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Lady bird [3.3K]
3 years ago
9

1. What is the acceleration of a point on the edge of a 0.30 m diameter grinding wheel rotating at 1600 rpm?

Physics
1 answer:
anygoal [31]3 years ago
8 0

Complete Question

1. What is the linear speed of a point on the edge of a 0.30 m diameter grinding wheel rotating at 1600 rpm?

2. What is the acceleration of the point?

Answer:

1

v  =  25.14 \  m/s

2

a =  4013.5 \  m/s^2

Explanation:

From the question we are told that

    The  diameter is  d =  0.30 \  m

    The  angular speed is  w =  1600 \  rpm  =  \frac{ 1600 * 2 *  \pi }{60 }  =  167.6 \  rad /s

Generally the is mathematically represented as

         r =  \frac{d}{2}  =  \frac{0.30}{2 }  =  0.15 \  m

Generally the linear speed is mathematically represented as

     v  =  wr

     v  =  167.6 * 0.15

      v  =  25.14 \  m/s

Generally the acceleration is mathematically represented as

       a =  \frac{v^2 }{r}

       a =  \frac{25.14^2}{0.15}

         a =  4013.5 \  m/s^2

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3 years ago
Ace Manufacturing ordered 17 boxes with 85 ball bearings each.They also ordered 15 boxes with 90 spring each. How many more ball
photoshop1234 [79]

<u>Answer</u>

95


<u>Explanation</u>

One box has 85 ball bearings.

17 boxes = 17 × 85

               = 1,445 ball bearing


One contains 90 springs

15 boxes ⇒ 15 × 90

               ⇒ 1,350 spring


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3 0
4 years ago
A 1.05 kg block slides with a speed of 0.865 m/s on a frictionless horizontal surface until it encounters a spring with a force
djyliett [7]

Answer:

a) U = 0 J    

k = 0.393 J

E = 0.393 J

b) U = 0.0229J

k = 0.370 J

E = 0.393 J

c) U = 0.0914 J

k = 0.302 J

E = 0.393 J

d) U = 0.206 J

k = 0.187 J

E = 0.393 J

e) U = 0.366 J

k = 0.027 J

E = 0.393 J

Explanation:

Hi there!

The equations of kinetic energy and elastic potential energy are as follows:

k = 1/2 · m · v²

U = 1/2 · ks · x²

Where:

m = mass of the block.

v = velocity.

ks = spring constant.

x = displacement of the string.

a) When the spring is not compressed, the spring potential energy will be zero:

U = 1/2 · ks · x²

U = 1/2 · 457 N/m · (0 cm)²

U = 0 J

The kinetic energy of the block will be:

k = 1/2 · m · v²

k = 1/2 · 1.05 kg · (0.865 m/s)²

k = 0.393 J

The mechanical energy will be:

E = k + U = 0.393 J + 0 J = 0.393 J

This energy will be conserved, i.e., it will remain constant because there is no work done by friction nor by any other dissipative force (like air resistance). This means that the kinetic energy will be converted only into spring potential energy (there is no thermal energy due to friction, for example).

b) The spring potential energy will be:

U = 1/2 · 457 N/m · (0.01 m)²

U = 0.0229 J

Since the mechanical energy has to remain constant, we can use the equation of mechanical energy to obtain the kinetic energy:

E = k + U

0.393 J = k + 0.0229 J

0.393 J - 0.0229 J = k

k = 0.370 J

c) The procedure is now the same. Let´s calculate the spring potential energy with x = 0.02 m.

U = 1/2 · 457 N/m · (0.02 m)²

U = 0.0914 J

Using the equation of mechanical energy:

E = k + U

0.393 J = k + 0.0914 J

k = 0.393 J - 0.0914 J = 0.302 J

d) U = 1/2 · 457 N/m · (0.03 m)²

U = 0.206 J

E = 0.393 J

k = E - U = 0.393 J - 0.206 J

k = 0.187 J

e) U = 1/2 · 457 N/m · (0.04 m)²

U = 0.366 J

E = 0.393 J

k = E - U = 0.393 J - 0.366 J = 0.027 J.

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Answer:

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F^k=MkN

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