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Arisa [49]
4 years ago
13

Which occurs when two musical instruments use resonance?

Physics
2 answers:
polet [3.4K]4 years ago
8 0
D, they are both loud
likoan [24]4 years ago
5 0
D). they are both loud
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Answer:

C. 1.3 x 10-11 F

I took the test and got this correct, hope this helps

Explanation:

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3 years ago
Varun travels 4 meters towards south and three meters towards east . Find his distance and displacement.
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Explanation:

Distance given to travel in south =4m

Distance given to travel in east=3 m

Total distance would be 4+3=7 M

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Total displacement would be also 7 because he walks in south and east

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4 years ago
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As an adult, what is the best way to remain healthy?
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1. Eat a balanced diet and regular exercise. This is the best way because eating right and exercising are is more beneficial than just exercising.
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Write this number in standard notation.<br> 4.702 x 10–4
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Move the decimal three places to the left -> .0004702 
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You launch a cannonball at an angle of 35° and an initial velocity of 36 m/s (assume y = y₁=
velikii [3]

Answer:

Approximately 4.2\; {\rm s} (assuming that the projectile was launched at angle of 35^{\circ} above the horizon.)

Explanation:

Initial vertical component of velocity:

\begin{aligned}v_{y} &= v\, \sin(35^{\circ}) \\ &= (36\; {\rm m\cdot s^{-1}})\, (\sin(35^{\circ})) \\ &\approx 20.6\; {\rm m\cdot s^{-1}}\end{aligned}.

The question assumed that there is no drag on this projectile. Additionally, the altitude of this projectile just before landing y_{1} is the same as the altitude y_{0} at which this projectile was launched: y_{0} = y_{1}.

Hence, the initial vertical velocity of this projectile would be the exact opposite of the vertical velocity of this projectile right before landing. Since the initial vertical velocity is 20.6\; {\rm m\cdot s^{-1}} (upwards,) the vertical velocity right before landing would be (-20.6\; {\rm m\cdot s^{-1}}) (downwards.) The change in vertical velocity is:

\begin{aligned}\Delta v_{y} &= (-20.6\; {\rm m\cdot s^{-1}}) - (20.6\; {\rm m\cdot s^{-1}}) \\ &= -41.2\; {\rm m\cdot s^{-1}}\end{aligned}.

Since there is no drag on this projectile, the vertical acceleration of this projectile would be g. In other words, a = g = -9.81\; {\rm m\cdot s^{-2}}.

Hence, the time it takes to achieve a (vertical) velocity change of \Delta v_{y} would be:

\begin{aligned} t &= \frac{\Delta v_{y}}{a_{y}} \\ &= \frac{-41.2\; {\rm m\cdot s^{-1}}}{-9.81\; {\rm m\cdot s^{-2}}} \\ &\approx 4.2\; {\rm s} \end{aligned}.

Hence, this projectile would be in the air for approximately 4.2\; {\rm s}.

8 0
2 years ago
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