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Jet001 [13]
4 years ago
14

A block with mass m = 4.5 kg is attached to two springs with spring constants kleft = 36 N/m and kright = 50 N/m. The block is p

ulled a distance x = 0.25 m to the left of its equilibrium position and released from rest. 1)What is the magnitude of the net force on the block (the moment it is released)?
Physics
1 answer:
DaniilM [7]4 years ago
6 0

Answer:

21.5 N

Explanation:

As the block is pulled to a distance of 0.25 m to the left of its equilibrium position, the left spring is compressed by x = 0.25 m and exerting an elastic force to the right, the right spring is stretched by x =0.25 m and also exerting an elastic force to the right. Knowing their spring constant, we can calculate the total net force at that instant

F = F_{left} + F_{right}

F = k_{left}x + k_{right}x

F = x(k_{left} + k_{right}) = 0.25(36 + 50) = 0.25*86 = 21.5 N

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Answer:

Explanation:

1.) What is the net force in the horizontal (x) direction?

Fnet = 8 - 3 = 5 N left

2.) what is the acceleration in the horizontal (x) direction?​

a = Fnet/m = 5/2 = 2.5 m/s² left

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Explain how a thumbtack is designed so that you do not have to use a lot of force to push it through paper or the surface of a b
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Thumbtacks have a flat top which gives you a bigger area to push on, and the end is tapered and sharp to make it easier to push it into a wall, etc.
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(b) Find the position, velocity, and acceleration of the mass at time t = 5π/6.
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Solution 
x(t) = 8 cos t, x(5π/6)= 8 cos(<span>5π/6)
</span>cos(5π/6)=cos(3π/6 + 2π/6 )=cos(π/3 +π/2)= - sin π/3  (cos (x+<span>π/2)= -sinx)
</span>x(t) = -8sin <span>π/3 = - 4 .sqrt3
</span>v(t) = -8sint = -8sin (π/3 +<span>π/2)= -8 cosπ/3 </span>(sin (x+π/2)= cosx)
v(t) =<span> -8 cosπ/3 = -8/2= - 4 
</span>a(5π/6) = - 8cost = -(- sin π/3)= 4 .<span>sqrt3
</span>a(5π/6) =  4 .<span>sqrt3</span>
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A second baseman tosses the ball to the first baseman, who catches it at the same level from which it was thrown. The throw is m
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 <span>(a) 

Taking the angle of the pitch, 37.5°, and the particle's initial velocity, 18.0 ms^-1, we get: 

18.0*cos37.5 = v_x = 14.28 ms^-1, the projectile's horizontal component. 

(b) 
To much the same end do we derive the vertical component: 

18.0*sin37.5 = v_y = 10.96 ms^-1 

Which we then divide by acceleration, a_y, to derive the time till maximal displacement, 

10.96/9.8 = 1.12 s 

Finally, doubling this value should yield the particle's total time with r_y > 0 

<span>2.24 s

I hope my answer has come to your help. Thank you for posting your question here in Brainly.
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(b) Can the speed of a rocket exceed the exhaust speed of the fuel? Explain.
muminat

<u>Yes. The speed of a rocket can exceed the exhaust speed of the fuel.</u>

How this is explained?

  • The thrust of the rocket does not depend on the relative speed of the gases or the relative speed of the rocket.
  • It depends on conservation of momentum.

What is conservation of momentum?

  • Conservation of momentum, general law of physics according to which the quantity called momentum that characterizes motion never changes in an isolated collection of objects; that is, the total momentum of a system remains constant.
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To know more about conservation of momentum, refer:

brainly.com/question/7538238

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