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wlad13 [49]
4 years ago
5

wo ladybugs of the same mass are on a record that is on a record player. The record player is turned on and it reaches a constan

t velocity. show answer No Attempt If the male ladybug is all the way at the edge and the female ladybug is halfway between the center and the edge, what can be said about their moments of inertia?
Physics
1 answer:
coldgirl [10]4 years ago
8 0

Answer:

The moment of inertia of the male ladybug is 4 times the moment of inertia of the female ladybug.

Explanation:

The moment of inertia of a single-point object rotating about an axis is given by

I=mr^2

where

m is the mass of the object

r is the distance of the object from the axis of rotation

In this problem, we have:

- The mass of the two ladybugs is the same, m_M=m_F

- The distance of the male ladybug from the axis of rotation is twice that of the female ladybug, r_M=2r_F

So, the moment of inertia of the female ladybug is

I_F=m_F r_F^2

While the moment of inertia of the male ladybug is

I_M = m_M r_M^2 = m_F (2r_F)^2=4 m_F r_F^2=4I_F

Therefore, the moment of inertia of the male ladybug is 4 times the moment of inertia of the female ladybug.

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U guys are the best I’m surprised I’m getting all of this right
ella [17]
The answer is B, 1/4 of the cake
5 0
3 years ago
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The force of the pulley is 652 newtons. It accelerates at 53 m/s. How much does the object weigh. a 1230 grams b 1.230 gram c 12
RoseWind [281]

Answer:

d. 12.30 grams

Explanation:

Given the following data;

Force = 652 N

Acceleration = 53 m/s²

To find how much the object weigh, we would need to determine its mass;

Force = mass * acceleration

652 = mass * 53

Mass = 652/53

<em>Mass = 12.30 grams.</em>

<em>Therefore, the object weighs 12.30 grams. </em>

5 0
3 years ago
A wheel is rotating freely at angular speed 420 rev/min on a shaft whose rotational inertia is negligible. A second wheel, initi
Mashcka [7]

Answer:60 rev/min

Explanation:

Given

angular speed of first shaft \omega _1=420\ rev/min

Moment of inertia of second shaft is seven times times the rotational speed of the first i.e. If I is the moment  of inertia of first wheel so moment of inertia of second is 7 I

As there is no external torque therefore angular momentum is conserved

L_1=L_2

I_1\omega _1=I_2\omega _2

I\times (420)=7 I\times (\omega _2)

\omega _2=\frac{420}{7}

\omega _2=60\ rev/min  

8 0
3 years ago
Q11) If you were standing at the top of a building and you dropped a rock.
Dafna1 [17]

Answer:

Part A

The distance travel by the rock is approximately 132.496 m

Part B

The speed when the rock hits the ground is approximately 50.96 m/s

Explanation:

Part A

The question is focused on the kinetics equation of a free falling object

The given parameter is the time it takes the rock to hit the ground, t = 5.2 s

For an object in free fall, we have;

h = 1/2·g·t²

Where;

h = The height from which the object is dropped

g = The acceleration due to gravity ≈ 9.8 m/s²

t = The time taken to travel the distance, h = 5.2 s

∴ h = 1/2 × 9.8 m/s² × (5.2 s)² ≈ 132.496 m

The distance travel by the rock, h ≈ 132.496 m

Part B

The speed, 'v', when the rock hits the ground, is given by the following kinematic equation,

v = g·t

∴ v = 9.8 m/s² × 5.2 s = 50.96 m/s

The speed when the rock hits the ground, v ≈ 50.96 m/s.

8 0
3 years ago
A container explodes and breaks into three fragments that fly off 120° apart from each other, with mass ratios 1: 4: 2. If the f
RSB [31]

Answer:

V₂ = 1.5 m/s

Explanation:

given,

speed of the first piece = 6 m/s

speed of the third piece = 3 m/s

speed of the second fragment = ?

mass ratios = 1 : 4 : 2

fragment break  fly off = 120°

α = β = γ  = 120°

sin α = sin β = sin γ = 0.866

using lammi's theorem

\dfrac{A}{sin\alpha}=\dfrac{B}{sin\beta}=\dfrac{C}{sin\gamma}

A,B and C is momentum of the fragments

\dfrac{m\times 6}{0.866}=\dfrac{4m\times v_2}{0.866}=\dfrac{2m\times 3}{0.866}

4 x V₂ = 2 x 3

V₂ = 1.5 m/s

3 0
3 years ago
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