s
v2 = ?
We can use a kinematic equation to solve this problem. We should use:
v2 = v1 + at
If we plug in the values listed above, we should get:
v2 = 0 + (6)(55)
Next, we should solve the problem by performing the multiplication on the right side of the equation.
v2 = 330 m/s
Therefore, the final velocity reached by the object is 330 m/s.
Hope this helps!
Answer:
a) τ = 0.672 N m
, b) θ = 150 rad
, c) W = 100.8 J
Explanation:
a) for this part let's start by finding angular acceleration, when the angular velocity stops it is zero (w = 0)
w = w₀ + α t
α = -w₀ / t
α = 120 / 2.5
α = 48 rad / s²
The moment of inertia of a cylinder is
I = ½ M R²
Let's calculate the torque
τ = I α
τ = ½ M R² α
τ = ½ 2.8 0.1² 48
τ = 0.672 N m
b) we look for the angle by kinematics
θ = w₀ t + ½ α t2
θ = ½ α t²
θ = ½ 48 2.5²
θ = 150 rad
c) work in angular movement
W = τ θ
W = 0.672 150
W = 100.8 J