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Yanka [14]
3 years ago
13

A gun has a muzzle speed of 90 meters per second. What angle of elevation should be used to hit an object 150 meters away? Negle

ct air resistance and use g=9.8m/sec2 as the acceleration of gravity.
Physics
1 answer:
Artist 52 [7]3 years ago
7 0

Answer:

θ₀ = 84.78° (OR) 5.22°

Explanation:

This situation can be treated as projectile motion. The parameters of this projectile motion are:

R = Range of Projectile = 150 m

V₀ = Launch Speed of Projectile = 90 m/s

g = 9.8 m/s²

θ₀ = Launch angle (OR) Angle of Elevation = ?

The formula for range of a projectile is given as:

R = V₀² Sin 2θ₀/g

Sin 2θ₀ = Rg/V₀²

Sin 2θ₀ = (150 m)(9.8 m/s²)/(90 m/s)²

2θ₀ = Sin⁻¹ (0.18)

θ₀ = 10.45°/2

<u>θ₀ = 5.22°</u>

Also, we know that for the same launch velocity the range will be same for complementary angles. Therefore, another possible value of angle is:

θ₀ = 90° - 5.22°

<u>θ₀ = 84.78°</u>

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Answer:

480J

Explanation:

Using the formula:

Delta U = Q - W

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Delta U: Changes in internal Energy (J)

W:Work (J)

We can plug in the give numbers, Q and W.

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6 0
2 years ago
What does special relativity reveal about the speed of light relative to its source?
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<h3>What makes special relativity so crucial?</h3>

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5 0
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3 years ago
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PtichkaEL [24]

Answer:

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u = 0

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Substituting in the equation,

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3 years ago
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