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3 years ago

What behavior of waves explains why someone can be looking away from you while talking and you still hear them?

Physics

1 answer:

hram777 [196]3 years ago

3 0

u might be right

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You might be interested in

Name two index fossils. Why are these fossils important to the practice of relative aging?

Inessa05 [86]

Dinosaur and bird so u can see how they developed and how they aged to become something popular<span />

8 0

3 years ago

A 65.0-kg woman steps off a 10.0-m diving platform and drops straight down into the water. 1) If she reaches a depth of 3.20 m,

timurjin [86]

Answer:

F=2627.6N

Explanation:

The work done by this resistive force while traveling a distance <em>d</em> underwater would be:

$W=F.d=-Fd$

where the minus sign appears because the force is upwards and the displacement downwards.

This work is equal to the change of mechanical energy. At the diving plataform and underwater, when she stops moving, the woman has no kinetic energy, so all can be written in terms of her total change of gravitational potential energy:

$W=\Delta E=U_f-U_i=mgh_f-mgh_i=mg(h_f-h_i)$

Putting all together:

$F=-\frac{W}{d}=-\frac{mg(h_f-h_i)}{d}=-\frac{(65kg)(9.8m/s^2)(-3.2m-10m)}{3.2m}=2627.6N$

7 0

3 years ago

A tank with a volume of 0.150 m3 contains 27.0oC helium gas at a pressure of 100 atm. How many balloons can be blown up if each

jekas [21]

Answer:

884 balloons

Explanation:

Assume ideal gas, since temperature is constant, then the product of pressure and volume is constant.

So if pressures reduces from 100 to 1.2, the new volume would be

$V_2 = \frac{P_1V_1}{P_2} = \frac{100*0.15}{1.2} = 12.5 m^2$

The spherical volume of each of the balloon of 30cm diameter (15 cm or 0.15 m in radius) is

$V_b = \frac{4}{3}\pir^3 = \frac{4}{3}\pi 0.15^3 = 0.014 m^3$

The number of balloons that 12.5 m3 can fill in is

$V_2/V_b = 12.5 / 0.014 = 884$

8 0

3 years ago

A round pipe of varying diameter carries petroleum from a wellhead to a refinery. At the wellhead, the pipe's diameter is 57.3 c

Mashcka [7]

Answer with Explanation:

We are given that

Diameter of pipe, $d_1=0.573 m$

$v_1=13.5 m/s$

$v_2=5.83 m/s$

Volume flow rate of the petroleum along the pipe= $Q_{refinery}=A_1v_1=v_(\frac{\pi d^2_1}{4})$

$Q_{refinery}=13.5\times (\pi\times \frac{0.573)^2}{4})=3.48 m^3/s$

By equation of continuity

$A_1v_1=A_2v_2$

$\frac{\pi d^2_1}{4}v_1=\frac{\pi d^2_2}{4}v_2$

$d^2_2=\frac{v_1}{v_2}d^2_1$

$d_2=\sqrt{\frac{v_1}{v_2}}d_1$

$d_2=0.573\sqrt{\frac{13.5}{5.83}}$

$d_2=0.87 m$

$d_2=0.87\time 100=87 cm$

1 m=100 cm

4 0

3 years ago

A particle that carries a net charge of -41.8 μc is held in a region of constant, uniform electric field. the electric field vec

miss Akunina [59]

The total work done by the electric field on the charge is given by the scalar product between the electric force acting on the charge and the displacement of the charge:
$W=F d cos \theta$
where the force is F=qE, d=0.556 and $\theta=55.2^{\circ}$ . Using the value of q and E given by the problem, we find
$W=qEdcos\theta = 6.39\cdot10^{-5}J$

3 0

4 years ago