A wheel has a radius of r = 2.0 m and it rolls down a smooth incline. The height of the incline is h = 8.0 m . What is the angular velocity ω of the wheel at the bottom of the incline?
Express your answer in radians per second.

Answers

Answer 1

The angular velocity of the wheel at the bottom of the incline is 4.429 rad/sec

The angular velocity (ω) of an object is the rate at which the object's angle position is changing in relation to time.

For a wheel attached to an incline angle, the angular velocity can be computed by considering the conservation of energy theorem.

As such the total kinetic energy (K.E) and rotational kinetic energy (R.K.E) at a point is equal to the total potential energy (P.E) at the other point.

i.e.

P.E = K.E + R.K.E

[tex]\mathbf{mgh = \dfrac{1}{2}m(r \times \omega)^2 + \dfrac{1}{2}\times I \times \omega^2}[/tex]

[tex]\mathbf{gh = \dfrac{1}{2}(r \times \omega)^2 + \dfrac{1}{2}\times r^2 \times \omega^2}[/tex]

[tex]\mathbf{2 \times \dfrac{gh}{r^2} =\omega^2 + \omega^2}[/tex]

[tex]\mathbf{2 \omega^2=2 \times \dfrac{9.81 \times 8 m }{2.0 ^2} }[/tex]

[tex]\mathbf{\omega^2=\dfrac{39.24 }{2}}[/tex]

[tex]\mathbf{\omega=\sqrt{19.62 } \ rad/sec}[/tex]

[tex]\mathbf{\omega=4.429 \ rad/sec}[/tex]

Therefore, we can conclude that the angular velocity of the wheel at the bottom of the incline is 4.429 rad/sec

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Answer 2

The angular velocity of the wheel depends on the mass, radius and the

mode of rotation of the wheel (with or without slipping).

The angle velocity at the bottom of the incline, ω ≈ 4.43 rad/sec

Reasons:

The given parameters are;

Radius of the wheel, r = 2.0 m

Height of the incline, h = 8.0 m

Required:

Angular velocity of the wheel at the bottom of the incline.

Solution:

The potential energy of the wheel at the top of the hill, P.E. = m·g·h

[tex]Sum \ of \ the \ kinetic \ energy \ of \ the \ wheel, \ K.E. = \mathbf{\displaystyle \frac{1}{2} \cdot m \cdot v^2 + \frac{1}{2} \cdot I \cdot \omega ^2}[/tex]

Where;

v = The translational velocity of the wheel = ω·r

I = The moment of inertia of the wheel = m·r²

Therefore'

[tex]Sum \ of \ K.E. = \displaystyle \frac{1}{2} \cdot m \cdot (\omega \cdot r)^2 + \frac{1}{2} \cdot m \cdot r^2 \cdot \omega ^2 = \mathbf{m \cdot r^2 \cdot \omega^2}[/tex]

At the bottom of the hill, the potential energy is converted to kinetic energy

Therefore;

P.E. = Sum of K.E.

m·g·h = m·r²·ω²

g·h = r²·ω²

[tex]\displaystyle \omega = \sqrt{ \frac{g \cdot h}{r^2} } = \mathbf{ \frac{\sqrt{g \cdot h} }{r}}[/tex]

Where;

g = Acceleration due to gravity ≈ 9.81 m/s²

Therefore;

[tex]\displaystyle \omega = \frac{\sqrt{9.81 \times 8} }{2} \approx \mathbf{ 4.43}[/tex]

The angular velocity of the of the wheel at the bottom of the incline, ω ≈ 4.43 rad/sec

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A Wheel Has A Radius Of R = 2.0 M And It Rolls Down A Smooth Incline. The Height Of The Incline Is H

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The half-life of the element Lokium is 4

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Answer:

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plz help ASAP I'll mark as brainliest ​

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Hi there!

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Answer:

Explanation:

The word "sheet" implies that the copper is quite thin.

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Repeat the dilution process until the mix is essentially sand and water, then drive the remaining water from the sand by drying.

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Crazy Wally Ok Ok ok hhahahaha

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Explanation:

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The relationship between the potential and the electric field allows to find the results for the value of the electric field as a function of the distance is:

In the attachment we see the graph of the electric field as a function of distance.

Electric potential is defined by the change in potential energy of a test charge between two points, between the value of the test charge.

          dV = - E . ds

          E = [tex]- \frac{dV}{ds} \ \hat s[/tex]  

Where the bold letters indicate vectors, V is the potential difference, E the electric field and s the path.

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        E = 0

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       V-V₀ = m (x- x₀)

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      E = - m i ^

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       [tex]m= \frac{V_f - V_o}{x_f- x_o}[/tex]  

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       [tex]m= \frac{-2 -2}{4-2} = \ -2 \ V/m[/tex]  

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       E =  [tex]2 \hat i \ V/m[/tex]  

         

3) From x₀ = 4 to x_f = 4.5 m, the potential varies from V₀ = -2 to V_f = 0.

We look for the equation of the line and we derive.

      E = - m i ^

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      [tex]m = \frac{0-(-2)}{4.5-4} = \ 4 V/m[/tex]  

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      E = 0

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We look for the equation of the line and we derive.

       E = - m i ^

       [tex]m = \frac{1-0}{8-6} = \ 0.5 \ V/m[/tex]  

      E = - 0.5 [tex]\hat i[/tex] V/m

6) From x₀ = 8m to x_f = 9m, the potential changes linearly from V₀ = 1 V to V_f = -1.

We look for the equation of the line and we derive.

       E = - m i ^

       [tex]m = \frac{-1-1}{9-8} = \ -2 \ V/m[/tex]

Let's substitute.

       E = 2 [tex]\hat i[/tex] V/m

7) From x₀ = 9m to x_f = 10 m, the potential changes linearly from V₀ = -1 V to V_f = -2 V

     

We look for the equation of the line and we derive.

       E = - m i ^

       [tex]m = \frac{-2+1}{10-9} = \ -1 \ V/m[/tex]

Let's substitute.

       E = 1 [tex]\hat i[/tex]  V/m

In the attachment we can see these Electric fields as a function of distance.

In conclusion, the relationship between the potential and the electric field we can find the results for the value of the electric field as a function of the distance is:

In the attachment we see the graph of the electric field as a function of distance.

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a)What angle θA, where 0∘≤θA<360∘, does A⃗ make with the +x-axis?
Express your answer in degrees.

b)What angle θB, where 0∘≤θB<360∘, does B⃗ make with the +x-axis?
Express your answer in degrees.

c)Vector C⃗ is the sum of A⃗ and B⃗ , so C⃗ =A⃗ +B⃗ . What angle θC, where 0∘≤θC<360∘, does C⃗ make with the +x-axis?
Express your answer in degrees.

Answers

Answer:

Explanation:

We can subtract directly the corresponding components and check using the parallelogram rule.

Explanation:

Have a look:

enter image source here

Where, graphically, I used the fact that:

A

B

=

A

+

(

B

)

For the magnitude we use Pythagoras (with the components) to get:

A

B

=

(

1

)

2

+

(

5

)

2

=

1

+

25

=

26

5.1

For the direction I can see that will be

90

from the

x

axis up to the

y

axis, plus the little bit passed the

y

axis given as:

θ

=

arctan

(

1

5

)

=

11.3

giving in total: angle

=

90

+

11.3

=

101.3

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Answers

Answer:

lithosphere

Explanation:

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Question

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Answer

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Answer:

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