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Sholpan [36]
2 years ago
14

In the popular game "Angry Birds" (refer Figure), for the red bird to hit the green bird with their range 55.0 m, what should be

the launch angle (with respect to the horizontal) of the slingshot if the time of flight is 2.50 s? ​

Physics
1 answer:
Inga [223]2 years ago
4 0

Answer:

Explanation:

What are we to use as gravity in this imaginary world? I will ASSUME 9.81 m/s² and assume no air resistance.

horizontal velocity component

vx = 55.0/2.50 = 22 m/s

vertical initial velocity component

(12.0 - 10.0) = 0 + vy₀(2.50) + ½(-9.81)2.50²

vy₀ = 13.0625 m/s

θ = arctan(vy₀/vx) = arctan(13.0625/22) = 30.6997225... = 30.7°

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What is the current through a 11v bulb with a power of 99w
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9.01amp

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3 years ago
A sound is recorded at 19 decibels. What is the intensity of the sound?
sp2606 [1]

1 \times 10^{-10.1} \mathrm{Wm}^{-2} is the intensity of the sound.

Answer: Option B

<u>Explanation:</u>

The range of sound intensity that people can recognize is so large (including 13 magnitude levels). The intensity of the weakest audible noise is called the hearing threshold. (intensity about 1 \times 10^{-12} \mathrm{Wm}^{-2}). Because it is difficult to imagine numbers in such a large range, it is advisable to use a scale from 0 to 100.

This is the goal of the decibel scale (dB).  Because logarithm has the property of recording a large number and returning a small number, the dB scale is based on a logarithmic scale. The scale is defined so that the hearing threshold has intensity level of sound as 0.

                     \text { Intensity }(d B)=(10 d B) \times \log _{10}\left(\frac{I}{I_{0}}\right)

Where,

I = Intensity of the sound produced

I_{0} = Standard Intensity of sound of 60 decibels = 1 \times 10^{-12} \mathrm{Wm}^{-2}

So for 19 decibels, determine I as follows,

                   19 d B=(10 d B) \times \log _{10}\left(\frac{I}{1 \times 10^{-12} W m^{-2}}\right)

                  \log _{10}\left(\frac{1}{1 \times 10^{-12} \mathrm{Wm}^{-2}}\right)=\frac{19}{10}

                  \log _{10}\left(\frac{1}{1 \times 10^{-12} \mathrm{Wm}^{-2}}\right)=1.9

When log goes to other side, express in 10 to the power of that side value,

                  \left(\frac{I}{1 \times 10^{-12} W m^{-2}}\right)=10^{1.9}

                  I=1 \times 10^{-12} \mathrm{Wm}^{-2} \times 10^{1.9}=1 \times 10^{-12-1.9}=1 \times 10^{-10.1} \mathrm{Wm}^{-2}

5 0
3 years ago
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