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2 years ago

Which of the following is the human ear able to detect?

Physics

2 answers:

Dmitry_Shevchenko [17]2 years ago

8 0

I just finished the quiz and musical note is correct :)

DaniilM [7]2 years ago

7 0

I believe it would be a musical note

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How much energy has 4×10^10m^3 of water collected in a reservoir at a hight of 100 m from the power house ?What kind of energy i

sveticcg [70]

Answer:

PE = 3.92x10^16J

potential energy

Explanation:

PE = m*g*h

mass of water = 1000kg/m³

(4*10^10m³)*1000kg = 4*10^13kg

PE = (4*10^13kg)*(9.81m/s²)*(100m)

PE = 3.92x10^16J

7 0

3 years ago

60 POINTS ANSWER CORRECTLY

ohaa [14]

Answer:

C ) 1.53

Explanation:

The critical angle of a material is given by the formula

$sin c = \frac{1}{n}$

where

c is the critical angle

n is the refractive index

This formula is valid if the second medium is air (which is the case of the problem).

In this problem, we know the critical angle:

$c=40.8^{\circ}$

Therefore we can rearrange the equation to find the refractive index:

$n=\frac{1}{sin c}=\frac{1}{sin 40.8^{\circ}}=1.53$

5 0

3 years ago

Read 2 more answers

(Figure 1) shows the angular-velocity-versus-time graph for a particle moving in a circle. How many revolutions does the object

Travka [436]

Answer: 10.34

Explanation:

Given

$\omega -t$ graph for a particle is given

angle turned by the particle in radians is given by the area under $\omega -t$ graph

The area is given by

$A=20\times (2-0)+10(4-2)+\dfrac{1}{2}\times (20-10)\times (3-2)\\A=40+20+5=65\ rad$

Revolutions(N) made by the object is given by

$N=\dfrac{65}{2\pi }=10.34$

4 0

2 years ago

The hydrogen stored inside a large weather balloon has a mass of 13.558 g. What is the volume of this balloon if the density of

sergejj [24]

(13.558 gm) · (1 L / 0.089 gm) = 152.34 L (rounded)

(fraction equal to ' 1 ') ^

6 0

3 years ago

Read 2 more answers

A binary star system consists of two equal mass stars that revolve in circular orbits about their center of mass. The period of

Vladimir79 [104]

Answer:

$m = 2.23 \times 10^{-32} kg$

Explanation:

Given data:

PERIOD OF MOTION IS T = 25.5 days

orbital speeds = 220 km/s

we know that

acceleration due to centripetal force is $a = \frac{F}{m} = \frac{V^2}{r}$

Gravitational force $F= \frac{Gm m}{d^2}$

we know that

$v = \frac{2\pi R}{T}$

solving for

$R = \frac{vT}{2\pi}$

$F = \frac{Gm^2}{4(\frac{vT}{2\pi})^2}$

$F = G\times \frac{\pi m}{(vT)^2}$

$a = \frac{v^2}{\frac{vT}{2\pi}}$

$a = \frac{2\pi v}{T}$

we know that

f =ma

$G\times \frac{\pi m}{(vT)^2} = a = \frac{2\pi m v}{T}$

solving for m

$m = \frac{2Tv^3}{\pi G}$

$m = \frac{2\times 25.5 \times 86400 \times 220000^3\ m/s}{\pi \times 6.67\times 10^{-11}}$

$m = 2.23 \times 10^{-32} kg$

5 0

3 years ago