A baseball of mass m = 0.145 kg is suspended vertically from a tree by a string of length L = 1.1 m and negligible mass. Take z
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Answer:
θ = 12.95º
Explanation:
For this exercise it is best to separate the process into two parts, one where they collide and another where the system moves altar the maximum height
Let's start by finding the speed of the bar plus clay ball system, using amount of momentum
The mass of the bar (M = 0.080 kg) and the mass of the clay ball (m = 0.015 kg) with speed (v₀ = 2.0 m / s)
Initial before the crash
p₀ = m v₀
Final after the crash before starting the movement
= (m + M) v
p₀ =
m v₀ = (m + M) v
v = v₀ m / (m + M)
v = 2.0 0.015 / (0.015 +0.080)
v = 0.316 m / s
With this speed the clay plus bar system comes out, let's use the concept of conservation of mechanical energy
Lower
Em₀ = K = ½ (m + M) v²
Higher
= U = (m + M) g y
Em₀ =
½ (m + M) v² = (m + M) g y
y = ½ v² / g
y = ½ 0.316² / 9.8
y = 0.00509 m
Let's look for the angle the height from the pivot point is
L = 0.40 / 2 = 0.20 cm
The distance that went up is
y = L - L cos θ
cos θ = (L-y) / L
θ = cos⁻¹ (L-y) / L
θ = cos⁻¹-1 ((0.20 - 0.00509) /0.20)
θ = 12.95º
To solve the exercise it is necessary to take into account the concepts of wavelength as a function of speed.
From the definition we know that the wavelength is described under the equation,
Where,
c = Speed of light (vacuum)
f = frequency
Our values are,
Replacing we have,
<em>Therefore the wavelength of this wave is </em>
Answer:
Explanation:
<u>Instant Velocity and Acceleration </u>
Give the position of an object as a function of time y(x), the instant velocity can be obtained by
Where y'(x) is the first derivative of y respect to time x. The instant acceleration is given by
We are given the function for y
Note we have changed the last term to be quadratic, so the question has more sense.
The velocity is
And the acceleration is