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Setler79 [48]
3 years ago
14

Four springs with the following spring constants, 113.0 N/m, 65.0 N/m, 102.0 N/m, and 101.0 N/m are connected in series. What is

their effective spring constant? 004393 Submit Answer Incorrect. Tries 4/99 Previous Tres If a mass of 0.31 kg is attached what will be the frequency of oscillation?
Physics
1 answer:
Llana [10]3 years ago
4 0

Answer:

K_e_q=22.75878093\frac{N}{m}

f=1.363684118Hz

Explanation:

In order to calculate the equivalent spring constant we need to use the next formula:

\frac{1}{K_e_q} =\frac{1}{K_1} +\frac{1}{K_2} +\frac{1}{K_3} +\frac{1}{K_4}

Replacing the data provided:

\frac{1}{K_e_q} =\frac{1}{113} +\frac{1}{65} +\frac{1}{102} +\frac{1}{101}

K_e_q=22.75878093\frac{N}{m}

Finally, to calculate the frequency of oscillation we use this:

f=\frac{1}{2(pi)} \sqrt{\frac{k}{m} }

Replacing m and k:

f=\frac{1}{2(pi)} \sqrt{\frac{22.75878093}{0.31} } =1.363684118Hz

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The speed at which an object falls and the acceleration at which it falls are
allochka39001 [22]

No, the speed at which an object falls is not equal to the acceleration at which it falls.

Answer:

Option B

Explanation:

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6 0
3 years ago
suppose a child walks from the outer edge of a rotating Merry-Go-Round to the inside does angular velocity of the Merry-Go-Round
Akimi4 [234]

Okay here initially child walks on the outer edge of the disc and then started to move inside

So here as the child and Merry go round is an isolated system so there is no external Torque on this system from outside

As here we can see there is external force acting on this system by the hinge of the Merry go round as well as due to gravity so we can not use momentum conservation to solve such type of questions.

But as we can say that there is no external torque on this system about the hinge point so we will use conservation of angular momentum for this system

Here as we know that

\tau = \frac{dL}{dt}

where L = angular momentum

since here torque is ZERO

0 = \frac{dL}{dt}

L = constant

so here we can write initial angular momentum of the system as

L = (I_1 + I_2)*\omega

here we know that

I_1 = moment of inertia of merry go round

I_2 = moment of inertia of child

so here we can say

(I_1 + I_2)* \omega_1 = (I_1 + I_2')\omega_2

so here as the child moves from edge to inside the disc it moment of inertia will decrease because as we know that moment of inertia of child is given as

I_2 = mr^2

here m = mass of child

r = distance of child from center

Since child is moving inside so his distance from center is decreasing

so here moment of inertia of child is decreasing as he starts moving inside

so final angular speed of merry go round will increase as child go inside

\omega_2 = \frac{(I_1 + I_2)*\omega}{(I_1 + I_2')}

so here as

I_2' < I_2

final angular speed will be more than initial speed as child moves inside

3 0
3 years ago
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