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baherus [9]
3 years ago
8

What type of wave is shown below?

Physics
1 answer:
Elena-2011 [213]3 years ago
5 0

Answer:

<u>B</u><u>.</u><u> </u><u>Transverse</u><u> </u><u>wave</u><u>.</u>

Explanation:

Because it has troughs and crests.

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A 1,70 atm, una muestra de gas ocupa 4,25 litros. Si la presión en el gas aumenta a 2.40 atm, ¿cuál será el nuevo volumen?
statuscvo [17]

At 1.70 atm, a gas sample occupies 4.25 liters. If the pressure in the gas increases to 2.40 atm, what will the new volume be?

Answer:

3.01L

Explanation:

Given parameters:

Initial pressure, P1  = 1.7atm

Initial volume, V1  = 4.25L

Final pressure, P2  = 2.4atm

Unknown:

Final or new volume, V2  = ?

Solution:

To solve this problem, we use Boyle's law which states that "the volume of a fixed mass of a gas varies inversely as the pressure changes, if the temperature is constant".

            P1 V1  = P2 V2

P1 is the initial pressure

V1 is the initial volume

P2 final pressure

V2 final volume

        1.7 x 4.25  = 2.4 x V2

             V2  = 3.01L

8 0
4 years ago
How i can change my bad habits and one is i always trust peopñe
Lelu [443]

what? I guess:

  • practice different habits. If you fail don't give up.
  • don't always trust people, some are not what they seem.

this question doesn't make any sense...

5 0
4 years ago
Read 2 more answers
Replacing an object attached to a spring with an object having 14 the original mass will change the frequency of oscillation of
Pachacha [2.7K]

Answer:

<em>The frequency changes by a factor of  0.27.</em>

<em></em>

Explanation:

The frequency of an object with mass m attached to a spring is given as

f = \frac{1}{2\pi } \sqrt{\frac{k}{m} }

where f is the frequency

k is the spring constant of the spring

m is the mass of the substance on the spring.

If the mass of the system is increased by 14 means the new frequency becomes

f_{n} = \frac{1}{2\pi } \sqrt{\frac{k}{14m} }

simplifying, we have

f_{n} = \frac{1}{2\pi \sqrt{14} } \sqrt{\frac{k}{m} }

f_{n} = \frac{1}{3.742*2\pi  } \sqrt{\frac{k}{m} }

if we divide this final frequency by the original frequency, we'll have

==> \frac{1}{3.742*2\pi  } \sqrt{\frac{k}{m} }  ÷  \frac{1}{2\pi } \sqrt{\frac{k}{m} }

==> \frac{1}{3.742*2\pi  } \sqrt{\frac{k}{m} }  x  2\pi \sqrt{\frac{m}{k} }

==> 1/3.742 = <em>0.27</em>

7 0
3 years ago
PLEASE HELP ME,, I WOULD BE SO HAPPY
Scrat [10]

Answer:

thanks for the points liar

Explanation:

4 0
3 years ago
Which forces tend to slow down an object
laila [671]
The answer is "friction and air resistance" gravity does some of the work by keeping the object from floating away, but friction and air resistance does the biggest part. Friction is how rough the ground it meaning on tile, dirt, grass, etc... that would slow down the object and air resistance is the gravity pushing on the object also making it stop. 

Hope this helps!
8 0
4 years ago
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