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Arada [10]
3 years ago
15

Which scenarios are examples of physical changes? Select three options

Physics
1 answer:
pashok25 [27]3 years ago
4 0

Answer:

A kid becoming an adult

A leg becoming bruised

A person's blood pressure raising because they are running

need a picture to answer specific questions.

Explanation:

All of these are physical changes. Hope that this helps you and have a great day :)

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!!!URGENT!!! Worth 100 points!!
slavikrds [6]

The snail would travel 100mm

Each minutes is 25mm and it took 4 minutes

4 x 25 = 100

or

25+25+25+25=100

7 0
3 years ago
Read 2 more answers
What is the wavelength (in nm) of radiation that has an energy content of 1.0x10^2 kJ/mol?
hammer [34]
<span>Energy = h nu, where nu is the frequency
h = 6.63 x 10^-34 J-s, Planck's constant
So nu = E/h = 1 x 10^5 J /h = 0.15 x 10^29 / s
nu lambda = c, the speed of light.
lambda = wavelength = c / nu =3 x 10^8 / 0.15 x 10^29 = 20 x 10^-21 m.

this can possibly be a gamma ray. Gamma rays are very penetrating. It's both matter and an energy. They are electromagnetic radiation that results from a radioactive material. 

</span><span>
</span>
7 0
3 years ago
Read 2 more answers
The example of basic reserch
damaskus [11]

Answer:

A study looking at how alcohol consumption impacts the brain. A study to discover the components making up human DNA. A study accessing whether stress levels 

Explanation:

For example, a scientist that tries to figure out how the body makes cholesterol, or what causes a particular disease, is performing basic scienc

3 0
3 years ago
A horizontal board of negligible thickness and area 2.0 m2 hangs from a spring scale that reads 80 N when a 4.0 m/s wind moves b
ki77a [65]

Answer:

Scale reading for no wind W'=60N

Explanation:

From the question we are told that

Area A= 2.0 m^2

Weight of board W=80

Velocity V=4.0m/s

Density of air \delta= 1.25 kg/m3 .

Generally the equation for pressure difference by Bernoulli equation is mathematically given by

  dP=\frac{1}{2}pv^2

  dP=10Pa

Generally force acting on the board by air is mathematically given by

F=\triangle PA

F=(10)2=>20N

Therefore

Scale reading for no wind W'

W'=W-F\\W'=80-20

W'=60N

Scale reading for no wind W'=60N

6 0
3 years ago
A pipe open only at one end has a fundamental frequency of 266 Hz. A second pipe, initially identical to the first pipe, is shor
Alika [10]

Answer:

1.16cm were cut off the end of the second pipe

Explanation:

The fundamental frequency in the first pipe is,

<em><u>Since the speed of sound is not given in the question, we would assume it to be 340m/s</u></em>

f1 = v/4L, where v is the speed of sound and L is the length of the pipe

266 = 340/4L

L = 0.31954 m = 0.32 m

It is given that the second pipe is identical to the first pipe by cutting off a portion of the open end. So, consider L’ be the length that was cut from the first pipe.

<u>So, the length of the second pipe is L – L’</u>

Then, the fundamental frequency in the second pipe is

f2 = v/4(L - L’)

<u>The beat frequency due to the fundamental frequencies of the first and second pipe is</u>

f2 – f1 = 10hz

[v/4(L - L’)] – 266 = 10

[v/4(L – L’)] = 10 + 266

[v/4(L – L’)] = 276

(L - L’) = v/(4 x 276)

(L – L’) = 340/(4 x 276)

(L – L’) = 0.30797

L’ = 0.31954 – 0.30797

L’ = 0.01157 m = 1.157 cm ≅ 1.16cm  

Hence, 1.16 cm were cut from the end of the second pipe

6 0
3 years ago
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