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Makovka662 [10]
3 years ago
14

The carnival ride has a 2.0m radius and rotates 1.1 times persecond.

Physics
1 answer:
Lady_Fox [76]3 years ago
3 0

Answer:

a) v = 13.8 m / s , b) a = 95.49 m / s² , c) a force that goes to the center of the carnival ride  and d)   μ = 0.10

Explanation:

For this exercise we will use the angular kinematics relationships and the equation that relate this to the linear kinematics

a) reduce the magnitudes to the SI system

     w = 1.1 rev / s (2pi rad / 1rev) = 6.91 rad / s

The equation that relates linear and angular velocity is

     v = w r

     v = 6.91  2

     v = 13.8 m / s

b) centripetal acceleration is given by

     a = v² / r = w² r

     a = 6.91² 2

     a = 95.49 m / s²

c) this acceleration is produced by a force that goes to the center of the carnival ride

d) Here we use Newton's second law

     fr -W = 0

     fr = W

     μ N = mg

Radial shaft

      N = m a

      N = m w² r

      μ m w²  r = m g

      μ = g / w² r

      μ = 9.8 / 6.91² 2

      μ = 0.10

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Answer:

I=2.67\times 10^{-6}\ W/m

Explanation:

Given that,

Frequency of a radio antenna is 1 MHz

Power, P = 21 kW

We need to find the the waves intensity 25 km from the antenna . The object emits intenisty evenly in all direction. It can be given by :

I=\dfrac{P}{4\pi r^2}\\\\I=\dfrac{21\times 10^3}{4\pi (25000)^2}\\\\I=2.67\times 10^{-6}\ W/m

So, the wave intensity 25 km from the antenna is 2.67\times 10^{-6}\ W/m.

5 0
3 years ago
Block A of mass M is on a horizontal surface of negligible friction. An identical block B is attached to block A by a light stri
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Answer:

T’= 4/3 T  

The new tension is 4/3 = 1.33 of the previous tension the answer e

Explanation:

For this problem let's use Newton's second law applied to each body

Body A

X axis

      T = m_A a

Axis y

     N- W_A = 0

Body B

Vertical axis

     W_B - T = m_B a

In the reference system we have selected the direction to the right as positive, therefore the downward movement is also positive. The acceleration of the two bodies must be the same so that the rope cannot tension

We write the equations

    T = m_A a

    W_B –T = M_B a

We solve this system of equations

     m_B g = (m_A + m_B) a

    a = m_B / (m_A + m_B) g

In this initial case

     m_A = M

     m_B = M

     a = M / (1 + 1) M g

     a = ½ g

Let's find the tension

    T = m_A a

    T = M ½ g

    T = ½ M g

Now we change the mass of the second block

    m_B = 2M

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We seek tension for this case

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Let's look for the relationship between the tensions of the two cases

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   T’/ T = 4/3

   T’= 4/3 T

The new tension is 4/3 = 1.33 of the previous tension the answer  e

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3 years ago
The newton is defined as the:
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The Newton is defined as the:

C. Force that can give a 1-kilogram mass an acceleration of 1m/sec squared.

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A 35.0 g bullet strikes a 5.3 kg stationary wooden block and embeds itself in the block. The block and bullet fly off together a
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Answer:

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