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vova2212 [387]
3 years ago
6

When hung from an ideal spring with spring constant k = 1.5 N/m, it bounces up and down with some frequency ω, if you stop the b

ouncing and let it swing from side to side through a small angle, the frequency of the pendulum is half the bounce frequency ω/2. What is the length of the un–stretched spring l =? Note ω should not be part of your answer.
Physics
1 answer:
Umnica [9.8K]3 years ago
7 0

Answer:

L = ¼ k g / m

Explanation:

This is an interesting exercise, in the first case the spring bounces under its own weight and in the second it oscillates under its own weight.

The first case angular velocity, spring mass system is

    w₁² = k / m

The second case the angular velocity is

    w₂² = L / g

They tell us

    w₂ = ½ w₁

Let's replace and calculate

     √ (L / g) = ½ √ (k / m)

      L / g = ¼ k / m

       

      L = ¼ k g / m

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Explanation:

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7 0
3 years ago
Please help me TT. I need this to be submitted soon TT. Thank you​
Verizon [17]

Answer:

1)

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a) v = √(  ( (-2)(-1.6 × 10^(-16))(3000V) ) / (2.84 × 10^(-20)kg) ) = 5.81227 × 10^3

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3 0
3 years ago
If the amplitude of a sound increases, which statement is true?
Aleksandr-060686 [28]
What are the options?
5 0
3 years ago
A school bus moves at speed of 35mi/he for 20miles. How long will it take for the bus to get to school
lara [203]

Answer:

Time, t = 0.57 hours

Explanation:

It is given that,

Speed of the school bus, v = 35 mi/hr

Distance covered by the bus, d = 20 miles

We need to find the time taken by the bus to get to school. Time taken by the bus is given by :

t = 0.57 hours

So, the time taken by the bus to reach school is 0.57 hours. Hence, this is the required solution.

8 0
2 years ago
An infinite line of charge with charge density λ1 = 0.6 μC/cm is aligned with the y-axis.
Ganezh [65]

Answer:

1440 × 10⁴ N/C

Explanation:

Data provided in the question:

Charge density λ1 = 0.6 μC/cm = 6 × 10⁻⁵ C/m

a = 7.5 cm = 0.075 m

Now,

Electric field due to a line charge at a distance 'a' is given as:

Ex(P) = \frac{\lambda}{2\pi\epsilon_0a}

also,

we know

\frac{1}{4\pi\epsilon_0} = 9 × 10⁹ Nm²/C²

Thus,

we have

Ex(P) = \frac{2}{2}\times\frac{\lambda}{2\pi\epsilon_0a}=\frac{2\lambda}{4\pi\epsilon_0a}

therefore,

Ex(P) = \frac{2\times6\times10^{-5}\times9\times10^9}{0.075}

= 1440 × 10⁴ N/C

8 0
3 years ago
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