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Yanka [14]
4 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]4 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 person with mass mp = 74 kg stands on a spinning platform disk with a radius of R = 2.31 m and mass md = 183 kg. The disk is i
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Answer:

1) 883 kgm2

2) 532 kgm2

3) 2.99 rad/s

4) 944 J

5) 6.87 m/s2

6) 1.8 rad/s

Explanation:

1)Suppose the spinning platform disk is solid with a uniform distributed mass. Then its moments of inertia is:

I_d = m_dR^2/2 = 183*2.31^2/2 = 488 kgm^2

If we treat the person as a point mass, then the total moment of inertia of the system about the center of the disk when the person stands on the rim of the disk:

I_{rim} = I_d + m_pR^2 = 488 + 74*2.31^2 = 883 kgm^2

2) Similarly, he total moment of inertia of the system about the center of the disk when the person stands at the final location 2/3 of the way toward the center of the disk (1/3 of the radius from the center):

I_{R/3} = I_d + m_p(R/3)^2 = 488 + 74*(2.31/3)^2 = 532 kgm^2

3) Since there's no external force, we can apply the law of momentum conservation to calculate the angular velocity at R/3 from the center:

I_{rim}\omega_{rim} = I_{R/3}\omega_{R/3}

\omega_{R/3} = \frac{I_{rim}\omega_{rim}}{I_{R/3}}

\omega_{R/3} = \frac{883*1.8}{532} = 2.99 rad/s

4)Kinetic energy before:

E_{rim} = I_{rim}\omega_{rim}^2/2 = 883*1.8^2/2 = 1430 J

Kinetic energy after:

E_{R/3} = I_{R/3}\omega_{R/3}^2/2 = 532*2.99^2/2 = 2374 J

So the change in kinetic energy is: 2374 - 1430 = 944 J

5) a_c = \omega_{R/3}^2(R/3) = 2.99^2*(2.31/3) = 6.87 m/s^2

6) If the person now walks back to the rim of the disk, then his final angular speed would be back to the original, which is 1.8 rad/s due to conservation of angular momentum.

3 0
4 years ago
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