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BabaBlast [244]
2 years ago
5

Assignment 1.4

Physics
1 answer:
Flauer [41]2 years ago
6 0

you can use the graph to estimate the value of x when y is required to start a doubles match with the sun and the sky very well from her childhood and a kind young women who lives with her wearing a black and a veil and shows to be in that position or they have a bad past or bad life that they want to take revenge on every one by murder goes on by them in the streets of St and St Mary's Church and its followers are known as Sikhs and the other children claim she

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A large fraction of the ultraviolet (UV) radiation coming from the sun is absorbed by the atmosphere. The main UV absorber in ou
Sergeeva-Olga [200]

Answer:

The wavelength of the radiation absorbed by ozone is 319.83 nm

Explanation:

Given;

frequency of absorbed ultraviolet (UV) radiation, f = 9.38×10¹⁴ Hz

speed of the absorbed ultraviolet (UV) radiation, equals speed of light, v = 3 x 10⁸ m/s

wavelength of the absorbed ultraviolet (UV) radiation, λ = ?

Apply wave equation for speed, frequency and wavelength;

v = fλ

λ = v / f

λ = (3 x 10⁸) / (9.38×10¹⁴)

λ = 3.1983 x 10⁻⁷ m

λ = 319.83 x  10⁻⁹ m

λ = 319.83 nm

Therefore, the wavelength of the radiation absorbed by ozone is 319.83 nm

5 0
4 years ago
List an example of each of the four classes polymers that living things make and use
slamgirl [31]


The four classes of polymers are:

1. Nucleic acids.  Examples are DNA and RNA

2. Protein. Examples are enzymes and hemoglobin

3. Carbohydrates. Examples as starch and glycogen

4. Lipids. Examples are triglycerides and phospholipids

The building blocks of nucleic acids are called bases and there are four types known as Guanine, Adenine, Thymine and Cytosine.

The building blocks of carbohydrates are glucose molecules.

The building blocks of protein are amino acids.

The building blocks of lipids are a combination of fatty acids and glycerol.

6 0
3 years ago
Two insulated copper wires of similar overall diameter have very different interiors. One wire possesses a solid core of copper,
Marrrta [24]

Answer:

a

 Solid Wire     I  =   0.01237 \  A      

  Stranded  Wire  I_2  =   0.00978 \  A

b

  Solid Wire   R  = 0.0149 \ \Omega

   Stranded  Wire  R_1  = 0.0189 \ \Omega

Explanation:

Considering the first question

From the question we are told that

  The  radius of the first wire is  r_1  = 1.53 mm = 0.0015 \  m

  The radius of  each strand is  r_0 =  0.306 \ mm =  0.000306 \ m

  The current density in both wires is  J  =  1750 \  A/m^2

Considering the first wire

     The  cross-sectional area of the first wire is

      A   = \pi  r^2

= >  A   = 3.142 *  (0.0015)^2

= >  A   = 7.0695 *10^{-6} \  m^2

Generally the current in the first wire is    

     I  =  J*A

=>  I  =  1750*7.0695 *10^{-6}

=>  I  =   0.01237 \  A

Considering the second wire  wire

The  cross-sectional area of the second wire is

     A_1  =  19 *  \pi r^2

=>     A_1  =  19 *3.142 *  (0.000306)^2

=>  A_1  =  5.5899 *10^{-6} \  m^2

Generally the current is  

      I_2  =  J  *  A_1

=>    I_2  =   1750  *  5.5899 *10^{-6}

=>    I_2  =   0.00978 \  A

Considering question two  

 From the question we are told that

     Resistivity is  \rho  =  1.69* 10^{-8} \Omega \cdot m

     The  length of each wire  is  l =  6.25 \  m

Generally the resistance of the first wire is mathematically represented as

    R  =  \frac{\rho *  l  }{A}

=> R  =  \frac{  1.69* 10^{-8} * 6.25 }{ 7.0695 *10^{-6} }

=> R  = 0.0149 \ \Omega

Generally the resistance of the first wire is mathematically represented as

    R_1  =  \frac{\rho *  l  }{A_1}

=> R_1  =  \frac{  1.69* 10^{-8} * 6.25 }{5.5899 *10^{-6} }

=> R_1  = 0.0189 \ \Omega

3 0
3 years ago
A student measured the specific heat of water to be 4.29 J/g.Co. The
leonid [27]

Answer:

2.63 %.

Explanation:

Given that,

The calculated value of the specific heat of water is 4.29 J/g.C

Original value of  specific heat of water is 4.18 J/g.C.

We need to find the student's percent error. The percentage error in any quantity is given by :

P=\dfrac{|\text{original value-calculated value}|}{\text{original value}}\times 100\\\\P=\dfrac{4.29-4.18}{4.18}\times 100\\\\P=2.63\%

So, the student's percent error is 2.63 %.

7 0
3 years ago
PLEASE ANSWER ASAP BEFORE MY TEACHER AND MY MOM KILLES ME PLEASE ASAP
aleksley [76]

Answer:

red is not at the bottom. its density is 0.9 not 9!

also I don't need the brainlest thanks

6 0
3 years ago
Read 2 more answers
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