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grandymaker [24]

3 years ago

14

Pendulum takes 1 second to move from x to y so it's frequency equal

1 answer:

lara [203]3 years ago

5 0

if it takes 1 second to move from x to y. You haven't specified if x and y are extreme point or x is extreme and y is equilibrium point

I.If both are extreme points the period is T=1*2=2s

f(frequency)=2π/T(period)

f=6.28/2=3.14Hz

II.If x is an extreme point and y is equilibrium

T=4*1=4s

f=6.28/4=1.57Hz

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If a sample of an unknown material with a mass of 0.68 g and a volume of 0.8 cm3 is dropped

Anna71 [15]

As the density of the unknown substance is 0.68g/0.8ml = 0.85g/ml, it is less dense than the maple syrup at 1.33g/ml and will float.

4 0

3 years ago

A fan at a rock concert is 50.0 m from the stage, and at this point the sound intensity level is 114 dB. Sound is detected when

Marianna [84]

Answer:

A) $P=13.92\ J.s^{-1}$

B) $v=3730.9912\ m.s^{-1}$

C) $v=74.44\ mm.s^{-1}$

D) mosquitoes speed in part B is very much larger than that of part C.

Explanation:

Given:

Distance form the sound source, $s=50\ m$

sound intensity level at the given location, $\beta=114\ dB$

diameter of the eardrum membrane in humans, $d=8.4 \times 10^{-3}\ m$

We have the minimum detectable intensity to the human ears, $I_0=10^{-12}\ W.m^{-2}$

(A)

<u>Now the intensity of the sound at the given location is related mathematically as:</u>

$\beta=10\ log(\frac{I}{I_0} )$ ..........................................(1)

$114=10\ log\ (\frac{I}{10^{-12}} )$

$11.4=log\ I+12\ log\ 10$

$I=0.2512\ W.m^{-2}$

<em>As we know :</em>

$I=\frac{P}{A}$

$0.2512=\frac{P}{\pi\times \frac{8.4^2}{4} }$

$P=13.92\ J.s^{-1}$ is the energy transferred to the eardrums per second.

(B)

mass of mosquito, $m=2\times 10^{-6}\ kg$

<u>Now the velocity of mosquito for the same kinetic energy:</u>

$KE=\frac{1}{2} m.v^2$

$13.92=\frac{1}{2}\times 2\times 10^{-6}\times v^2$

$v=3730.9912\ m.s^{-1}$

(C)

Given:

Sound intensity, $\beta = 20\ dB$

<u>Using eq. (1)</u>

$20=10\ log\ (\frac{I}{10^{-12}} )$

$2=log\ I+12\ log\ 10$

$I=10^{-10}\ W.m^{-2}$

Now, power:

$P=I.A$

$P=10^{-10}\times \pi\times \frac{8.4^2}{4}$

$P=5.54\times 10^{-9}\ J.s^{-1}$

Hence:

$KE=\frac{1}{2} m.v^2$

$5.54\times 10^{-9}=0.5\times 2\times 10^{-6}\times v^2$

$v=0.07444\ m.s^{-1}$

$v=74.44\ mm.s^{-1}$

(D)

mosquitoes speed in part B is very much larger than that of part C.

7 0

3 years ago

An air bubble of volume 20 cm³ is at the bottom of a lake 40 m deep, where the temperature is 4.0°C. The bubble rises to the sur

soldi70 [24.7K]

Answer:

100 cm³

Explanation:

Use ideal gas law:

PV = nRT

where P is absolute pressure, V is volume, n is number of moles, R is ideal gas constant, and T is absolute temperature.

n and R are constant, so:

P₁V₁/T₁ = P₂V₂/T₂

If we say point 1 is at 40m depth and point 2 is at the surface:

P₂ = 1.013×10⁵ Pa

T₂ = 20°C + 273.15 = 293.15 K

P₁ = ρgh + P₂

P₁ = (1000 kg/m³ × 9.8 m/s² × 40 m) + 1.013×10⁵ Pa

P₁ = 4.933×10⁵ Pa

T₁ = 4.0°C + 273.15 = 277.15 K

V₁ = 20 cm³

Plugging in:

(4.933×10⁵ Pa) (20 cm³) / (277.15 K) = (1.013×10⁵ Pa) V₂ / (293.15 K)

V₂ = 103 cm³

Rounding to 1 sig-fig, the bubble's volume at the surface is 100 cm³.

6 0

3 years ago

Find the total electric charge of 2.5 kg of electrons. Express your answer using two significant figures.

zimovet [89]

Answer : The total electric charge of electrons is, $-4.4\times 10^{11}C$

Explanation:

Answer : The number of electrons transferred are, $4.68\times 10^{20}$

Explanation :

First we have to calculate the number of electrons.

Number of electrons = $\frac{\text{Total mass of electrons}}{\text{Mass of one electron}}$

Mass of 1 electron = $9.1\times 10^{-31}kg$

Total mass of electron = 2.5 kg

Number of electrons = $\frac{2.5kg}{9.1\times 10^{-31}kg}$

Number of electrons = $2.75\times 10^{30}$

Now we have to calculate the total electric charge of electrons.

Formula used :

$Q=ne\\\\n=\frac{Q}{e}$

where,

n = number of electrons transferred = $2.75\times 10^{30}$

Q = charge on electrons = ?

e = charge on 1 electron = $-1.602\times 10^{-19}C$

Now put all the given values in the above formula, we get:

$2.75\times 10^{30}=\frac{Q}{-1.602\times 10^{-19}C}$

$Q=-4.4\times 10^{11}C$

Thus, the total electric charge of electrons is, $-4.4\times 10^{11}C$

4 0

3 years ago

PLEASE HELP ASAP!!! CORRECT ANSWER ONLY PLEASE!!!

Helga [31]

The third hour was positive acceleration and the second hour was negative acceleration

7 0

3 years ago