Answer:
a. A = 0.735 m
b. T = 0.73 s
c. ΔE = 120 J decrease
d. The missing energy has turned into interned energy in the completely inelastic collision
Explanation:
a.
4 kg * 10 m /s + 6 kg * 0 m/s = 10 kg* vmax
vmax = 4.0 m/s
¹/₂ * m * v²max = ¹/₂ * k * A²
m * v² = k * A² ⇒ 10 kg * 4 m/s = 100 N/m * A²
A = √1.6 m ² = 1.26 m
At = 2.0 m - 1.26 m = 0.735 m
b.
T = 2π * √m / k ⇒ T = 2π * √4.0 kg / 100 N/m = 1.26 s
T = 2π *√ 10 / 100 *s² = 1.99 s
T = 1.99 s -1.26 s = 0.73 s
c.
E = ¹/₂ * m * v²max =
E₁ = ¹/₂ * 4.0 kg * 10² m/s = 200 J
E₂ = ¹/₂ * 10 * 4² = 80 J
200 J - 80 J = 120 J decrease
d.
The missing energy has turned into interned energy in the completely inelastic collision
Answer:
why would you waste points
Explanation:
I can guarantee you that it is not
C.<span>the angle that the incident ray makes with a line drawn perpendicular to the reflecting surface I hope this somewhat helps</span>
Answer:
t = 1.77 s
Explanation:
The equation of a traveling wave is
y = A sin [2π (x /λ -t /T)]
where A is the oscillation amplitude, λ the wavelength and T the period
the speed of the wave is constant and is given by
v = λ f
Where the frequency and period are related
f = 1 / T
we substitute
v = λ / T
let's develop the initial equation
y = A sin [(2π / λ) x - (2π / T) t +Ф]
where Ф is a phase constant given by the initial conditions
the equation given in the problem is
y = 5.26 sin (1.65 x - 4.64 t + 1.33)
if we compare the terms of the two equations
2π /λ = 1.65
λ = 2π / 1.65
λ = 3.81 m
2π / T = 4.64
T = 2π / 4.64
T = 1.35 s
we seek the speed of the wave
v = 3.81 / 1.35
v = 2.82 m / s
Since this speed is constant, we use the uniformly moving ratios
v = d / t
t = d / v
t = 5 / 2.82
t = 1.77 s
From the case we know that:
- The moment of inertia Icm of the uniform flat disk witout the point mass is Icm = MR².
- The moment of inerta with respect to point P on the disk without the point mass is Ip = 3MR².
- The total moment of inertia (of the disk with the point mass with respect to point P) is I total = 5MR².
Please refer to the image below.
We know from the case, that:
m = 2M
r = R
m2 = 1/2M
distance between the center of mass to point P = p = R
Distance of the point mass to point P = d = 2R
We know that the moment of inertia for an uniform flat disk is 1/2mr². Then the moment of inertia for the uniform flat disk is:
Icm = 1/2mr²
Icm = 1/2(2M)(R²)
Icm = MR² ... (i)
Next, we will find the moment of inertia of the disk with respect to point P. We know that point P is positioned at the arc of the disk. Hence:
Ip = Icm + mp²
Ip = MR² + (2M)R²
Ip = 3MR² ... (ii)
Then, the total moment of inertia of the disk with the point mass is:
I total = Ip + I mass
I total = 3MR² + (1/2M)(2R)²
I total = 3MR² + 2MR²
I total = 5MR² ... (iii)
Learn more about Uniform Flat Disk here: brainly.com/question/14595971
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