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exis [7]
3 years ago
9

Three balls, with masses of 3m,2m and m, are fastened to a massless rod of length L. The rotational inertias about the ledt

Physics
1 answer:
Oduvanchick [21]3 years ago
8 0

I = MR^2

The Attempt at a Solution:::

I total = (3M)(0)^2 + (2M)(L/2)^2 + (M)(L)^2

I total = 3ML^2/2

It says the answer is 3ML^2/4 though.

⛔⛔⛔⛔⛔⛔⛔⛔⛔⛔⛔⛔⛔⛔⛔⛔⛔

mark it as brainliest.... ✌✌✌

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Can someone explain how they got their answer or how I get the change in number? :(
enyata [817]

Answer:

See Below

Explanation:

Okay, I thinkkk what it is asking by what you summarzied for me issss:

They split the total time into four quarters. They then took (for the first quarter) the start time. Then when the first quarter ends and the second quarter starts is the "end" time.

They then subtract the start time of the second quarter from the end time of the first quarter.

I hope this helps, good luck! :D

5 0
2 years ago
Rita is a salon owner. She notices that her salon charged one of her clients, Linda, extra for a service that the clent did not
Aloiza [94]

Answer:

C hope it helps

call the client and inform her that she was incorrectly charged.

7 0
2 years ago
When a mass M hangs from a vertical wire of length L, waves travel on this wire with a speed V. What will be the speed of these
Zigmanuir [339]

Answer:

a)  v = 0.7071 v₀, b) v= v₀, c)  v = 0.577 v₀, d)   v = 1.41 v₀, e)  v = 0.447 v₀

Explanation:

The speed of a wave along an eta string given by the expression

          v = \sqrt{ \frac{T}{ \mu } }

where T is the tension of the string and μ is linear density

a) the mass of the cable is double

          m = 2m₀

let's find the new linear density

          μ = m / l

iinitial density

          μ₀ = m₀ / l

final density

          μ = 2m₀ / lo

          μ = 2 μ₀

we substitute in the equation for the velocity

initial            v₀ = \sqrt{ \frac{T_o}{ \mu_o} }

with the new dough

                    v = \sqrt{ \frac{T_o}{ 2 \mu_o} }

                    v = 1 /√2  \sqrt{ \frac{T_o}{ \mu_o} }

                    v = 1 /√2 v₀

                    v = 0.7071 v₀

b) we double the length of the cable

If the cable also increases its mass, the relationship is maintained

              μ = μ₀

   in this case the speed does not change

c) the cable l = l₀ and m = 3m₀

we look for the density

           μ = 3m₀ / l₀

           μ = 3 m₀/l₀

           μ = 3 μ₀

            v = \sqrt{ \frac{T_o}{ 3 \mu_o} }

            v = 1 /√3  v₀

            v = 0.577 v₀

d) l = 2l₀

            μ = m₀ / 2l₀

            μ = μ₀/ 2

           v = \sqrt{ \frac{T_o}{ \frac{ \mu_o}{2} } }

           v = √2 v₀

            v = 1.41 v₀

e) m = 10m₀ and l = 2l₀

we look for the density

             μ = 10 m₀/2l₀

             μ = 5 μ₀

we look for speed

             v = \sqrt{ \frac{T_o}{5 \mu_o} }

             v = 1 /√5  v₀

             v = 0.447 v₀

5 0
3 years ago
Explain what happens to the pitch of a cell phone ring when the amplitude of a sound wave increases. Justify your reasoning usin
dalvyx [7]
<span>Nothing happens to the pitch of a cell phone ring when the amplitude
of a sound wave increases. 

Pitch and amplitude are both characteristics of a wave, but they're not
connected, and they don't influence each other.</span>
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