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Anna35 [415]
3 years ago
11

A spring-mass system has a spring constant k = 383 N/m, length of the spring L = 0.5 m, and the mass attached to it is M = 3.8 k

g. What will be its frequency of oscillation if it is pulled by 5.0 cm from its equilibrium position and released?
Physics
1 answer:
sergiy2304 [10]3 years ago
4 0

Answer:

Frequency = f = 10.0394 (1/s)

Explanation:

The frequency of oscillation of the system is given by the action:

f= √(k/m)

f= system count

k = spring constant

m = mass connected to the spring

Therefore the frequency will be:

f= √(k/m) = √(383(N/m) / (3.8kg))= √( 100.7895 (kg×m/s²)/(kg ) =

= √( 100.7895 (1/s²) = 10.0394 (1/s)

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What is E of a hydrogen atom in the 3p state?
notsponge [240]

Answer:

E=-1.51 eV.

L=\hbar\sqrt{2}

Explanation:

The nth level energy of a hydrogen atom is defined by the formula,

E_{n}=-\frac{13.6}{n^{2} }

Given in the question, the hydrogen atom is in the 3p state.

Then energy of n=3 state is,

E_{n}=-\frac{13.6}{(3)^{2} }\\E_{n}=-1.51eV

Therefore, energy of the hydrogen atom in the 3p state is -1.51 eV.

Now, the value of L can be calculated as,

L=\hbar\sqrt{l(l+1)}

For 3p state, l=1

L=\hbar\sqrt{1(1+1)}\\L=\hbar\sqrt{2}

Therefore, the value of L of a hydrogen atom in 3p state is L=\hbar\sqrt{2}.

4 0
3 years ago
2) A skier stands at rest and begins to ski downhill with an acceleration of 3.0 m/s² {downhill). What is
olga2289 [7]

Answer:

her displacement <em>s=337.5m</em>

Explanation:

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7 0
1 year ago
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A 26 foot ladder is lowered down a vertical wall at a rate of 3 feet per minute. The base of the ladder is sliding away from the
lakkis [162]

Answer:

(i) 7.2 feet per minute.

(ii) No, the rate would be different.

(iii) The rate would be always positive.

(iv) the resultant change would be constant.

(v) 0 feet per min

Explanation:

Let the length of ladder is l, x be the height of the top of the ladder from the ground and y be the length of the bottom of the ladder from the wall,

By making the diagram of this situation,

Applying Pythagoras theorem,

l^2 = x^2 + y^2-----(1)

Differentiating with respect to t ( time ),

0=2x\frac{dx}{dt} + 2y\frac{dy}{dt}  ( l = 26 feet = constant )

\implies 2y\frac{dy}{dt} = -2x\frac{dx}{dt}

\implies \frac{dy}{dt}=-\frac{x}{y}\frac{dx}{dt}

We have,

y = 10, \frac{dx}{dt}= -3\text{ feet per min}

\frac{dy}{dt}=\frac{3x}{10}-----(X)

(i) From equation (1),

26^2 = x^2 + 10^2

676=x^2 + 100

576 = x^2

\implies x = 24\text{ feet}

From equation (X),

\frac{dy}{dt}=\frac{3\times 24}{10}=7.2\text{ feet per min}

(ii) From equation (X),

\frac{dy}{dt}\propto x

Thus, for different value of x the value of \frac{dy}{dt} would be different.

(iii) Since, distance = Positive number,

So, the value of y will always a positive number.

Thus, from equation (X),

The rate would always be a positive.

(iv) The length of the ladder is constant, so, the resultant change would be constant.

i.e. x = increases ⇒ y = decreases

y = decreases ⇒ y = increases

(v) if ladder hit the ground x = 0,

So, from equation (X),

\frac{dy}{dt}=0\text{ feet per min}

3 0
3 years ago
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julia-pushkina [17]

Answer: C

Explanation: Air

Hope this helps! Brainlist Plz?

3 0
3 years ago
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What elements compose the stars and planets in the universe
Savatey [412]

Well hydrogen would be the main element, as a process called nuclear fusion with both helium and hydrogen atoms occurs within stars. And planets are the products of dead stars that have burned through their supplies of hydrogen, helium, and carbon. Planets are a product of this.

6 0
3 years ago
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