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2 years ago

Help pls, answer for brainlist

Physics

1 answer:

diamong [38]2 years ago

4 0

Answer:

C for the first question. C for the second ( i think ) I searched the third up it should be C again ( ikr CRAZY 3 c's ) B I think for the fourth one and D for the last question.

IF ANY OF THESE ARE WRONG I DEEPLY APOLOGIZE MY HEAD IS HURTING FROM THIS ALREADY

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How do the three types of boundary's work together to keep the Earth at equilibrium?

Maurinko [17]

There are three types: divergent, convergent, and transform boundaries. I hope this helps.

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2 years ago

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1. What is the bond angle between the hydrogen atoms in an ammonia (NH3) molecule? 2. What is the bond angle between the oxygen

Klio2033 [76]

The answer is 107 degrees. The geometric shape for ammonia is Trigonal Pyramidal, even though its electron geometry is “Tetrahedral”. This is because ammonia has a lone pair of electrons that occupy its space like the other 3 hydrogens in the geometric structure.

The answer 180 degrees. This is because of the linear geometric structure of carbon dioxide. The oxygen atom is on either side of the carbon atom, each is bound by a double covalent bond. All the atoms are involved in the bond and there are no one pair electrons.

The answer is tetrahedral geometry. This is because all the 4 valence electrons of the carbon are involved in a bond with a hydrogen atom. The angles in a tetrahedral geometric arrangement, such as in methane, is 109.5 degrees, where the hydrogen atoms are as far apart, from each other, as possible .

7 0

3 years ago

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A train at rest emits a sound at a frequency of 1057 Hz. An observer in a car travels away from the sound source at a speed of 2

il63 [147K]

Answer:

993.52 Hz

Explanation:

The frequency of sound emitted by the stationery train is 1057 Hz.

The car travels away from the train at 20.6 m/s.

The frequency the observer hears is given by the formula:

$f_o = \frac{v - v_o}{v}f$

where v = velocity of sound = 343 m/s

vo = velocity of observer

f = frequency from source

This phenomenon is known as Doppler's effect.

Therefore:

$f_o = \frac{343 - 20.6}{343} * 1057\\ \\f_o = 322.4 / 343 * 1057\\\\f_o = 993.52 Hz$

The frequency heard by the observer is 993.52 Hz.

4 0

3 years ago

You are singing a song at karaoke and reach a part that requires a louder, more intense sound. What must you do to produce a lou

FrozenT [24]

More energy is used in creating a louder sound from the mouth.

<h3>What must you do to produce a louder sound?</h3>

We use more energy in order to produce a louder sound because energy is the thing that helps in the formation of loud sound. Loudness is dependent on the energy that creates loud sound in the mouth. More energy we apply, more loud sound will produce. Energy is the main factor which gives us a loud sound in each and every instruments. Without, we can't imagine loud sound.

So we can conclude that more energy is used in creating a louder sound from the mouth.

Learn more about sound here: brainly.com/question/1199084

#SPJ1

6 0

1 year ago

If you travel from Yakima to Ellensburg (Yakima to Ellensburg is 50 miles) with a speed of 60 miles/hour for half of the

koban [17]

Answer:

$\displaystyle \frac{480}{7}\approx 68.6\; \rm mph$ .

Explanation:

The average speed of an object is equal to total distance over total time.

Distance traveled: $\rm 50 \; mi$ .

How much time is taken? This trip is divided into two halves, each of distance $\displaystyle \frac{50}{2} = 25\;\rm mi$ .

Time spent on the first half of the trip:

$\displaystyle t_1 = \frac{s_1}{v_1} = \frac{25}{60} = \frac{5}{12}\; \rm hours$ .

Similarly, time spent on the second half of the trip:

$\displaystyle t_2 = \frac{s_2}{v_2} = \frac{25}{80} = \frac{5}{16}\; \rm hours$ .

In total:

$\displaystyle \frac{5}{12} + \frac{5}{16} = \frac{35}{48} \; \rm hours$ .

Average speed:

$\begin{aligned} \text{Average speed} &= \frac{\text{Total Distance}}{\text{Total Time}}\\ &= 50 \left/\frac{35}{48}\right.\\ &= 50 \cdot \frac{48}{35} \\&= \frac{480}{7}\approx 68.6\; \rm mph \end{aligned}$ .

This value turned out to be slightly different from the average of the speed during the two halves of the journey. The reason is that the object traveled at each speed for a different amount of time. It spent more time at the slower speed, which gives that speed a greater weight in the average. That explains why the average speed is closer to $\rm 60\; mph$ rather than $\rm 80\; mph$ .

3 0

2 years ago