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Maksim231197 [3]

2 years ago

14

Let this oscillator have the same energy as a mass on a spring, with the same k and m, released from rest at a displacement of 5

.00cm from equilibrium. what is the quantum number n of the state of the harmonic oscillator? express the quantum number to three significant figures.

1 answer:

allochka39001 [22]2 years ago

3 0

There are a lot of same examples that you may have worked before, where the mass on a spring uses a classics when it comes to mechanics. So in this system, always put in your mind that there is an enormous quantum standard that one can use in the equation. It should be 2.10x10 raise to a negative sixth. J.

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Your boss asks you to design a room that can be as soundproof as possible and provides you with three samples of material. The o

earnstyle [38]

The correct answer is Option C) Sample C would be best, because the percentage of the energy in an incident wave that remains in a reflected wave from this material is the smallest.

As the coefficient of absorption would define the energy present in the reflected wave, the material C has the highest percentage of absorption i.e. 62% and would be best suitable to make a sound proof room.

4 0

2 years ago

Read 2 more answers

Case 1: A DJ starts up her phonograph player. The turntable accelerates uniformly from rest, and takes t₁ = 11.9 seconds to get

olga_2 [115]

Answer:

Part a)

$\omega = 8.17 rad/s$

Part b)

$N = 7.74 rev$

Part c)

$\alpha = 0.69 rad/s^2$

Part d)

$\alpha = 0.48 rad/s^2$

Part e)

$t = 9.14 s$

Explanation:

Part a)

Angular speed is given as

$\omega = 2\pi f$

$\omega = 2\pi(\frac{78}{60})$

$\omega = 8.17 rad/s$

Part b)

Since turn table is accelerating uniformly

so we will have

$\theta = \frac{\omega_f + \omega_i}{2} t$

$\theta = \frac{8.17 + 0}{2}(11.9)$

$2N\pi = 48.6$

$N = 7.74 rev$

Part c)

angular acceleration is given as

$\alpha = \frac{\omega_f - \omega_i}{t}$

$\alpha = \frac{8.17 - 0}{11.9}$

$\alpha = 0.69 rad/s^2$

Part d)

When its angular speed changes to 120 rpm

then we will have

$\omega_2 = 2\pi (\frac{120}{60})$

$\omega_2 = 12.56 rad/s$

number of turns revolved is 15 times

so we have

$\omega_f^2 - \omega_i^2 = 2 \alpha \theta$

$12.56^2 - 8.17^2 = 2\alpha (2\pi\times 15)$

$\alpha = 0.48 rad/s^2$

Part e)

now for uniform acceleration we have

$\omega_f - \omega_i = \alpha t$

$12.56 - 8.17 = 0.48 t$

$t = 9.14 s$

7 0

2 years ago

Two friends, barbara and neil, are out rollerblading. with respect to the ground, barbara is skating due south at a speed of 5.9

Semmy [17]

<span>As seen by Barbara, Neil is traveling at a velocity of 6.1 m/s at and angle of 76.7 degrees north from due west. Let's assume that both Barbara and Neil start out at coordinate (0,0) and skate for exactly 1 second. Where do they end up? Barbara is going due south at 5.9 m/s, so she's at (0,-5.9) Neil is going due west at 1.4 m/s, so he's at (-1.4,0) Now to see Neil's relative motion to Barbara, compute a translation that will place Barbara back at (0,0) and apply that same translation to Neil. Adding (0,5.9) to their coordinates will do this. So the translated coordinates for Neil is now (-1.4, 5.9) and Barbara is at (0,0). The magnitude of Neil's velocity as seen by Barbara is sqrt((-1.4)^2 + 5.9^2) = sqrt(1.96 + 34.81) = sqrt(36.77) = 6.1 m/s The angle of his vector relative to due west will be atan(5.9/1.4) = atan(4.214285714) = 76.7 degrees So as seen by Barbara, Neil is traveling at a velocity of 6.1 m/s at and angle of 76.7 degrees north from due west.</span>

5 0

2 years ago

Read 2 more answers

A meteoroid is in a circular orbit 600 km above the surface of a distant planet. The planet has the same mass as Earth but has a

Inga [223]

<h2>Answer:</h2><h2>The acceleration of the meteoroid due to the gravitational force exerted by the planet = 12.12 m/ $s^{2}$ </h2>

Explanation:

A meteoroid is in a circular orbit 600 km above the surface of a distant planet.

Mass of the planet = mass of earth = 5.972 x $10^{24}$ Kg

Radius of the earth = 90% of earth radius = 90% 6370 = 5733 km

The acceleration of the meteoroid due to the gravitational force exerted by the planet = ?

By formula, g = $\frac{GM}{r^{2} }$

where g is the acceleration due to the gravity

G is the universal gravitational constant = 6.67 x $10^{-11}$ $m^{3} kg^{-1} s^{-2}$

M is the mass of the planet

r is the radius of the planet

Substituting the values, we get

g = $\frac{(6.67 * 10^{-11}) (5.972 * 10^{24}) }{5733^{2} }$

g = 12.12 m/ $s^{2}$

The acceleration of the meteoroid due to the gravitational force exerted by the planet = 12.12 m/ $s^{2}$

4 0

2 years ago

An 80-kg quarterback jumps straight up in the air right before throwing a 0.43-kg football horizontally at 15 m/s . How fast wil

Oksanka [162]

Answer:

vBxf = 0.08625m/s

Explanation:

This is a problem about the momentum conservation law. The total momentum before equals the total momentum after.

$p_f=p_i$

pf: final momentum

pi: initial momentum

The analysis of the momentum conservation is about a horizontal momentum (x axis). When the quarterback jumps straight up, his horizontal momentum is zero. Then, after the quarterback throw the ball the sum of the momentum of both quarterback and ball must be zero.

Then, you have:

$m_Qv_{Qxi}+m_{Bxi}v_{Bxi}=m_Qv_{Qxf}+m_{Bxf}v_{Bxf}$ (1)

mQ: the mass of the quarterback = 80kg

mB: the mass of the football = 0.43kg

(vQx)i: the horizontal velocity of quarterback before throwing the ball = 0m/s

(vBx)i: the horizontal velocity of football before being thrown = 0m/s

(vQx)f: the horizontal velocity of quarterback after throwing the ball = ?

(vBx)f: the horizontal velocity of football after being thrown = 15 m/s

You replace the values of the variables in the equation (1), and you solve for (vBx)f:

$0\ kgm/s=-(80kg)(v_{Bxf})+(0.46kg)(15m/s)\\\\v_{Bxf}=\frac{(0.46kg)(15m/s)}{80kg}=0.08625\frac{m}{s}$

Where you have taken the speed of the quarterback as negative because is in the negative direction of the x axis.

Hence, the speed of the quarterback after he throws the ball is 0.08625m/s

6 0

2 years ago