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Alborosie
3 years ago
13

A disk-shaped merry-go-round of radius 2.13 m and mass 175 kg rotates freely with an angular speed of 0.651 rev/s. A 55.4 kg per

son running tangentially to the rim of the merry-go-round at 3.51 m/s jumps onto its rim and holds on. Before jumping on the merry-go-round, the person was moving in the same direction as the merry-go-round's rim.
Calculate the final kinetic energy for this system.
Physics
1 answer:
GREYUIT [131]3 years ago
8 0

Answer:

Explanation:

To find out the angular velocity of merry-go-round after person jumps on it , we shall apply law of conservation of ANGULAR momentum

I₁ ω₁ + I₂ ω₂ = ( I₁  + I₂ ) ω

I₁ is moment of inertia of disk , I₂ moment of inertia of running person , I is the moment of inertia of disk -man system , ω₁ and ω₂ are angular velocity of disc and man .

I₁ = 1/2 mr²

= .5 x 175 x 2.13²

= 396.97 kgm²

I₂ = m r²

= 55.4 x 2.13²

= 251.34 mgm²

ω₁ = .651 rev /s

= .651 x 2π rad /s

ω₂ = tangential velocity of man / radius of disc

= 3.51 / 2.13

= 1.65 rad/s

I₁ ω₁ + I₂ ω₂ = ( I₁  + I₂ ) ω

396.97 x  .651 x 2π + 251.34 x 1.65 = ( 396.97 + 251.34 ) ω

ω = 3.14 rad /s

kinetic energy = 1/2 I ω²

= 3196 J

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a truck drives up a hill with a 15 incline. if the truck has a constant speed of 22m/s, what are the horizontal and vertical com
Kamila [148]

Answer:

speed of truck (v) =  22 m/s ,

angle of hill (Θ) =15°

Find

Vertical component (Fv) = ?

Harizontal component (Fh) =?

               Vertical component (Fh) = V cosΘ

                                                        = 22. cos 15

                                                        = 21.25 m/s.

               Harizontal component (Fv) = V sinΘ

                                                            = 22. sin 15

                                                            = 5.69 m/s.

3 0
3 years ago
Which illustrates the law of conservation of matter? A. producers because they create new matter B. decomposers because they des
Klio2033 [76]

D. compost bins because they recycle matter into a new form

Explanation:

  • The law of conservation of matter is about the creation and how matter is being transferred. According to the law, the matter cannot be destroyed. The matter should always be transferred from one form to another in the universe. There is never destruction of matter happens. There is also one more point to it, as it cannot be destroyed it also cannot be created.
  • Here in the options, option A tells us the creation which is not possible, option B says about the destruction of matter which is not true according to the law, C is about storing the matter which will not happen because its get transferred and D is the correct option because it talks about the recycle/ transfer of matter.
7 0
3 years ago
What is (Fnet3)x, the x-component of the net force exerted by these two charges on a third charge q3
Pachacha [2.7K]

This question is incomplete, the complete question is;

Coulomb's law for the magnitude of the force F between two particles with charges Q and Q' separated by a distance d is

|FI = |QQ'I / d²

where K = 1/4π∈0, and

∈0 = 8.854 × 10⁻¹² C²/(N.m²) is the permittivity of free space.

Consider two point charges located on the x-axis:

one charge, q₁ = -18.5 nC, is located at

x₁ = -1.715m; the charge q₂ = 30.5 nC, is at the origin ( x₂=0 )

What is (Fnet3)x, the x-component of the net force exerted by these two charges on a third charge q₃ = 51.0 nC placed between q₁ and q₂ at x₃ = -1.085 m ?

Answer: (Fnet3)x = -3.3287 × 10⁻⁵ N

Explanation:

Given that;

Q₁ = -18.5 nC       Q₃ = 51 nC        Q₂ = 30.5 nC

x₁ = - 1.715m         x₃ = - 1.085m     x₂ = 0

Now

x - component of Net force on charge Q₃ is

(Fnet3)x = -K|Q₁I|Q₃I / r₁3² - -K|Q₂I|Q₃I / r₂3²

(Fnet3)x = -(9×10⁹)(51×10⁻⁹) [ 18.5 / ((-1.085 + 1.715)²) + (30.5 / (-1.085)² ] × 10⁻⁹

(Fnet3)x = -3.3287 × 10⁻⁵ N

6 0
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Aleonysh [2.5K]

Answer:

r = 0.0548 m

Explanation:

Given that,

Singly charged uranium-238 ions are accelerated through a potential difference of 2.20 kV and enter a uniform magnetic field of 1.90 T directed perpendicular to their velocities.

We need to find the radius of their circular path. The formula for the radius of path is given by :

r=\dfrac{1}{B}\sqrt{\dfrac{2mV}{q}}

m is mass of Singly charged uranium-238 ion, m=3.95\times 10^{-25}\ kg

q is charge

So,

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So, the radius of their circular path is equal to 0.0548 m.

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