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bonufazy [111]
3 years ago
12

A projectile is shot at an angle 45 degrees to the horizontalnear the surface of the earth but in the absence of air resistance.

When it reaches the highest point of its trajectory, its speed is150 m/s. In a second trial with the same projectile, the initialspeed is the same but the angle is now 37 degree with thehorizontal. At its highest point in this trajectory, the velocityof the projectile would be what?
Physics
1 answer:
ivann1987 [24]3 years ago
8 0

Answer:

v₂ = 176.24 m/s

Explanation:

given,

angle of projectile = 45°

speed = v₁ = 150 m/s

for second trail

speed = v₂ = ?

angle of projectile = 37°

maximum height attained formula,

H_{max}= \dfrac{v^2 sin^2(\theta)}{g}

now,

H_{max}= \dfrac{v_1^2 sin^2(\theta_1)}{g}

H_{max}= \dfrac{v_2^2 sin^2(\theta_2)}{g}

now, equating both the equations

\dfrac{v_2^2}{v_1^2}=\dfrac{sin^2(\theta_1)}{sin^2(\theta_2)}

\dfrac{v_2^2}{150^2}=\dfrac{sin^2(45^0)}{sin^2(37^0)}

   v₂² = 31061.79

   v₂ = 176.24 m/s

velocity of projectile would be equal to v₂ = 176.24 m/s

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A 3000-N force gives an object an acceleration of 15 m/s2. The mass of the object is
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<h3>Answer:</h3>

200 kg

<h3>General Formulas and Concepts:</h3>

<u>Math</u>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

  1. Brackets
  2. Parenthesis
  3. Exponents
  4. Multiplication
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  • Left to Right  

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  • Division Property of Equality
  • Addition Property of Equality
  • Subtraction Property of Equality<u> </u>

<u>Physics</u>

<u>Newton's Law of Motions </u>

Newton's 1st Law of Motion: An object at rest remains at rest and an object in motion stays in motion

Newton's 2nd Law of Motion: F = ma (Force is equal to [constant] mass times acceleration)

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<h3>Explanation:</h3>

<u>Step 1: Define</u>

[Given] F = 3000 N

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<u>Step 2: Solve for </u><em><u>m</u></em>

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5 0
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4. Describe how the velocity of an object changes if it undergoes uniformly acceleration motion. Can its direction change?
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Explanation:

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4 0
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