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3 years ago

What is the value of Vo in the

Physics

1 answer:

notsponge [240]3 years ago

5 0

Answer: 60m/s

Explanation:

From the diagram:

Θ = 30°

Vertical resolution (y-axis) :

Voy = VoSinΘ

g in the upward direction = negative (-) = - g

Vfinal = 0

Distance (H) traveled along y =

Time taken to reach maximum height :

From v = u + at

0 = usinΘ - gt

gt = usinΘ

t = usinΘ / g

Horizontal resolution:

S = ut + 1/2at^2

Substituting t = usinΘ / g ; Voy = usinΘ

S = (usinΘ × usinΘ / g) - 1/2 g × (usinΘ /g)^2

S = (u^2sin^2Θ / g) - (u^2sin^2Θ / 2g)

S = (u^2sin^2Θ) / 2g

Now if S = maximum height = 45m

Then,

45 = [Vo^2sin^2(30°)] / 2(10)

45 =[ Vo^2 * (0.5)^2] / 20

45 =( Vo^2 * 0.25) / 20

20 * 45 = Vo^2 * 0.25

900 / 0.25 = Vo^2

3600 = Vo^2

Vo = sqrt(3600)

Vo = 60m/s

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A block of mass 2 kg slides down a frictionless ramp of length 1.3 m tilted at an angle 25o to the horizontal. At the bottom of

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Answer:

Diagrams in pictures

Explanation:

Using energy I can get

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So the velocity at the end of the ramp is the squareroot of two times the initial height of the box times the gravity constant.

(H= 1,3m sin25)

V=2,32m/s

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And

X= v t +1/2 a t^2

Knowing v=2,32 m/s and x= 1,3 m

I can get

a= 6,21m/s2

F= m a

I can get the force of the box when it collides with the spring

F= 12, 425 N

The force the spring makes on the box then is

F = -12,425N = -k d

Then the spring's constant is k= 51,75N/m

To make the two diagrams I need the functions of time when the box slows down

I use the same two equations

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And

X= v t + 1/2 a t^2

Being now 2,32 my initial velocity and 0 my final velocity, and my distance 0,24 m.

I get there the time t=0,0689 seconds and the acceleration a= -33,67 m/s2 (negative because it's slowing down).

Then,

V(t)= - 33,67 m/s2 t for time between 0 and 0,689 sec

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for time between 0 and 0,689 sec

Diagrams and equations are in the pictures

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3 years ago