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3 years ago

When large amplitude sound vibrates the ear significantly, a _______ results.

2 answers:

7 0

The answer is D. The loudness of a sound, or volume of a sound wave, can be dictated by the abundance of the sound wave. Both pitch and volume can influence how "boisterous" a sound is and how it influences human ears particularly.

kondaur [170]3 years ago

4 0

D loud sound

Hope this helped:)

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A baseball player (mass = 75 kg) is running north towards a base. In order to avoid

Marina86 [1]

it's 1727 +822 just kidding

7 0

2 years ago

On a straight road (taken to be in the x direction) you drive for an hour at 60 km per hour, then quickly speed up to 120 km per

Luda [366]

Answer:

The average velocity is 180 km/hr

Explanation:

Given;

initial velocity, u = 60 km per hour

final velocity, v = 120 km per hour

initial time = 1 hour

final time = 2 hour

Initial position = 60 km/h x 1 hour = 60 km

final position = 120 km/h x 2 hour = 240 km

The average velocity is given by;

$V_{avg} = \frac{Final \ position\ - \ Initial \ position}{final \ time\ - \ initial \ time}\\\\V_{avg} = \frac{240km \ - \ 60km}{2hr\ - \ 1hr} \\\\V_{avg} = \frac{180 \ km}{1hr} \\\\V_{avg}= 180 \ km/hr$

Therefore, the average velocity is 180 km/hr

3 0

3 years ago

A 100-lb child stands on a scale while riding in an elevator. What does the scale read while the elevator slows to stop at the l

Lelechka [254]

Answer: A 100-lb child stands on a scale while riding in an elevator. Then, the scale reading approaches to 100lb, while the elevator slows to stop at the lowest floor

Explanation: To find the correct answer, we need to know more about the apparent weight of a body in a lift.

<h3>What is the apparent weight of a body in a lift?</h3>

Consider a body of mass m kept on a weighing machine in a lift.
The readings on the machine is the force exerted by the body on the machine(action), which is equal to the force exerted by the machine on the body(reaction).
The reaction we get as the weight recorded by the machine, and it is called the apparent weight.

<h3>How to solve the question?</h3>

Here we have given with the actual weight of the body as 100lbs.
This 100lb child is standing on the scale or the weighing machine, when it is riding .
During this condition, the acceleration of the lift is towards downward, and thus, a force of ma .
There is also<em> mg </em>downwards and a normal reaction in the upward direction.
when we equate both the upward force and downward force, we get,

$ma=mg-N\\N=mg-ma$ i.e. during riding the scale reads a weight less than that of actual weight.

When the lift goes slow and stops the lowest floor, then the acceleration will be approaches to zero.

Thus, from the above explanation, it is clear that ,when the elevator moves to the lowest floor slowly and stops, then the apparent weight will become the actual weight.

Learn more about the apparent weight of the body in a lift here:

brainly.com/question/28045397

#SPJ4

7 0

1 year ago

Find the moments of inertia Ix, Iy, I0 for a lamina that occupies the part of the disk x2 y2 ≤ 36 in the first quadrant if the d

Tasya [4]

Answer:

I(x) = 1444×k × ${\pi}$

I(y) = 1444×k × ${\pi}$

I(o) = 3888×k × ${\pi}$

Explanation:

Given data

function = x^2 + y^2 ≤ 36

function = x^2 + y^2 ≤ 6^2

to find out

the moments of inertia Ix, Iy, Io

solution

first we consider the polar coordinate (a,θ)

and polar is directly proportional to a²

so p = k × a²

so that

x = a cosθ

y = a sinθ

dA = adθda

I(x) = ∫y²pdA

take limit 0 to 6 for a and o to $\pi /2$ for θ

I(x) = $\int_{0}^{6}\int_{0}^{\pi/2}$ y²p dA

I(x) = $\int_{0}^{6}\int_{0}^{\pi/2}$ (a sinθ)²(k × a²) adθda

I(x) = k $\int_{0}^{6}a^(5)$ da × $\int_{0}^{\pi/2}$ (sin²θ)dθ

I(x) = k $\int_{0}^{6}a^(5)$ da × $\int_{0}^{\pi/2}$ (1-cos2θ)/2 dθ

I(x) = k $({r}^{6}/6)^(5)_0$ × ${θ/2 - sin2θ/4}^{\pi /2}_0$

I(x) = k × $({6}^{6}/6)$ × ( ${\pi /4} - sin\pi /4)$

I(x) = k × $({6}^{5})$ × ${\pi /4}$

I(x) = 1444×k × ${\pi}$ .....................1

and we can say I(x) = I(y) by the symmetry rule

and here I(o) will be I(x) + I(y) i.e

I(o) = 2 × 1444×k × ${\pi}$

I(o) = 3888×k × ${\pi}$ ......................2

3 0

3 years ago

A car traveling at 60 mph has how much more energy than a car going at 15 mph?

VikaD [51]

Answer:

No, if a car is going faster. The RPM is obviously higher. If that is higher, you can burn through gas and energy much faster. A car going at 15mph would be cruising and wouldn't have to worry too much about burning our your vehicle.

Explanation:

7 0

3 years ago