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dedylja [7]
1 year ago
5

The steel plate is 0.3 m thick and has a density of 7850 kg>m3 . determine the location of its center of mass. also find the

reactions at the pin and roller support.

Physics
1 answer:
cestrela7 [59]1 year ago
3 0

answer  with full explanation is attached below


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A disk of mass m and radius r is initially held at rest just above a larger disk of mass M and radius R that is rotating at angu
Leto [7]

Answer:

Explanation:

Moment of inertia of larger disk   I₁ = 1/2 MR²

Moment of inertia of smaller  disk   I₂ = 1/2 m r ²

Initial angular velocity

We shall apply law of conservation of angular momentum .

initial total momentum = final angular momentum

I₁ X ωi  = ( I₁ + I₂ )ωf

1/2 MR² x ωi = 1/2 ( m r² + MR² ) ωf

ωf =  ωi   / ( 1 + m r²/MR² )

6 0
1 year ago
A train goes up a hill with a 15º incline. If the train has constant speed of 22 m/s, what are the vertical and horizontal compo
Kobotan [32]
Any two-dimensional vector in cartesian (x,y) coordinates can be broken down into individual horizontal and vertical components using trigonometry. If a train goes up a hill with 15 degree incline at a speed of 22 m/s, the horizontal component is 22cos(15)=21.3 m/s and the vertical component is 22sin(15)=5.5 m/s. 
8 0
2 years ago
A child on a playground swing is swinging back and forth (one complete oscillation) once every four seconds, as seen by her fath
Sati [7]

Answer:

A

Explanation:

Solution:-

- According to the law of relativity the relative speed between two moving objects is inversely proportional to the the time taken.

- Ignoring Doppler Effect.

- So if the relative speeds of two objects in motion i.e ( swing and spaceship) are positive then the time frame of reference for both object relative to other other decreases. So in other words if spaceship approaches the swing i.e relative velocity is positive then the time period of oscillation observed would be less than actual i.e less than 4 seconds.

- Similarly, if spaceship moves away from the swing i.e relative velocity is negative then the time period of oscillation observed would be more than actual i.e more than 4 seconds.

4 0
1 year ago
The mass m1 enters from the left with velocity v0 and strikes a mass m2 > m1 which is initially at rest. The collision betwee
enot [183]

Answer:

1. False 2) greater than. 3) less than 4) less than

Explanation:

1)

  • As the collision is perfectly elastic, kinetic energy must be conserved.
  • The expression for the final velocity of the mass m₁, for a perfectly elastic collision, is as follows:

        v_{1f} = v_{10} *\frac{m_{1} -m_{2} }{m_{1} +m_{2}}

  • As it can be seen, as m₁ ≠ m₂, v₁f ≠ 0.

2)

  • As total momentum must be conserved, we can see that as m₂ > m₁, from the equation above the final momentum of m₁ has an opposite sign to the initial one, so the momentum of m₂ must be greater than the initial momentum of m₁, to keep both sides of the equation balanced.

3)    

  • The maximum energy stored in the in the spring is given by the following expression:

       U =\frac{1}{2} *k * A^{2}

  • where A = maximum compression of the spring.
  • This energy is always the sum of the elastic potential energy and the kinetic energy of the mass (in absence of friction).
  • When the spring is in a relaxed state, the speed of the mass is maximum, so, its kinetic energy is maximum too.
  • Just prior to compress the spring, this kinetic energy is the kinetic energy of m₂, immediately after the collision.
  • As total kinetic energy must be conserved, the following condition must be met:

       KE_{10} = KE_{1f}  + KE_{2f}

  • So, it is clear that KE₂f  < KE₁₀
  • Therefore, the maximum energy stored in the spring is less than the initial energy in m₁.

4)

  • As explained above, if total kinetic energy must be conserved:

        KE_{10} = KE_{1f}  + KE_{2f}

  • So as kinetic energy is always positive, KEf₂ < KE₁₀.
4 0
1 year ago
A 0.450 kg soccer ball has a kinetic energy of 119 J.
Anastaziya [24]

Answer:

V is approximately = 23m/s

Explanation:

Kinetic energy = ½ mv²

Where m= mass = 0.450kg

V= velocity =?

K. E = 119J

Therefore

K. E = ½ mv²

Input values given

119= ½ × 0.450 × v²

Multiply both sides by 2

119 ×2  = 2 × 1/2 × 0.450 × v²

238= 0.450v²

Divide both sides by 0.450

238/0.450 = 0.450v²/0.450

v² = 528.89

Square root both sides

Sq rt v² = sq rt 528.89

V = 22.998m/s

V is approximately = 23m/s

I hope this was helpful, please rate as brainliest

8 0
2 years ago
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