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zloy xaker [14]

3 years ago

8

We know the frequency range of certain sounds are: 400-560 Hz, what are the ranges of wavelength in meters when the signal trans

mits in air?

1 answer:

Ksivusya [100]3 years ago

8 0

Answer:

Range of wavelength will be $5.35\times 10^5m$ to $7.5\times 10^5m$

Explanation:

We have given range of frequency is 400-560 Hz

Speed of the light $c=3\times 10^8m/sec$

We have to find the range of the wavelength of signal transmitted

Ween know that velocity is given by $v=\lambda f$ , here $\lambda$ is wavelength and f is frequency

So for 400 Hz frequency wavelength will be $\lambda =\frac{c}{f}=\frac{3\times 10^8}{400}=7.5\times 10^5m$

And wavelength for frequency 560 Hz $\lambda =\frac{c}{f}=\frac{3\times 10^8}{560}=5.35\times 10^5m$

So range of wavelength will be $5.35\times 10^5m$ to $7.5\times 10^5m$

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3. A concrete highway is built of slabs 12 m long (20o C). How wide should the expansion cracks between the slabs be (at 20o C)

lesantik [10]

Answer:

$\Delta x=2.304\times 10^{-2}\ m=2.304\ cm$ is the minimum gap between the slabs

Explanation:

Given:

length of the concrete highway, $l=12\ m$

coefficient of thermal expansion, $\alpha=12\times 10^{-6}\ ^{\circ}C^{-1}$

range of temperature variation, $(-30^{\circ}C\ to\ 50^{\circ}C) \Rightarrow \Delta T=80^{\circ}C$

<u>Now form the equation of thermal expansion:</u>

$\Delta l=l\times \alpha\times \Delta T$

$\Delta l=12\times (12\times 10^{-6})\times 80$

$\Delta l=1.152\times 10^{-2}\ m$

Since each slab of the highway expands by the above length so the minimum gap between the slabs to prevent buckling:

$\Delta x=2\times \Delta l$

$\Delta x=2\times( 1.152\times 10^{-2})$

$\Delta x=2.304\times 10^{-2}\ m=2.304\ cm$ is the minimum gap between the slabs

4 0

3 years ago

Hoover Dam on the Colorado River is the highest dam in the United States at 221m, with a power output of 680 MW. The dam generat

DiKsa [7]

Answer:

The power in this flow is $9.56\times10^{8}\ W$

Explanation:

Given that,

Distance = 221 m

Power output = 680 MW

Height =150 m

Average flow rate = 650 m³/s

Suppose we need to calculate the power in this flow in watt

We need to calculate the pressure

Using formula of pressure

$Pressure=\rho g h$

Where, $\rho$ = density

h = height

g = acceleration due to gravity

Put the value into the formula

$Pressure=1000\times9.8\times150$

$Pressure=1470000\ Nm^2$

We need to calculate the power

Using formula of power

$P=Pressure\times flow\ rate$

Put the value into the formula

$P=1470000\times650$

$P=9.56\times10^{8}\ W$

Hence, The power in this flow is $9.56\times10^{8}\ W$

5 0

3 years ago

How does a force fields work how do you make your own force field

Irina18 [472]

A force field works the same as a shield. I protects what ever is inside, or behind the field.

7 0

3 years ago

A briefcase sits stationary in an elevator. The mass of the briefcase if 4.5 kg. The elevator then begins accelerating upwards a

Korvikt [17]

Answer:

D. 48.985 N

Explanation:

Newton's second law states that:

$\sum F = ma$

which means that the net force acting on an object is equal to the product between the object's mass and its acceleration.

The equation of the forces for the briefcase in the elevator therefore is given by:

$N-mg=ma$

where

N is the normal reaction exerted on the briefcase

(mg) is the weight of the briefcase, with

m = 4.5 kg being its mass

g = 9.8 m/s^2 is the acceleration of gravity

a = 1.10 m/s^2 is the acceleration

Here we chose upward as positive direction.

Solving for N, we find the normal force:

$N=mg+ma=m(g+a)=(4.5)(9.81+1.10)=49.095 N$

So the closest answer is

D. 48.985 N

3 0

3 years ago

A tennis ball of mass 44.0 g is held just above a basketball of mass 594 g. With their centers vertically aligned, both are rele

ZanzabumX [31]

Answer:

u = 4.6 m/s

h = 8.01 m

Explanation:

Given:

Mass of the tennis ball, m = 44.0 g

Mass of the basket ball, M = 594 g

Height of fall, h = 1.08m

Now,

we have

$u^2-u'^2 = 2as$

where, s = distance = h

a = acceleration

u = final speed before the collision

u' = initial speed

since it is free fall case

thus,

a = g = acceleration due to gravity

u' = 0

thus we have

$u^2-0^2 = 2\times9.8\tiimes1.08$

or

$u = \sqrt{21.168}$

or

u = 4.6 m/s

b) Now after the bounce, the ball moves with the same velocity

thus, v = v₂

thus,

final speed ( $v_f$ ) = v = 4.6 m/s

Then conservation of energy says

$\frac{1}{2}mu_1^2+\frac{1}{2}Mu_2^2 = \frac{1}{2}mv_1^2+\frac{1}{2}Mv_2^2$

also

applying the concept of conservation of momentum

we have

mu₁ + Mu₂ = mv₁ + Mv₂

u₁ =velocity of the tennis ball before collision = -4.6 m/s

u₂ = velocity of the basketball before collision= 4.6 m/s

v₁ = velocity of the tennis ball after collision

v₂ = velocity of the basketball after collision

substituting the values in the equation, we get

Now,

solving both the equations simultaneously we get

$v = (\frac{2M}{m+M})u_1+(\frac{m-M}{m+M})u_2$

substituting the values in the above equation we get

$v = (\frac{2\times594}{44+594})(-4.6)+(\frac{44-594}{44+594})4.6$

or

$v = -8.565-3.965$

or

$v = -12.53m/s$

here negative sign depicts the motion of the ball in the upward direction

now the kinetic energy of the tennis ball

$K.E = \frac{1}{2}mv^2$

or

$K.E = \frac{1}{2}44\times 10^{-3}kg\times 12.53^2$

or

K.E = 3.45 J

also at the height the K.E will be the potential energy of the tennis ball

thus,

3.45 J = mgh

or

3.45 = 44 × 10⁻³ × 9.8 × h

h = 8.01 m

5 0

3 years ago