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3 years ago

The loudness of sound is determined by what?

2 answers:

Alex3 years ago

6 0

The amplitude of a sound wave determines its loudness or volume. A larger amplitude means a louder sound, and a smaller amplitude means a softer sound. In Figure 1 sound C is louder than sound B. The vibration of a source sets the amplitude of a wave.

baherus [9]3 years ago

5 0

The amplitude of a sound wave determines its loudness or volume. A larger amplitude means a louder sound, and a smaller amplitude means a softer sound

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A leaky 10-kg bucket is lifted from the ground to a height of 10 m at a constant speed with a rope that weighs 0.9 kg/m. Initial

sladkih [1.3K]

Answer:

The work done shall be 14715 Joules

Explanation:

The work done by a force 'F' in a displacement 'dy' is given by

$W=m(y)g\times dy$

At any position 'y' the weight shall be sum of weft of water and weight of string

$\therefore m(y)=m_{water}(y)+m_{string}(y)\\\\m(y)=30(1-\frac{y}{10})+0.9y$

Thus applying values we get

$W=\int m(y)g\times dy\\\\W=\int_{0}^{10}(30(1-\frac{y}{10}+0.9y)\times g)dy\\\\Solving\\W=294.3\int_{0}^{10}(1-\frac{y}{10}+0.9y)dy\\\\W=14715J$

8 0

3 years ago

In diving to a depth of 248 m, an elephant seal also moves 296 m due east of his starting point. What is the magnitude of the se

Vinvika [58]

Answer:

The displacement is 386.16m

Explanation:

A seal dives to a depth of 248m. To find displacement, we must calculate the resultant vectors which will give us the displacement

R= sqrt(vector1+vector2)

Since this is a right angle triangle

R= sqrt(248^2 + 296^2)

R= sqrt(149120)

R= 386.16m

Displacement = 386.16m

4 0

3 years ago

Sound is a type of mechanical wave and must have a medium through which to travel. Sound will move at different speeds when whic

Alex787 [66]

Hey there!

The answer would be B. The sound moves from air to water.

Sound travels through different mediums. It goes fastest in solids, a little slower in liquids, and slowest in air. Sound is a very fast wave, but remember that mediums can differ that. In a vacuum space, there is no sound at all. (ex. outer space)

Hope this helps !

5 0

3 years ago

Mike is standing on the roof of a building looking at the roof of the neighboring building that is 15 meters away and 10 meters

amid [387]

Answer:

Part a)

$t = 1.65 s$

Part b)

$x = 40.4 m$

Since the distance of other building is 15 m so YES it can make it to other building

Part c)

$v = 27.3 m/s$

direction of velocity is given as

$[tex]\theta = 26.35 degree$

Explanation:

Part a)

acceleration due to gravity on this planet is 3/4 times the gravity on earth

So the acceleration due to gravity on this new planet is given as

$a = \frac{3}{4}(9.81)$

$a = 7.36 m/s^2$

now the vertical displacement covered by the canister is given as

$y = 10 m$

now by kinematics we have

$y = \frac{1}{2}gt^2$

$10 = \frac{1}{2}(7.36)t^2$

$t = 1.65 s$

Part b)

Horizontal speed of the canister is given as

$v_x = 24.5 m/s$

now the distance moved by it

$x = v_x t$

$x = 24.5 (1.65)$

$x = 40.4 m$

Since the distance of other building is 15 m so YES it can make it to other building

Part c)

Final velocity in X direction will remains the same

$v_x = 24.5 m/s$

final velocity in Y direction

$v_y = v_i + at$

$v_y = 0 + (7.36)(1.65)$

$v_y = 12.14 m/s$

now magnitude of velocity is given as

$v = \sqrt{v_x^2 + v_y^2}$

$v = \sqrt{24.5^2 + 12.14^2}$

$v = 27.3 m/s$

direction of velocity is given as

$\theta = tan^{-1}\frac{v_y}{v_x}$

$\theta = tan^{-1}\frac{12.14}{24.5}$

$[tex]\theta = 26.35 degree$

6 0

3 years ago

For an object in uniform circular motion, what can you say about the directions of the velocity, acceleration, and net force vec

zavuch27 [327]

Answer: b) The velocity vector is perpendicular to the acceleration vector; the acceleration vector is parallel to the net force vector.

Explanation: A change in velocity creates an acceleration. As the object rotates through the circular path it is constantly changing direction, and hence accelerating, which causes a constant force to act upon the object. This Force acts towards the center of curvature, directly toward the axis of rotation in a direction parallel to the acceleration of the body along the path. Because the object is moving perpendicular to the force, the path followed by the object is a circular one. Hence the velocity of the object is perpendicular to the acceleration.

3 0

3 years ago