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polet [3.4K]
2 years ago
12

g A projectile is launched with speed v0 from point A. Determine the launch angle ! which results in the maximum range R up the

incline of angle " (where 0 ≤ " ≤ 90°). Evaluate your results for " = 0, 30°, and 45°
Physics
1 answer:
Svetlanka [38]2 years ago
5 0

Answer:

The range is maximum when the angle of projection is 45 degree.

Explanation:

The formula for the horizontal range of the projectile is given by

R = \frac{u^{2}Sin2\theta }{g}

The range should be maximum if the value of Sin2θ is maximum.

The maximum value of Sin2θ is 1.

It means 2θ = 90

θ = 45

Thus, the range is maximum when the angle of projection is 45 degree.

If the angle of projection is 0 degree

R = 0

It means the horizontal distance covered by the projectile is zero, it can move in vertical direction.

If the angle of projection is 30 degree.

R = \frac{u^{2}Sin60 }{9.8}

R = 0.088u^2

If the angle of projection is 45 degree.

R = \frac{u^{2}Sin90 }{g}

R = u^2 / g

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For the crowbar of the problem, the force applied in input is 40 N, while the force produced in output is equal to the weight of the rock that is lifted, so 400 N. Therefore, the mechanical advantage is
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3 0
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A chemist identifies compounds by identifying bright lines in their spectra. She does so by heating the compounds until they glo
padilas [110]

Answer:

 a)    λ = 189.43 10⁻⁹ m  b)    λ = 269.19 10⁻⁹ m

Explanation:

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Where m corresponds to the diffraction order

Let's use trigonometry to find the breast

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The diffraction spectrum is measured at very small angles, therefore

      tan θ = sin θ / cos θ = sin θ

We replace

      d y / L = m λ

Let's place in the first order m = 1

Let's look for the separation of the lines (d)

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     d = 501 10⁻⁹ 9.95 10⁻² / 15 10⁻²

     d = 332.33 10⁻⁹ m

Now we can look for the wavelength of the other line

     λ  = d y / L

    λ  = 332.33 10⁻⁹ 8.55 10⁻²/15 10⁻²

    λ = 189.43 10⁻⁹ m

Part B

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      λ  = 332.33 10⁻⁹ 12.15 10⁻² / 15 10⁻²

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4 0
2 years ago
A plane initially traveling at 200 m/s due west experiences a 10 m/s head wind coming from the opposite direction. A). What will
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Two balls with charges +Q and +4Q are separated by 3R. Where should you place another charged ball Q0 on the line between the tw
azamat

Answer:

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You want to move a heavy box with mass 30.0 kg across a carpeted floor. You pull hard on one of the edges of the box at an angle
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Answer:

a=5.54m/s^{2}

Explanation:

The net force, F_{net} of the box is expressed as a product of acceleration and mass hence

F_{net}=ma where m is mass and a is acceleration

Making a the subject, a= \frac {F_{net}}{m}

From the attached sketch,  

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Substituting ∑ F_{net} as F_{net} in the equation where we made a the subject

a= \frac {Fcos\theta-F_{f}}{m}

Since we’re given the value of F as 240N, F_{f} as 41.5N, \theta as 30^{o} and mass m as 30kg

a= \frac {240cos30-41.5}{30.0}=\frac {166.346}{30.0}=5.54m/s^{2}

6 0
2 years ago
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