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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.
Physics
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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A 4-N object object swings on the end of a string as a simple pendulum. At the bottom of the swing, the tension in the string is
Amanda [17]

Answer:

Explanation:

Given mg = 4N .

m = 4 / g

At the bottom of the swing let centripetal acceleration be a

T - mg = ma

9 - 4 = ma

5 = 4 a  / g

a =  5g / 4

6 0
2 years ago
block of mass 0.5kg on a horizontal surface is attached to a horizontal spring of negligible mass and spring constant 50N/m . Th
Alisiya [41]

Answer:

Explanation:

The mass of the block is 0.5kg

m = 0.5kg.

The spring constant is 50N/m

k =50N/m.

When the spring is stretch to 0.3m

e=0.3m

The spring oscillates from -0.3 to 0.3m

Therefore, amplitude is A=0.3m

Magnitude of acceleration and the direction of the force

The angular frequency (ω) is given as

ω = √(k/m)

ω = √(50/0.5)

ω = √100

ω = 10rad/s

The acceleration of a SHM is given as

a = -ω²A

a = -10²×0.3

a = -30m/s²

Since we need the magnitude of the acceleration,

Then, a = 30m/s²

To know the direction of net force let apply newtons second law

ΣFnet = ma

Fnet = 0.5 × -30

Fnet = -15N

Fnet = -15•i N

The net force is directed to the negative direction of the x -axis

8 0
2 years ago
What units are given to the right of the equals sign
zhuklara [117]
The answer

2y + 14 = 17

The 17 is to the right of the = sign
It is also the answer
7 0
2 years ago
Your town is installing a fountain in the main square. If the water is to rise 26.0 m (85.3 feet) above the fountain, how much p
Brums [2.3K]

Answer:

P = 3.55 \times 10^5 Pa

Explanation:

As we know that water from the fountain will raise to maximum height

H = 26.0 m

now by energy conservation we can say that initial speed of the water just after it moves out will be

\frac{1}{2}mv^2 = mgH

v = \sqrt{2gH}

v = \sqrt{2(9.81)(26)}

v = 22.6 m/s

Now we can use Bernuolli's theorem to find the initial pressure inside the pipe

P = P_0 + \frac{1}{2}\rho v^2

P = 10^5 + \frac{1}{2}(1000)(22.6^2)

P = 3.55 \times 10^5 Pa

6 0
2 years ago
A ball of mass m and radius R is both sliding and spinning on a horizontal surface so that its rotational kinetic energy equals
spin [16.1K]

Answer:

\frac{v_{cm}}{\omega} = 1.122\cdot R

Explanation:

According to the statement of the problems, the following identity exists:

K_{t} = K_{r}

\frac{1}{2}\cdot m \cdot v_{cm}^{2} = 0.63\cdot m \cdot R^{2} \cdot \omega^{2}

After some algebraic handling, the ratio is obtained:

\frac{v_{cm}^{2}}{\omega^{2}}=1.26\cdot R^{2}

\frac{v_{cm}}{\omega} = 1.122\cdot R

4 0
2 years ago
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