How could you make a sound wave have a lower pitch

1 answer:

3 0

It depends on the length of the item you're making noise with. If you hit the lowest part of the item you get a low pitch. If you hit the top part of the item you get a high pitch.
"AB84"

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Blue bumper car, traveling east with a force of 100 N, runs into a yellow bumper car traveling west with a force of 300 N. Becau

Pani-rosa [81]

Answer:

the blue bumper car

Explanation:

because the yellow bumper car is coming in with much more force

and according to Newton, actions and reactions are equal and opposite, so the yellow bumper car will five out 300N while the blue car will give out only 100N

then it almost because a fight between 300N vs 100N

and obviously, the 300N will win and the 100N will be pushed back

4 0

3 years ago

You and your friends are doing physics experiments on a frozen pond that serves as a frictionless, horizontal surface. Sam, with

Misha Larkins [42]

Answer:

Sam's and Abigail speeds before colliding were a. 12.34 m/s and b. 2.86 m/s, respectively. Their total kinetic energy was diminished by c. 1484.42 J, approximately

Explanation:

By conservation of momentum, we have

$\Delta \vec{p} = 0\\\\m\vec{v}_i = m\vec{v}_f\\m_S v_S\hat{i} + m_A v_A\hat{j} = m_S \times 7 \cos{(40)} \hat{i} + m_S \times 7 \sin{(40)} \hat{j} + m_A \times 9.4 \cos{(18)} \hat{i} - m_A \times 9.4 \sin{(18)} \hat{j}.$

Writing for each direction at a time,

$m_S v_S = 7m_S \cos{(40)} +9.4 m_A \cos{(18)}\\v_S = 7 \cos{(40)} + 9.4 \frac{m_A}{m_S} \cos{(18)} = 7 \cos{(40)} + 9.4\times\frac{57}{73} \cos{(18)} \approx \mathbf{12.34}\\m_A v_A = 7m_S\sin{(40)} - 9.4m_A\sin{(18)}\\v_A = 7\frac{m_S}{m_A}\sin{(40)} - 9.4\sin{(18)} = 7\times\frac{73}{57}\sin{(40)} - 9.4\sin{(18)} \approx \mathbf{2.86}.$

Their kinetic energy changed by

$K_f - K_i = \left( \frac{1}{2}m_s v_{fs}^2 + \frac{1}{2}m_a v_{fa}^2 \right) - \left( \frac{1}{2}m_s v_{is}^2 + \frac{1}{2}m_a v_{ia}^2 \right) = \left( \frac{1}{2}\times 73 \times 7^2 + \frac{1}{2}\times 57 \times 9.4^2 \right) - \left( \frac{1}{2}\times 73 \times 12.34^2 + \frac{1}{2}\times 57 \times 2.86^2 \right) \approx \mathbf{-1484.418 J}.$