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algol [13]
2 years ago
14

PLease help with this this is pretty hard for me i kinda get it but not fully

Physics
1 answer:
vodka [1.7K]2 years ago
5 0

Explanation:

Force = mass x acceleration

F= ma (formula)

15N = o.5kg x a

Divide both sides by o.5

a = 30 (greater acceleration)

15N = 4.5kg x a

Divide both sides by 4.5

a= 3.33 (great acceleration)

15N = 0.05kg x a

Divide both sides by 0.05

a = 300 (greatest acceleration)

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A ball has a diameter of 3.79 cm and average density of 0.0838 g/cm3.
suter [353]

Answer: 0.258 N

Explanation:

As the density of the object is much less than the density of water, it’s clear that the buoyant force, is greater than the weight of the object, which means that in normal conditions, it would float in water.

So, in order to get the ball submerged in water, we need to add a downward force, that add to the weight, in order to compensate the buoyant force, as follows:

F = Fb – Fg

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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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