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enot [183]
3 years ago
9

A rock is thrown horizontally from the top of a radio tower lands 17.0 m from the base of the tower. if the speed at which the o

bject was projected was 9.50 m/s, how high is the tower?
Physics
1 answer:
VikaD [51]3 years ago
4 0
For this problem, we use the equations derived from the projectile motion. Particularly, we use the trajectory of the projectile equation written below:

y = xtanθ + gx/2v₀²cosθ
where
y is the vertical height
x is the horizontal distance
θ is the launch angle
g is 9.81 m/s²
v₀ is the initial velocity

Since it was thrown horizontally, θ = 0°. Substituting the values,

y = 17tan0° + (9.81)(17)/2(9.50)²cos0°
y = 0.924 m
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Answer:

F > W * sin(α)

Explanation:

The force needed for the box to start sliding up depends on the incline (α).

The external forces acting on the box would be the weight, the normal reaction and the lifting force that is applied to make it slide up.

These forces can be decomposed on their normal and tangential (to the slide plane) components.

The weight will be split into

Wn = W * cos(α) (in normal direction)

Wt = W * sin(α) (in tangential direction)

The normal reaction will be alligned with the normal axis, and will be equal to -Wn

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To mke the box slide up, a force must be applied, that is opposite to the tangential component of the weight and at least a little larger

F > |-W * sin(α)| (in tangential direction)

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What type of evidence is needed for a hypothesis to be supported or not supported?
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Is a paper clip a conducted or insulator
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A proton is released from rest at the origin in a uniform electric field in the positive x direction with magnitude 850 N/C. Wha
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A proton is released from rest at the origin in a uniform electric field in the positive x direction with magnitude 850 N/C. The change in the electric potential energy of the proton-field system when the proton travels to x = 2.50m is -3.40 × 10⁻¹⁶ J (Option B)

<h3 /><h3>How is the change in electric potential energy of the proton-field system calculated?</h3>

  • Work done on the proton =Negative of the change in the electric potential energy of the proton field
  • In the given case, W = -qΔV
  • -W = qΔV
  • = qEcosθ
  • Therefore, work done on the proton = -e(8.50×10^2 N/C)(2.5m)(1)
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  • Any change in the potential energy indicates the work done by the proton.
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To learn more about electric potential energy, refer

brainly.com/question/14306881

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