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gtnhenbr [62]
4 years ago
6

25) A 10 kg box is sitting on a shelf 15 feet in the air. How could you increase the potential energy of the box? (check all tha

t apply)  Raise the box to 20 feet.  Lower the box to 10 feet  Increase the mass of the box to 15 kg  Decrease the bass of the box to 5 kg  Drop the box off of the shelf.
Physics
1 answer:
irakobra [83]4 years ago
7 0
Potential energy is the energy possessed by a body or an object while at rest while kinetic energy is the energy while the body is at motion.
Potential energy is given by Mgh where m is the mass of the object, g is the gravitational acceleration (a constant) and h is the height.
Therefore, potential energy varies directly proportional to the mass of the object and the height.
Thus in this case, an increase in the mass of the box by 15 kg and raising the height of the box by 20 feet, will result to an increase in potential energy of the box.
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Although 0 dB is often referred to as the lower threshold of human hearing, it is important to realize that the human ear is not
IRISSAK [1]

Answer:

20 Hz

15.8 times

Explanation:

A

Although the range of frequency for any human's ear is usually said to be between 20 Hz and 20 kHz. And since the question asked for the least intense frequency, that has to be 20 Hz. Essentially the frequency most people perceive the least intense sound is 20 Hz.

B

A 100-Hz sound must be 10^1.2 times or 15.8 times more intense compared to a 1000-Hz sound to be perceived as equal to 60 phons of loudness

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Kaylis [27]

Answer:

 Δt'/ T% = 90.3%

Explanation:

Simple harmonic movement is described by the expression

         x = A cos (wt)

we find the time for the two points of motion

x = - 0.3 A

        -0.3 A = A cos (w t₁)

         w t₁ = cos -1 (-0.3)

         

remember that angles are in radians

        w t₁ = 1.875 rad

x = 0.3 A

        0.3 A = A cos w t₂

        w t₂ = cos -1 (0.3)

         w t₂ = 1,266 rad

         

Now let's calculate the time of a complete period

x= -A

        w t₃ = cos⁻¹ (-1)

        w t₃ = π rad

this angle for the forward movement and the same time for the return movement in the oscillation to the same point, which is the definition of period

         T = 2 t₃

         T = 2π / w     s

now we can calculate the fraction of time in the given time interval

        Δt / T = (t₁ -t₂) / T

        Δt / T = (1,875 - 1,266) / 2pi

        Δt / T = 0.0969

 

This is the fraction for when the mass is from 0 to 0.3, for regions of oscillation of greater amplitude the fraction is

         Δt'/ T = 1 - 0.0969

         Δt '/ T = 0.903

         Δt'/ T% = 90.3%

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