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ki77a [65]
3 years ago
7

A ball is thrown with an angle of 25.0 to the horizon with a speed of 18.0 m/s. What are its horizontal and vertical components?

Physics
1 answer:
blondinia [14]3 years ago
6 0
      Best Answer:  <span> vi=24 m/s
θ=43°
x=16m

0) Use this equation because you have x component and not y.
You need time or part c so you solve it like this
x=(vi*cosθ)t
16=(24cos43°)t
t=.911552s=.912s

a) This is a trajectory and you need y so you use this equation.
y=(tanθ)x-(gx^2/(2(vi*cosθ)^2)
y=tan43°*16m-((-9.8*16m)^2)/(2(24*cos43°...
y=14.920m-39.901m
y=-24.9808m= -25.0m

b) Break it down into it's components.
vx=cosθ*vi
vx=cos43°*24
vx=17.5525m/s = 17.5m/s

c) vy= sin 43°*24-.5(-9.8)(.912s)
vy=22.2m/s

I'm not 100% positive on some of these but it should look something like the work above. </span>
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Answer:

F = GMmx/[√(a² + x²)]³

Explanation:

The force dF on the mass element dm of the ring due to the sphere of mass, m at a distance L from the mass element is

dF = GmdM/L²

Since the ring is symmetrical, the vertical components of this force cancel out leaving the horizontal components to add.

So, the horizontal components add from two symmetrically opposite mass elements dM,

Thus, the horizontal component of the force is

dF' = dFcosФ where Ф is the angle between L and the x axis

dF' = GmdMcosФ/L²

L² = a² + x² where a = radius of ring and x = distance of axis of ring from sphere.

L = √(a² + x²)

cosФ = x/L

dF' = GmdMcosФ/L²

dF' = GmdMx/L³

dF' = GmdMx/[√(a² + x²)]³

Integrating both sides we have

∫dF' = ∫GmdMx/[√(a² + x²)]³

∫dF' = Gm∫dMx/[√(a² + x²)]³    ∫dM = M

F = GmMx/[√(a² + x²)]³  

F = GMmx/[√(a² + x²)]³

So, the force due to the sphere of mass m is

F = GMmx/[√(a² + x²)]³

3 0
2 years ago
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Answer:

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A = 0.09 m^2, dB/dt = 0.190 T/s

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e = 0.09 x 0.190 = 0.0171 V

(b)

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A 3.00-kg ball swings rapidly in a complete vertical circle of radius 2.00 m by a light string that is fixed at one end. The bal
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Answer

given,

mass of the ball = 3 kg

swing in vertical circle with radius = 2 m

   work done by the gravity = ?          

   work done by the tension = ?            

Work done by the gravity = - m g Δh            

 Δ h = 2 + 2 = 4 m                                                                

Work done by the gravity =- 3 \times 9.8 \times 4

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work done by gravity is equal to -117.6 J            

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work done by tension is equal to 0 J                          

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Answer:

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