With which of the following does the growth rate decrease as the population gets bigger?

How will a wave change as it moves from a layer of higher velocity into a lower layer of lower velocity?

denis-greek [22]

Explanation:

Wave is defined as a disturbance or oscillation that travels through space-time, accompanied by a transfer of energy. Wave motion transfers energy from one point to another, often with no permanent displacement of the particles of the medium.

The velocity of wave is equal to the product of its wavelength and frequency (number of vibrations per second). Longitudinal waves like sound waves travel through a medium.

Therefore, a wave move from a layer of high velocity to that of a lower velocity the wavelength changes (that is, decreases) as it moves.

6 0

3 years ago

2) A 0.4kg ball moves in horizontal circle of radius 3 m at speed of 100m/s. What

sweet-ann [11.9K]

Answer:

F = 1300 N

Explanation:

F = mv²/R = 0.4(100²)/3 = 1333.3333...

6 0

3 years ago

In which part of the wave are the particles of the medium closer together?

Lesechka [4]

B.) Compressions
HOPPE IT HELPED

6 0

3 years ago

Read 2 more answers

How much work is required to compress 5.05 mol of air at 19.5°C and 1.00 atm to one-eleventh of the original volume by an isothe

Rus_ich [418]

Explanation:

(a) For an isothermal process, work done is represented as follows.

W = $-nRT ln(\frac{V_{2}}{V_{1}})$

Putting the given values into the above formula as follows.

W = $-nRT ln(\frac{V_{2}}{V_{1}})$

= $- 5.05 mol \times 8.314 J/mol K \times (19.5 + 273) K \times ln (\frac{\frac{V_{1}}{11}}{V_{1}})$

= $-12280.82 \times ln (0.09)$

= $-12280.82 \times -2.41$

= 29596.78 J

or, = 29.596 kJ (as 1 kJ = 1000 J)

Therefore, the required work is 29.596 kJ.

(b) For an adiabatic process, work done is as follows.

W = $\frac{P_{1}V^{\gamma}_{1}(V^{1-\gamma}_{2} - V(1-\gamma)_{1})}{(1 - \gamma)}$

= $\frac{-nRT_{1}(11^{\gamma - 1} - 1)}{1 - \gamma}$

= $\frac{-5.05 \times 8.314 J/mol K \times 292.5 (11^{1.4 - 1} - 1)}{1 - 1.4}$

= 49.41 kJ

Therefore, work required to produce the same compression in an adiabatic process is 49.41 kJ.

(c) We know that for an isothermal process,

$P_{1}V_{1} = P_{2}V_{2}$

or, $P_{2} = \frac{P_{1}V_{1}}{V_{2}}$

= $1 atm (\frac{V_{1}}{\frac{V_{1}}{11}})$

= 11 atm

Hence, the required pressure is 11 atm.

(d) For adiabatic process,

$P_{1}V^{\gamma}_{1} = P_{2}V^{\gamma}_{2}$

or, $P_{2} = P_{1} (\frac{V_{1}}{V_{2}})^{1.4}$

= $1 atm (\frac{V_{1}}{\frac{V_{1}}{11}})^{1.4}$

= 28.7 atm

Therefore, required pressure is 28.7 atm.

6 0

4 years ago

Sally takes two bar magnets and randomly tapes one end of each bar magnet. she labels the magnets A and B. She brings the taped

vlabodo [156]

Since the two taped poles of the magnets labeled A and B attracts one to each other, we know that the two taped poles are oppsosite.

So, you can predict with total certainty that when she brings the taped end of the third magnet (magnet C) near each of the first two magntes, in one case they will attract each other and in the other case they will repele mutually.

You are certain of that because, since the taped poles of the first two magnets are opposite, the pole of the third magnet has to be equal to one of the two first taped poles and opposite to the other of the two firest taped poles.

8 0

3 years ago