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kifflom [539]
2 years ago
5

A 1640 kg merry-go-round with a radius of 7.50 m accelerates from rest to a rate of 1.00 revolution per 8.00 s. Estimate the mer

ry-go-round as a solid cylinder and determine the net work needed for this acceleration.
Physics
1 answer:
son4ous [18]2 years ago
6 0

Solution :

Given data :

Mass of the merry-go-round, m= 1640 kg

Radius of the merry-go-round, r = 7.50 m

Angular speed, $\omega = \frac{1}{8}$  rev/sec

                             $=\frac{2 \pi \times 7.5}{8}$  rad/sec

                              = 5.89 rad/sec

Therefore, force required,

$F=m.\omega^2.r$

   $$=1640 \times (5.89)^2 \times 7.5  

   = 427126.9 N

Thus, the net work done for the acceleration is given by :

W = F x r

   = 427126.9 x 7.5

   = 3,203,451.75 J

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Three beads are placed on the vertices of an equilateral triangle of side d = 3.4cm. The first bead of mass m1=140gis placed on
Vlad [161]

Answer:

Xcm = 1.95 cm  and Ycm = 1.76 cm

Explanation:

The very useful concept of mass center is

     R cm = 1/M  ∑ m_{i}  r_{i}

Where ri, mi are the mass positions of the bodies from some reference point by selecting and M is the total mass of the body.

Let's look for the total mass

     M = m₁ + m₂ + m₃

     M = 140 + 45 + 85

     M = 270 g

Let's look for the position of each point

Point 1. top vertex, if the triangle has as side d

      R₁ = d / 2 i ^ + d j ^

      R₁ = (1.7 cm i ^ + 3.4 j ^) cm

Point 2. left vertex. What is the origin of the system?

      R₂ = 0

Point 3. Right vertex

      R₃ = d i ^

      R₃ = 3.4 i ^ cm

a) The x component of the massage center

      Xcm = 1 / M (m₁ x₁ + m₂ x₂ + m₃ x₃)

      Xcm = 1 / M (m₁ d / 2 + 0 + m₃ d)

      Xcm = d / M (m₁ / 2 + m₃)

b)   Let's write the mass center component x

      Xcm = 1/270 (1.7 140 + 0 + 3.4 85)

      Xcm = 238/270

      Xcm = 1.95 cm

c) let's find the component and center of mass

     Ycm = 1 / M (m₁ y₁ + m₂ y₂ + m₃ y₃)

    Ycm = 1 / M (m₁ d + 0 + 0)

    Ycm = m₁ / M d

d) let's calculate

    Y cm = 1/270 (140 3.4 + 0 + 0)

    Ycm = 1.76 cm

8 0
3 years ago
If you walk 30 meters forwards, and then turn around and walk 25 meters backwards, what is the distance that you walked? What di
xeze [42]

Given :

Walk in forward direction is 30 m .

Walk in backward direction is 25 m .

To Find :

The distance and displacement .

Solution :

We know , distance is total distance covered and displacement is distance between final and initial position .

So , distance travelled is :

D = 30 + 25 m = 55 m .

Now , we first move 30 m in forward direction and then 25 m in backward direction .

So , displacement is :

D = 30 - 25 m = 5 m .

Therefore , distance and displacement covered is 55 m and 5 m respectively .

Hence , this is the required solution .

5 0
2 years ago
How much power does it take to lift a 30.0 n box 10.0 m high in 5.00 s, if you must apply a 62n force to lift the box?
Illusion [34]
Power is defined as the rate at which the body is doing work:
P=\frac{W}{t}
Work is defined as displacement done by the force times that displacement:
W=F\cdot h
We know that we need 62N to move the box, so when we apply this force along the path of 10m we have done:
W=62N\cdot10m=620J
of work.
Now we just divide that by 5s to get how much power is required:
P=\frac{620J}{5s}=124W
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kakasveta [241]
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2 years ago
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Pachacha [2.7K]

Answer:

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