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Andre45 [30]
3 years ago
15

A bicycle wheel rotates counterclockwise at a constant rate. You can use one of the wheel’s spokes to measure the wheel’s angula

r displacement, Delta theta. How is this angular displacement related to the radius of the wheel, r, and the arc length, s, along the circumference of the wheel?
Physics
1 answer:
nevsk [136]3 years ago
4 0

Answer: s=delta/360° ×2πr

Explanation:

The center of the wheel is located.The distance from the centre spoke to the outside wheel is the radius r...the distance between two spokes in the wheels is an arc which forms an angle(theta) at the centre... It is a fraction of the angle formed at the middle to the total angle of the wheel(360° as in circle) to the product of the circumference of the wheel(circle)

Length of arc(s)=theta/360 ×2πr

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A vessel at rest at the origin of an xy coordinate system explodes into three pieces. Just after the explosion, one piece, of ma
andreyandreev [35.5K]

Answer:

Magnitude of the third piece of the vessel after collision is (v√2) m/s

Explanation:

According To Newton's 2nd law, in such a system, momentum is conserved

Momentum before explosion = Momentum after explosion

Total mass of the vessel before explosion = m + m + 3m = 5m

Velocity = (0î + 0j) m/s since it was at rest.

Momentum = 5m (0î + 0j)

Momentum of the first piece after the explosion = m(-vî)

Momentum of the 2nd piece after explosion = m(-vj)

For the momentum of the 3rd piece,

Mass = 3m, let the velocity be (Vₓî + Vᵧj) m/s

Momentum of 3rd piece after collision = 3m (Vₓî + Vᵧj)

Momentum before explosion = Momentum after explosion

5m (0î + 0j) = m(-vî) + m(-vj) + 3m (Vₓî + Vᵧj)

The term m can be factories out,

Grouping the x and y components together,

0î + 0j = (-vî + Vₓî) + (-vj + Vᵧj)

-v + Vₓ = 0, Vₓ = v

-v + Vᵧ = 0, Vᵧ = v

So, the velocity of the 3rd fragment,

(Vₓî + Vᵧj) = (vî + vj) m/s

Magnitude = √(v² + v²) = √(2v²) = v√2 m/s

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4 years ago
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The 2-kg collar is attached to a spring that has an un-stretched length of 3.0 m. If the collar is drawn to point B and releases
Bezzdna [24]

Complete Question

The image for this question is shown on the first uploaded image

Answer:

v =  3.4 \ m/s

Explanation:

From the question we are told that

   The mass of the collar is  m =  2 \ kg

    The original length is  L =  3.0 \ m

     The spring constant is  k =  3.0 \ N/m

     

Generally the extension of the spring  is  mathematically evaluated as

        e =  4 -3 =  1 \ m

Now with Pythagoras theorem we can obtain the length from A to B as

        AB  =  \sqrt{5 ^2  + 4^2}

       AB  = 6.4 \ m

The  extension of the spring at B is  

     e_b  =  6.4 -  3 =  3.4 \ m

According to the law of energy conservation

   The energy stored in the spring at point A +  the kinetic energy of the  spring =  The  energy stored on the spring at B

So

     \frac{1}{2}  *  k * e  +  \frac{1}{2}  * m* v^2  =  \frac{1}{2}  *  k * e_b

substituting values

    \frac{1}{2}  *  3 * 1^2  +  \frac{1}{2}  * 2* v^2  =  \frac{1}{2}  *  3 *  3.4^2

=>   v =  3.4 \ m/s

8 0
4 years ago
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