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stich3 [128]
3 years ago
12

A 1kg object loses 20 j of gpe as it falls how far does it fall

Physics
1 answer:
jarptica [38.1K]3 years ago
4 0
Gravitational potential energy (GPE) is energy within an object with respect to its height. The equation of gpe is:

GPE = mgh

where:
m = mass of object
g = gravitational constant, 9.81 m/s^2
h = height

For a given gpe of 20 J, the height can be calculated as follows:
*Remember to be consistent with units. Since Joule is derived from SI units, make sure to use SI units for the other variables

GPE = 20 J = 1 (9.81) (h)
h = 2.04 meters
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Consider the following cyclic process carried out in two steps on a gas. Step 1: 44 J of heat is added to the gas, and 20. J of
notka56 [123]

Answer:37 J

Explanation:

Given

Step :1

Heat added Q=44 J

Work done=-20 J

\Delta E_1=Q+W=44-20=24 J

Step :2

Heat added Q=-61 J

work done W_2

\Delta E_2=Q+W_2

\Delta E_2=61+W_2

\Delta E_1+\Delta E_2=0

as the process is cyclic

44-20-61+W_2=0

W_2=37 J

work done in compression is 37 J

3 0
3 years ago
A concert loudspeaker suspended high off the ground emits 34 W of sound power. A small microphone with a 1.0 cm2 area is 44 m fr
rjkz [21]

Answer:

<u>Part A</u>

I = 1.4 mW/m²  

<u>Part B</u>

β = 91.46 dB

Explanation:

<u>Part A</u>

Sound intensity is the power per unit area of sound waves in a direction perpendicular to that area. Sound intensity is also called acoustic intensity.

For a spherical sound wave, the sound intensity is given by;

                                            I = \frac{P}{A}

                                            I = \frac{P}{4\pi r^{2}}

Where;

P is the source of power in watts (W)

I is the intensity of the sound in watt per square meter (W/m2)

r is the distance r away

Given:

P = 34 W,

A = 1.0 cm²

r = 44 m

The sound intensity at the position of the microphone is calculated to be;

                                     I = \frac{34}{4\pi (44)^{2}}

                                     I = \frac{34}{4\pi (44)^{2}}

                                     I = 0.0013975 W/m²

                                 ≈  I = 0.0014 W/m² = 1.4 × 10⁻³ W/m²

                                     I = 1.4 mW/m²

The sound intensity at the position of the microphone is 1.4 mW/m².

<u>Part B</u>

Sound intensity level or acoustic intensity level is the level of the intensity of a sound relative to a reference value.  It is a a logarithmic quantity. It is denoted by β and expressed in nepers, bels, or decibels.

Sound intensity level is calculated as;  

                                    β = 10log_{10}\frac{I}{I_{0}}  dB

Where,

β is the Sound intensity level in decibels (dB)

I is the sound intensity;

I₀ is the reference sound intensity;

By pluging-in, I₀ is 1.0 × 10⁻¹² W/m²

           ∴        β = 10log_{10}\frac{1.4 * 10^{-3} W/m^{2}}{1.0 * 10^{-12} W/m^{2}}

                      β = 10log_{10} (1.4 * 10^{9})

                      β = 91.46 dB

The sound intensity level at the position of the microphone is 91.46 dB.                

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