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Amiraneli [1.4K]
3 years ago
9

What measures the distance between two consecutive crests of a wave?

Physics
1 answer:
Marina86 [1]3 years ago
8 0
Answer A is incorrect
A crest is just one point. It is not the distance between 2 crests.

B  is incorrect
A trough is just 1 point. It is not the distance between 2 troughs.

C is incorrect.
the amplitude measures the height of a crest from the middle of the wave to the crest (or trough).

D is the correct answer. That is the distance between 2 crests or 2 troughs 
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Which of the following describes the way a population can be dispersed?
zubka84 [21]

Answer:

option b. <em>uniform</em><em> </em><em>population</em>

5 0
3 years ago
Explain why atmospheric pressure changes as atmospheric depth changes.
brilliants [131]
The lower you go <span>in relation to the top of the atmosphere the larger the column of air is that is pressing down on you. </span>
4 0
3 years ago
A car leaves an intersection traveling west. Its position 4 sec later is 21 ft from the intersection. At the same time, another
Alex_Xolod [135]

Answer:

15.8 ft/s

Explanation:

\frac{da}{dt} = Velocity of car A = 9 ft/s

a = Distance car A travels = 21 ft

\frac{db}{dt} = Velocity of car B = 13 ft/s

b = Distance car B travels = ft

c = Distance between A and B after 4 seconds = √(a²+b²) = √(21²+28²) = √1225 ft

From Pythagoras theorem

a²+b² = c²

Now, differentiating with respect to time

2a\frac{da}{dt}+2b\frac{db}{dt}=2c\frac{dc}{dt}\\\Rightarrow a\frac{da}{dt}+b\frac{db}{dt}=c\frac{dc}{dt}\\\Rightarrow \frac{dc}{dt}=\frac{a\frac{da}{dt}+b\frac{db}{dt}}{c}\\\Rightarrow \frac{dc}{dt}=\frac{21\times 9+28\times 13}{\sqrt{1225}}\\\Rightarrow \frac{dc}{dt}=15.8\ ft/s

∴ Rate at which distance between the cars is increasing three hours later is 15.8 ft/s

6 0
4 years ago
In midair an M = 145 kg bomb explodes into two pieces of m1 = 115 kg and another, respectively. Before the explosion, the bomb w
Daniel [21]

Answer:

v_2=-133.17m/s, the minus meaning west.

Explanation:

We know that linear momentum must be conserved, so it will be the same before (p_i) and after (p_f) the explosion. We will take the east direction as positive.

Before the explosion we have p_i=m_iv_i=Mv_i.

After the explosion we have pieces 1 and 2, so p_f=m_1v_1+m_2v_2.

These equations must be vectorial but since we look at the instants before and after the explosions and the bomb fragments in only 2 pieces the problem can be simplified in one dimension with direction east-west.

Since we know momentum must be conserved we have:

Mv_i=m_1v_1+m_2v_2

Which means (since we want v_2 and M=m_1+m_2):

v_2=\frac{Mv_i-m_1v_1}{m_2}=\frac{Mv_i-m_1v_1}{M-m_1}

So for our values we have:

v_2=\frac{(145kg)(24m/s)-(115kg)(65m/s)}{(145kg-115kg)}=-133.17m/s

5 0
3 years ago
Identify the characteristic of the transverse wave that halved from wave A (black) to wave B (green).
kolbaska11 [484]

Answer:

i believe that the answer is D: wavelength

5 0
3 years ago
Read 2 more answers
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