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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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(a) How many fringes appear between the first diffraction-envelope minima to either side of the central maximum in a double-slit
Ainat [17]

Answer:

a

The number of fringe is  z  = 3 fringes

b

The  ratio is I = 0.2545I_o

Explanation:

a

 From the question we are told that

        The wavelength is  \lambda = 600 nm

        The distance between the slit is  d = 0.117mm = 0.117 *10^{-3} m

        The width of the slit is  a = 35.7 \mu m = 35.7 *10^{-6}m

let  z be the number of fringes that appear between the first diffraction-envelope minima to either side of the central maximum in a double-slit pattern is  and this mathematically represented as

             z = \frac{d}{a}

Substituting values

             z = \frac{0.117*10^{-3}}{35.7 *10^{-6}}  

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b

   From the question  we are told that the order  of the bright fringe is  n = 3

   Generally the intensity of  a pattern  is mathematically represented as

                 I = I_o cos^2 [\frac{\pi d sin \theta}{\lambda} ][\frac{sin (\pi a sin \frac{\theta}{\lambda } )}{\pi a sin \frac{\theta}{\lambda} } ]

Where I_o is the intensity  of the  central fringe

 And  Generally  sin \theta = \frac{n \lambda }{d}

               I = I_o co^2 [ \frac{\pi (\frac{n \lambda}{d} )}{\lambda} ] [\frac{\frac{sin (\pi a (\frac{n \lambda}{d} ))}{\lambda} }{\frac{\pi a (\frac{n \lambda}{d} )}{\lambda} } ]

               I = I_o cos^2 (n \pi)[\frac{\frac{sin(\pi a (\frac{n \lambda}{d} ))}{\lambda} )}{ \frac{ \pi a (\frac{n \lambda }{d} )}{\lambda} } ]

               I = I_o cos^2 (3 \pi) [\frac{sin (\frac{3 \pi }{6} )}{\frac{3 \pi}{6} } ]

                I = I_o (1)(0.2545)

                  I = 0.2545I_o

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