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Alisiya [41]
3 years ago
6

A bowling ball is far from uniform. Lightweight bowling balls are made of a relatively low-density core surrounded by a thin she

ll with much higher density. A 7.0 lb (3.2 kg) bowling ball has a diameter of 0.216 m; 0.196 m of this is a 1.6 kgcore, surrounded by a 1.6 kg shell. This composition gives the ball a higher moment of inertia than it would have if it were made of a uniform material. Given the importance of the angular motion of the ball as it moves down the alley, this has real consequences for the game.
(a)Model a real bowling ball as a 0.196-m-diameter core with mass 1.6 kg plus a thin 1.6 kg shell with diameter 0.206 m (the average of the inner and outer diameters). What is the total moment of inertia?

Express your answer with the appropriate units.

(b)Find the moment of inertia of a uniform 3.2 kg ball with diameter 0.216 m.

Express your answer with the appropriate units.
Physics
1 answer:
tester [92]3 years ago
7 0

Answer:

a)  I = 1,75 10-² kg m²  and b)  I = 1.49 10⁻² kg m²

Explanation:

The expression for the moment of inertia is

    I = ∫ r² dm

The moment of inertia is a scalar by which an additive magnitude, we can add the moments of inertia of each part of the system, taking into account the axis of rotation.

    I = I core + I shell

The moment of inertia of a solid sphere is

    I sphere = 2/5 MR²

The moment of inertia of a thin spherical shell is

    I shell = 2/3 M R²

a) Let's apply to our system, first to the core of weight 1.6 kg and diameter 0.196m, the radius is half the diameter

     R = d / 2

     R= 0.196 m / 2 = 0.098 m

     I core = 2/5 1.6 0.098²

     I core = 6.147 10-3 kg m²

Let's calculate the moment of inertia of the shell of mass 1.6 kg with a diameter of 0.206 m

    R = 0.206 / 2

    R = 0.103 m

    I shell = 2/3 1.6 0.103²

    I shell = 1,132 10-2 kg m²

The moment of inertia of the ball is the sum of these moments of inertia,

    I = I core + I shell

    I = 6,147 10⁻³ + 1,132 10⁻² = 6,147 10⁻³ + 11.32 10⁻³

    I = 17.47 10⁻³ kg m²

    I = 1,747 10-² kg m²

b) Now the ball is report with mass 3.2kg and diameter 0.216 m

    R = 0.216 / 2

    R = 0.108 m

It is a uniform sphere

    I = 2/5 M R²

    I = 2/5 3.2 0.108²

    I = 1.49 10⁻² kg m²

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Answer:

A physical change

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4 years ago
How much heat is absorbed by a 27g iron skillet when its temperature rises from 13oC to 23oC?
solniwko [45]
The equation for heat would be q = m*Delta T* S

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4 years ago
12. An organ pipe that is 1.75 m long and open at both ends produces sound of
podryga [215]

Answer:

354 m/s

Explanation:

For the second overtune (Third harmonic) of an open pipe,

λ = 2L/3................................ Equation 1

Where L = Length of the open pipe, λ = Wave length.

Given: L = 1.75 m.

Substitute into equation 1

λ = 2(1.75)/3

λ = 1.17 m.

From the question,

V = λf.......................... Equation 2

V = speed of sound in the room, f = frequency

Given: f = 303 Hz.

Substitute into equation 2

V = 1.17(303)

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Tpy6a [65]

Answer:

The time taken by the personal computer to 'surf' the internet is 11.13 h.

Explanation:

The expression for the energy in terms of current and time is as follows;

E= Vit

Here, V is the potential, i is the current and t is the time.

According to question, in doing a load of clothes, a clothes drier uses 18 A of current at 240 V for 41 min.

Convert time from minute to second.

t=\frac{41}{60}s

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Put V= 240 V, t=\frac{41}{60}s and i=18 A in the expression for the energy.

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Calculate the time taken by the computer to 'surf' the internet.

E'=V'i't'

Here, V' is the voltage used by the a personal computer, i' is the current and t' is the time taken by the personal computer.

Put E'= 2937.6 W, V= 120 V and i'= 2.2 A.

2937.6 = (120)(2.2)t'

t'=\frac{2937.6}{264}

t'= 11.13 h

Therefore, the time taken by the personal computer to 'surf' the internet is 11.13 h.

4 0
4 years ago
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