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Katarina [22]
3 years ago
6

A large, 34.0 kg bell is hung from a wooden beam so it can swing back and forth with negligible friction. The bell's center of m

ass is 0.70 m below the pivot. The bell's moment of inertia about an axis at the pivot is 18.0 kg*m^2. The clapper is a small, 1.8 kg mass attached to one end of a slender rod that has length L and negligible mass. The other end of the rod is attached to the inside of the bell; the rod can swing freely about the same axis as the bell.
(A) What should bethe length L of the clapper rod for the bell to ring silently, thatis, for the period of oscillation for the bell to equal that forthe clapper?
Physics
1 answer:
Lorico [155]3 years ago
7 0

Answer:

length L of the clapper rod for the bell to ring silently = 0.756m

Explanation:

We are given;

Mass of Bell;m_b = 34 kg

Distance of centre of mass from pivot;d = 0.7m

The bells moment of inertia about an axis at the pivot;I = 18 kg.m²

Mass of clapper;m_c = 1.8 kg

Length of slender rod is L

Now, the formula for period of physical pendulum having small amplitude is given as;

T_b = 2π√(I/mgd)

Where;

I is moment of inertia

m is mass

g is acceleration due to gravity = 9.8 m/s²

d is distance from rotation axis to centre of gravity

Plugging in the relevant values and using mass of bell, we have;

T_b = 2π√(18/(34*9.81*0.7)

T_b = 2π√(18/(34*9.81*0.7)

T_b = 1.745 s

Now, the formula for period for a simple pendulum which is essentially what the clapper rod is would be;

T_c = 2π√(L/g)

Now, we want to find length of clapper L.

Thus, let's make it the subject;

L = g(T_c/2π)²

Now, we are told that for the bell to ring silently, T_b = T_c.

Thus, T_c = 1.745 s.

So,

L = 9.8(1.745/2π)²

L = 0.756m

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Answer:

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Explanation:

Heat transfer by convection is the transfer of heat by fluid transport from one place to another, such that convection takes place when the heat that comes in contact of fluid containing body is moved to other parts of the container by the transporting fluid

Heat is transferred within a fluid medium mainly by convection (movement of heat by the transfer of fluid particles in the medium)

Convection heat transfer is a combination of conduction and advection heat transfer

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2 years ago
A small 16 kilogram canoe is floating downriver at a speed of 4 m/s. What is the canoe's kinetic energy?
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Kinetic Energy,K.E=1/2MV²

mass,m=16kg

velocity,v=4m/s

K.E=1/2×16×4²

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5 0
3 years ago
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A force of 6600 N is exerted on a piston that has an area of 0.010 m2
sveticcg [70]

Answer:

Choice A: approximately 0.015\; \rm m^2, assuming that the two pistons are connected via some confined liquid to form a simple machine.

Explanation:

Assume that the two pistons are connected via some liquid that is confined. Pressure from the first piston:

\displaystyle P_1 = \frac{F_1}{A_1} = \frac{6.600\times 10^3\; \rm N}{1.0\times 10^{-2}\; \rm m^{2}} = 6.6\times 10^{5}\; \rm N \cdot m^{-2}.

By Pascal's Principle, because the first piston exerted a pressure of 6.6\times 10^{5}\; \rm N \cdot m^{-2} on the liquid, the liquid will now exert the same amount of pressure on the walls of the container.

Assume that the second piston is part of that wall. The pressure on the second piston will also be 6.6\times 10^{5}\; \rm N \cdot m^{-2}. In other words:

P_2 = P_1 = 6.6\times 10^{5}\; \rm N \cdot m^{-2}.

To achieve a force of 9.900 \times 10^3\; \rm N, the surface area of the second piston should be:

\displaystyle A_2 = \frac{F_2}{P_2} = \frac{9.900\times 10^{3}\; \rm N}{6.6\times 10^5\; \rm N \cdot m^{-2}} \approx 0.015\; \rm m^{2}.

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3 years ago
A position vector in the first quadrant has an x-component of 18 m and a magnitude of 30 m. What is the value of its y-component
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Answer:

The value is 24meters

Explanation:

Using

r= xi+yj

To get the magnitude of vector x

We say

/r/= √x²+y²

So

30²= √18² + y²

y= √576

Y= 24m

7 0
3 years ago
A capacitor is connected across an ac source. Suppose the frequency of the source is doubled. What happens to the capacitive rea
fgiga [73]

The capacitive reactance is reduced by a factor of 2.

<h3>Calculation:</h3>

We know the capacitive reactance is given as,

Xc = \frac{1}{2\pi fC}

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Xc\\ = capacitive reactance

f = frequency

C = capacitance

It is given that frequency is doubled, i.e.,

f' = 2f

To find,

Xc =?

Xc' = \frac{1}{2\pi f'C}

      = \frac{1}{2\pi 2f C}

      = \frac{1}{2} (\frac{1}{2\pi fC} )\\

Xc' = \frac{1}{2} Xc

Therefore, the capacitive reactance is reduced by a factor of 2.

I understand the question you are looking for is this:

A capacitor is connected across an AC source. Suppose the frequency of the source is doubled. What happens to the capacitive reactant of the inductor?

  1. The capacitive reactance is doubled.
  2. The capacitive reactance is traduced by a factor of 4.
  3. The capacitive reactance remains constant.
  4. The capacitive reactance is quadrupled.
  5. The capacitive reactance is reduced by a factor of 2.

Learn more about capacitive reactance here:

brainly.com/question/23427243

#SPJ4

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