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Lorico [155]
2 years ago
15

1. If the angle of the ramp were increased from 30˙ to 45˙, how would this change the weight of the box? Explain.

Physics
1 answer:
yawa3891 [41]2 years ago
4 0

When the angle of the ramp increases, the weight of the box acting perpendicular to the ramp decreases.

<h3>Normal reaction of the box</h3>

The normal reaction of the box is due to weight of the box acting perpendicular to the ramp.

Fn = Wcosθ

<h3>when the angle of the ramp = 30⁰</h3>

Fn = Wcos(30)

Fn = 0.866W

<h3>when the angle of the ramp = 45⁰</h3>

Fn = W x cos(45)

Fn = 0.7071W

Thus, when the angle of the ramp increases, the weight of the box acting perpendicular to the ramp decreases.

Learn more about normal reaction here: brainly.com/question/18292235

#SPJ1

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Answer: h = 3.34 m

Explanation:

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PE = mgh

h = PE/mg = 4.92 / (0.150(9.81)) = 3.34352... ≈3.34 m

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One drawback of dams is that they can flood land upstream from the dam and reduce water flow downstream from the dam. Question 2
vivado [14]

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True

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7 0
3 years ago
Suppose that the sound level of a conversation is initially at an angry 70 db and then drops to a soothing 50 db. assuming that
stepan [7]
Angry sound level = 70 db
Soothing sound level = 50 db
Frequency, f = 500 Hz
Assuming speed of sound = 345 m/s
Density (assumed) = 1.21 kg/m^3
Reference sound intensity, Io = 1*10^-12 w/m^2

Part (a): Initial sound intensity (angry sound)
10log (I/Io) = Sound level
Therefore,
For Ia = 70 db
Ia/(1*10^-12) = 10^(70/10)
Ia = 10^(70/10)*10^-12 = 1*10^-5 W/m^2

Part (b): Final sound intensity (soothing sound)
Is = 50 db
Therefore,
Is = 10^(50/10)*10^-12 = 18*10^-7 W/m^2

Part (c): Initial sound wave amplitude
Now,
I (W/m^2) = 0.5*A^2*density*velocity*4*π^2*frequency^2

Making A the subject;
A = Sqrt [I/(0.5*density*velocity*4π^2*frequency^2)]

Substituting;
A_initial = Sqrt [(1*10^-5)/(0.5*1.21*345*4π^2*500^2)] = 6.97*10^-8 m = 69.7 nm

Part (d): Final sound wave amplitude
A_final = Sqrt [(1*10^-7)/(0.5*1.21*345*4π^2*500^2)] = 6.97*10^-9 m = 6.97 nm
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Answer:

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