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Iteru [2.4K]
3 years ago
9

With a quick flick of her wrist, an Ultimate Frisbee player can give a Frisbee an angular velocity of 4.00 revolutions per secon

d. To do this, the player accelerates the Frisbee from rest through an angle of 55.0° before letting it go. (a) What is the Frisbee's angular velocity, in units of rad/s, when the Ultimate Frisbee player releases the Frisbee? rad/s (b) Through what angle, in radians, does the player rotate the Frisbee? rad (C) What is the Frisbee's angular acceleration while the Frisbee player is flicking her wrist? Assume the angular acceleration is constant. rad/s2 (d) How much time does this process take?
Physics
1 answer:
Ad libitum [116K]3 years ago
8 0

Answer:

a) w = 25.1 rad/s, b) θ  = 0.9599 rad , c) α = 328.1 rad/s²  d) t=  0.0765 s

Explanation: Let's work on this exercise with the equations of angular kinematics

a) The angular velocity is

    w = 4.00 rev / s (2π rad / 1 rev)  

     w = 25.1 rad/s

b) let's reduce the angle of degrees to radians

    θ = 55 ° (π rad / 180 °)

    θ  = 0.9599 rad

c) Let's use the initial angular velocity as the system part of the rest is zero

    w² = w₀² + 2 α θ

    α = w² / 2 θ

   α = 25.1²/2 0.9599

   α = 328.1 rad / s²

d)

   w = w₀ + α t

    t = w / α

    t = 25.1 / 328.1

    t=  0.0765 s

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Given that,

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We need to calculate the value \vec{v}\times\vec{B}

(\vec{v}\times\vec{B})=3i+4j+11k

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Using formula of magnetic force

\vec{F}=q(\vec{v}\times\vec{B})

Put the value into the formula

\vec{F}=1.6\times10^{-19}\times(3i+4j+11k)

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3 years ago
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2 years ago
What is the order of magnitude of the distance of Sun to nearest star in meters?
neonofarm [45]

Answer:

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Explanation:

Since we are making an order of magnitude estimate, we will make a series of simplifying assumptions to get an answer that is roughly right.

Let's model the Milky Way galaxy as a disk.

The volume of a disk is:

V

=

π

⋅

r

2

⋅

h

Plugging in our numbers (and assuming that

π

≈

3

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V

=

π

⋅

(

10

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m

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2

⋅

(

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19

m

)

V

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3

×

10

61

m

3

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Now, all we need to do is find how many stars per cubic meter (

ρ

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Let's look at the neighborhood around the Sun. We know that in a sphere with a radius of

4

×

10

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ρ

=

n

V

Using the volume of a sphere

V

=

4

3

π

r

3

ρ

=

1

4

3

π

(

4

×

10

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m

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3

ρ

=

1

256

10

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m

3

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ρ

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n

V

n

=

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V

Plugging in the density of the solar neighborhood and the volume of the Milky Way:

n

=

(

1

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10

−

48

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−

3

)

⋅

(

3

×

10

61

m

3

)

n

=

3

256

10

13

n

=

1

×

10

11

stars (or 100 billion stars)

Is this reasonable? Other estimates say that there are are 100-400 billion stars in the Milky Way. This is exactly what we found.

4 0
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