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OleMash [197]
3 years ago
8

Now, let's see what happens when the cannon is high above the ground. Click on the cannon, and drag it upward as far as it goes

(15 m above the ground). Set the initial velocity to 14 m/s, and fire several pumpkins while varying the angle. For what angle is the range the greatest?
choices:


A. 45∘

B. 20∘

C. 30∘

D. 40∘

E. 50∘
Physics
1 answer:
kipiarov [429]3 years ago
5 0

Answer:

<h2>B. 20°</h2>

Explanation:

Range in projectile is defined as the distance covered in the horizontal direction. It is expressed as R = U²sin2Ф/g

U is the initial velocity of the body (in m/s)

Ф is the angle of projection

g is the acceleration due to gravity.

Given U = 14m/s, g = 9.8m/s and range R = 15 m

we will substitute this value into the formula to get the projection angle Ф as shown;

15 = 15²sin2Ф/9.8

15*9.8 = 15²sin2Ф

147 = 225sin2Ф

sin2Ф = 147/225

sin2Ф = 0.6533

2Ф = sin⁻¹0.6533

2Ф = 40.79°

Ф = 40.79°/2

Ф = 20.39° ≈ 20°

Hence, the range is greatest at angle 20°

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The speed of a wave is 40 m/s. If the wavelength is 80 centimeters, what is the frequency of the wave?
Vedmedyk [2.9K]
<h2><u>Q</u><u>u</u><u>e</u><u>s</u><u>t</u><u>i</u><u>o</u><u>n</u>:-</h2>

The speed of a wave is 40 m/s. If the wavelength is 80 centimeters, what is the frequency of the wave ?

<h2><u>A</u><u>n</u><u>s</u><u>w</u><u>e</u><u>r</u>:-</h2>

<h3>Given:-</h3>

Velocity (V) = 40 m/s

Wavelength (\lambda) = 80 cm = 0.8 m

<h3>To Find:-</h3>

The frequency (F) of the wave.

<h2>Solution:-</h2>

We know,

\bf V \: = \: F \: × \: \lambda

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F = \frac{40}{0.8}

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