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o-na [289]
3 years ago
13

A merry-go-round spins freely when Diego moves quickly to the center along a radius of the merry-go-round. As he does this, it i

s true to say that Group of answer choices the moment of inertia of the system increases and the angular speed increases. the moment of inertia of the system increases and the angular speed decreases. the moment of inertia of the system decreases and the angular speed increases. the moment of inertia of the system decreases and the angular speed decreases.
Physics
1 answer:
kaheart [24]3 years ago
5 0

Answer:

The moment of inertia of the system decreases and the angular speed increases.

Explanation:

This very concept might not seem to be interesting at first, but in combination with the law of the conservation of angular momentum, it can be used to describe many fascinating physical phenomena and predict motion in a wide range of situations.

In other words, the moment of inertia for an object describes its resistance to angular acceleration, accounting for the distribution of mass around its axis of rotation.

Therefore, in the course of this action, it is said that the moment of inertia of the system decreases and the angular speed increases.

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What do all group 2 elements have in common? They all gain two electrons to form a stable outer energy level. They form covalent
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Group elements that will become +2 ions
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3 years ago
Read 2 more answers
Rony fills a bucket with water and whirls it in a vertical circle to demonstrate that the water will not spill from the bucket a
Romashka-Z-Leto [24]

Answer:

0 N, 3.49 m/s

Explanation:

Draw a free body diagram for the bucket at the top of the swing.  There are two forces acting on the bucket: weight and tension, both downwards.

If we take the sum of the forces in the radial direction, where towards the center is positive:

∑F = ma

W + T = m v² / r

The higher the velocity that Rony swings the bucket, the more tension there will be.  The slowest he can swing it is when the tension is 0.

W = m v² / r

mg = m v² / r

g = v² / r

v = √(gr)

Given that r = 1.24 m:

v = √(9.8 m/s² × 1.24 m)

v = 3.49 m/s

8 0
3 years ago
Radiation measured in emissions/sec is called the curie. <br> a. True<br> b. False
iragen [17]
True is the correct anwser


3 0
3 years ago
Problem 20.75 A thin rectangular coil 2 cm by 9 cm has 44 turns of copper wire. It is made to rotate with angular frequency 110
NeTakaya

Answer:

(a) 16.5528 V

(b) 827.64 W

Explanation:

(a)

The formula for maximum emf produced in a coil is given as

E₀ = BANω.................. Equation 1

Where E₀ = maximum emf produced in the coil, B = magnetic Field, A = Area of the rectangular coil, N = number of turns, ω = angular velocity.

Given: B = 1.9 T, N = 44 turns, ω = 110 rad/s, A = 2×9 = 18 cm² = 18/10000 = 0.0018 m²

Substitute into equation 1

E₀ = 1.9(44)(110)(0.0018)

E₀ = 16.5528 V

(b)

Maximum power

P₀ = E₀²/R...................... Equation 2

Where R = Resistance.

Given: R = 50 ohms, E₀ = 16.5528 V

Substitute into equation 2

P₀ = 50(16.5528)

P₀ = 827.64 W

8 0
3 years ago
You will begin with a relatively standard calculation.Consider a concave spherical mirror with a radius of curvature equal to 60
Strike441 [17]

a) 30.0 cm

For any mirror, the radius of curvature is twice the focal length:

r = 2f

where

r is the radius of curvature

f is the focal length

For the mirror in this problem, we have

r = 60.0 cm is the radius of curvature

Therefore, solving the equation above for f, we find its focal length:

f=\frac{r}{2}=\frac{60.0 cm}{2}=30.0 cm

b) 90 cm

The mirror equation is:

\frac{1}{s'}=\frac{1}{f}-\frac{1}{s}

where

s' is the distance of the image from the mirror

f is the focal length

s is the distance of the object from the mirror

For the situation in the problem, we have

f = +30.0 cm is the focal length (positive for a concave mirror)

s = 45.0 cm is the object distance from the mirror

Solving the formula for s', we find

\frac{1}{s'}=\frac{1}{30.0 cm}-\frac{1}{45.0 cm}=0.011 cm^{-1}

s'=\frac{1}{0.011 cm^{-1}}=90 cm

c) -2

The magnification of the mirror is given by

M=-\frac{s'}{s}

where in this problem we have

s' = 90 cm is the image distance

s = 45.0 cm is the object distance

Solving the equation, we find:

M=-\frac{90 cm}{45 cm}=-2

So, the magnification is -2.

d) -12.0 cm

The magnification can also be rewritten as

M=\frac{y'}{y}

where

y' is the height of the image

y is the heigth of the object

In this problem, we know

y = 6.0 cm is the height of the object

M = -2 is the magnification

Solving the equation for y', we find

y'=My=(-2)(6.0 cm)=-12.0 cm

and the negative sign means that the image is inverted.

Part e and f are exactly identical as part b) and c).

6 0
4 years ago
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