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Kipish [7]
3 years ago
12

Three wheels each of radius 1 have their centers at respective verticies of an equilateral triangle of side length 4. A belt is

wrapped continuously around the wheels. Find the length of the belt.
Physics
1 answer:
olya-2409 [2.1K]3 years ago
4 0

Answer:

L  = 6 * ( π + 1 )

Explanation:

The side of the equilateral triangle is  4, and each one of the circles is of radius 1. Then

Triangle vertex   A  B  and C

Trajectory of the belt, beginning in vertex A

1.-First circle  A  one turn

L₁  = 2*π*1   = 2*π

2.-Length between the circle A and B

L₂  = 2

3.- To wrap this circle we need to wrap the circle and to run the belt through the radius twice. The firs over the side AB and the second over the side BC, therefore

L₃ = 2*π*1 + 2*(1)

L₃ = 2*π + 2

4.-Length between two nearest points of circles B and C is 2 and length of the circle in C is 2*π*1. Then

L₄  = 2 + 2*π

Total length of the belt  is:

L  =  L₁ + L₂ + L₃ + L₄

L = 2*π + 2 + (2*π + 2 ) + ( 2 + 2*π )

L  = 6*π + 6

L  = 6 * ( π + 1 )

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The freezing point is the same as the melting point.

If it freezes at -58°C, hence the melting point is also <span>-58°C.</span>
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Assume that the upward direction is positive and the downward direction is negative. What is the ball's velocity (in m/s) when i
LenaWriter [7]

The given question is incomplete. The complete question is as follows.

You throw a ball vertically upward, and as it leaves your hand, its speed is 26.0 m/s.

(a) How high (in m) does it rise above the level where it leaves your hand?

(b) How long (in s) does it take to reach its highest point?

(c) How long (in s) does the ball take to return to the level where it left your hand after it reaches its highest point?

(d) Assume that the upward direction is positive and the downward direction is negative. What is the ball's velocity (in m/s) when it returns to the level where it left your hand? (Indicate the direction with the sign of your answer.)

Explanation:

(a) For maximum height, the formula will be as follows.

           v^{2} = u^{2} + 2as

                 a = v^{2} - 2gh

or,                 h = \frac{v^{2}}{2g}

                        = \frac{(26)^{2}}{2 \times 9.8}

                        = \frac{676}{19.6}

                        = 34.5 m/s

Hence, it rises 34.5 m/s above the level where it leaves your hand.

(b) Time to reach maximum height is as follows.

            v = u + at

or,           v - gt = 0

                 t = \frac{v}{g}

                   = \frac{26}{10}

                   = 2.6 sec

Therefore, it will take 2.6 sec to reach its highest point.

(c)  Time taken by the ball to ascent is equal to the time it has taken to descent.

Therefore, time taken by the ball to return to the level where it left your hand after it reaches its highest point? is also 2.6 sec.

(d)  Speed of the ball will be 26 m/s in the downward direction. Hence, the velocity will be -26 m/s.

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3 years ago
When the three primary pigments are mixed , the resulting is
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You would end up with a brown/black color depending on how much of each pigment was added! Hope this helps.
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A 0.850 kg block is attached to a spring with spring constant 18 N/m . While the block is sitting at rest, a student hits it wit
Sveta_85 [38]

The block's speed at the point where x=0.25A is v = 31.95 cm/s.

<h3>What is Spring constant?</h3>

The spring stiffness is quantified by the spring constant, or k. For various springs and materials, it varies. The stiffer the spring is and the harder it is to stretch, the bigger the spring constant.

question is incomplete, this is the remaining statement

What is the amplitude of the subsequent oscillations? And What is the block's speed at the point where x=0.25A?

x = Asin(wt)

v = Aw coswt

at t = 0

w = sqrt(k/m)

v = Aw

A = v/w

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part b )

E = 1/2mv^2 + 1/2kx^2 = 1/2kA^2

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mv^2 = kA^2 - kA^2/16

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v^2 = 15/16 * (k/m) * A^2

v^2 = 15/16 *w^2A^2

v = sqrt(15/16) * wA

v = 31.95 cm/s

to learn more about spring constant go to -

brainly.com/question/23885190

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