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EleoNora [17]
1 year ago
5

 A baseball is thrown at a speed of Vo at an angle of elevation θ. Determine its range (R) in terms of Vo and θ.​

Physics
1 answer:
aksik [14]1 year ago
4 0

The range of baseball thrown at a speed of Vo at an angle of elevation theta is Vo^2sin2theta/g.

<h3>What is range and how it comes out to be so ?</h3>
  1. See we have already studied about projectile in so many instances and in our earlier classes also.
  2. Projectile is simply a thing or an object thrown at a speed by the means of external force under the influence of gravity.
  3. Range is the maximum height that can be attained by the projectile , that is thrown up.
  4. We can also sometimes depict it as the horizontal displacement of the projectile.
  5. Whenever we throw a baseball up at sky , there is a launch point , the distance between the launch point and the end point is range.

To know more about range visit:

brainly.com/question/17016688

#SPJ13

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Answer:

resistor R₂ has the lowest current density

Explanation:

The current density is

          j = I / A

now let's analyze each case

a) R₂ has an area 2A₀ and a length L₀ that R₁

b) R₃ has an area Ao and a length 3L₀ what R₁

we can see that all the area is given in relation to the resistance R₁

 

the current density in R₁ is

         j₁ = I / A₀

the current density in R₂

         j₂ = I / 2A₀

         j₂ 2 = ½ I/A₀

the current density in R₃

         j₃ = I / A₀

         j₂ < j₁ = j₃

therefore resistor R₂ has the lowest current density

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The diagram shows two charged objects, X and Y.<br> Answer<br> X: positive<br> Y: positive
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Which of the following statements is true? A. X-rays have higher frequencies than microwaves. B. UV rays have higher frequencies
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A block with a mass of 55 kg is pulled with a force of 220 N. Assuming a frictionless surface, what will be the acceleration of
dexar [7]

Answer:

B) 4 m/s²

Explanation:

  • Since the surface it's kept on is frictionless, the net external force acting on the block will be 220 N.
  • Mass of the block = 55 kg

\boxed{ \sf{ F= m \times a}}

Here,

  • F is the net external force applied to an object,
  • m is the mass of that object, and
  • a is the acceleration it acquires.

Substituting the respective values for F and m:

==> 220 = 55 × a

==> 220 = 55 × a

<em>dividing the equation by 55</em><em>:</em>

==> 4 = a

Therefore, an object of 55 kg when pulled across a <u>frictionless surface</u> by a force of 220 N accelerates at <u>4</u><u> (m/s)/s</u>

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