Answer:
<em><u>solution</u></em>
<em>3</em><em>0</em><em>8</em><em>=</em><em>2</em><em>0</em><em> </em><em>swings</em><em> </em>
<em> </em><em>?</em><em>:</em><em>:</em><em>:</em><em> </em><em>=</em><em>1</em>
<em>(</em><em> </em><em>3</em><em>0</em><em>8</em><em>×</em><em>1</em><em>)</em><em>÷</em><em>2</em><em>0</em>
<em>3</em><em>0</em><em>8</em><em>÷</em><em>2</em><em>0</em>
<em>1</em><em>5</em><em>4</em><em>÷</em><em>1</em><em>0</em>
<em>=</em><em>1</em><em>5</em><em>.</em><em>4</em>
<em>=</em>15.4
Answer:
330.24 Hz
Explanation:
Given:
Frequency, f = 320 Hz
L1 = 25.8 cm
L2 = 78.4 cm
L3 = 131.1 cm
Let the wavelength be λ
Then, L1 which is the length of the column of air is λ/4.
λ/4 = 25.8 cm
λ = 25.8 × 4 = 103.2 cm = 1.032 m
Then, speed of sound in air is:
v = λ f
⇒ v = 1.032 × 320 Hz
⇒ v = 330.24 m/s
Answer:
<em>The range is 35.35 m</em>
Explanation:
<u>Projectile Motion</u>
It's the type of motion that experiences an object projected near the Earth's surface and moves along a curved path exclusively under the action of gravity.
Being vo the initial speed of the object, θ the initial launch angle, and 
the acceleration of gravity, then the maximum horizontal distance traveled by the object (also called Range) is:
The projectile was launched at an angle of θ=30° with an initial speed vo=20 m/s. Calculating the range:
The range is 35.35 m
