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Vera_Pavlovna [14]
3 years ago
6

Heptane is always composed of 84.0% carbon and 16.0% hydrogen. This illustrates the law of multiple proportions. conservation of

mass. definite proportions. all of the above
Physics
1 answer:
BigorU [14]3 years ago
5 0

Heptane is always composed of 84.0% carbon and 16.0% hydrogen. This illustrates the "law of definite proportions".

Answer: Option C

<u>Explanation:</u>

Proust's law states that every chemical compound used to made up of element constituents with constant proportions in terms of its mass and also independent from its sources and synthesis method. In 1779, Joseph Proust gave other names to the Proust's law as, the law of composition or definite proportions or constant compositions.

This can understood from given example like: Oxygen is composed of 8/9 of the mass of any sample of pure water while the hydrogen fills up the remaining 1/9 of the mass. The basis of stoichiometry is structured with the law of multiple proportions along the law of definite proportions.

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Two isolated, concentric, conducting spherical shells have radii R1 = 0.500 m and R2 = 1.00 m, uniform charges q1=+2.00 µC and q
scZoUnD [109]

Complete Question

The diagram for this question is shown on the first uploaded image  

Answer:

a E =1.685*10^3 N/C

b E =36.69*10^3 N/C

c E = 0 N/C

d V = 6.7*10^3 V

e   V = 26.79*10^3V

f   V = 34.67 *10^3 V

g   V= 44.95*10^3 V

h    V= 44.95*10^3 V

i    V= 44.95*10^3 V

Explanation:

From the question we are given that

       The first charge q_1 = 2.00 \mu C = 2.00*10^{-6} C

       The second charge q_2 =1.00 \muC = 1.00*10^{-6}

      The first radius R_1 = 0.500m

      The second radius R_2 = 1.00m

 Generally \ Electric \ field = \frac{1}{4\pi\epsilon_0}\frac{q_1+\ q_2}{r^2}

And Potential \ Difference = \frac{1}{4\pi \epsilon_0}   [\frac{q_1 }{r}+\frac{q_2}{R_2} ]

The objective is to obtain the the magnitude of electric for different cases

And the potential difference for other cases

Considering a

                      r  = 4.00 m

           E = \frac{((2+1)*10^{-6})*8.99*10^9}{16}

                = 1.685*10^3 N/C

Considering b

           r = 0.700 m \ , R_2 > r > R_1

This implies that the electric field would be

            E = \frac{1}{4\pi \epsilon_0}\frac{q_1}{r^2}

             This because it the electric filed of the charge which is below it in distance that it would feel

            E = 8*99*10^9  \frac{2*10^{-6}}{0.4900}

               = 36.69*10^3 N/C

   Considering c

                      r  = 0.200 m

=>   r

 The electric field = 0

     This is because the both charge are above it in terms of distance so it wont feel the effect of their electric field

       Considering d

                  r  = 4.00 m

=> r > R_1 >r>R_2

Now the potential difference is

                  V =\frac{1}{4\pi \epsilon_0} \frac{q_1 + \ q_2}{r} = 8.99*10^9 * \frac{3*10^{-6}}{4} = 6.7*10^3 V

This so because the distance between the charge we are considering is further than the two charges given  

          Considering e

                       r = 1.00 m R_2 = r > R_1

                V = \frac{1}{4\pi \epsilon_0} [\frac{q_1}{r} +\frac{q_2}{R_2}  ] = 8.99*10^9 * [\frac{2.00*10^{-6}}{1.00} \frac{1.00*10^{-6}}{1.00} ] = 26.79 *10^3 V

          Considering f

              r = 0.700 m \ , R_2 > r > R_1

                      V = \frac{1}{4\pi \epsilon_0} [\frac{q_1}{r} +\frac{q_2}{R_2}  ] = 8.99*10^9 * [\frac{2.00*10^{-6}}{0.700} \frac{1.0*10^{-6}}{1.00} ] = 34.67 *10^3 V

          Considering g

             r =0.500\m , R_1 >r =R_1

   V = \frac{1}{4\pi \epsilon_0} [\frac{q_1}{r} +\frac{q_2}{R_2}  ] = 8.99*10^9 * [\frac{2.00*10^{-6}}{0.500} \frac{1.0*10^{-6}}{1.00} ] = 44.95 *10^3 V

          Considering h

                r =0.200\m , R_1 >R_1>r

  V = \frac{1}{4\pi \epsilon_0} [\frac{q_1}{R_1} +\frac{q_2}{R_2}  ] = 8.99*10^9 * [\frac{2.00*10^{-6}}{0.500} \frac{1.0*10^{-6}}{1.00} ] = 44.95 *10^3 V

           Considering i    

   r =0\ m \ , R_1 >R_1>r

  V = \frac{1}{4\pi \epsilon_0} [\frac{q_1}{R_1} +\frac{q_2}{R_2}  ] = 8.99*10^9 * [\frac{2.00*10^{-6}}{0.500} \frac{1.0*10^{-6}}{1.00} ] = 44.95 *10^3 V

8 0
3 years ago
A string of length 100 cm is held fixed at both ends and vibrates in a standing wave pattern. The wavelengths of the constituent
azamat

The wavelengths of the constituent travelling waves CANNOT be 400 cm.

The given parameters:

  • <em>Length of the string, L = 100 cm</em>

<em />

The wavelengths of the constituent travelling waves is calculated as follows;

L = \frac{n \lambda}{2} \\\\n\lambda = 2L\\\\\lambda = \frac{2L}{n}

for first mode: n = 1

\lambda = \frac{2\times 100 \ cm}{1} \\\\\lambda = 200 \ cm

for second mode: n = 2

\lambda = \frac{2L}{2} = L = 100 \ cm

For the third mode: n = 3

\lambda = \frac{2L}{3} \\\\\lambda = \frac{2 \times 100}{3} = 67 \ cm

For fourth mode: n = 4

\lambda = \frac{2L}{4} \\\\\lambda = \frac{2 \times 100}{4} = 50  \ cm

Thus, we can conclude that, the wavelengths of the constituent travelling waves CANNOT be 400 cm.

The complete question is below:

A string of length 100 cm is held fixed at both ends and vibrates in a standing wave pattern. The wavelengths of the constituent travelling waves CANNOT be:

A. 400 cm

B. 200 cm

C. 100 cm

D. 67 cm

E. 50 cm

Learn more about wavelengths of travelling waves here: brainly.com/question/19249186

5 0
2 years ago
Which of the following statements are true? Check all that apply. Check all that apply. The gravitational force between two obje
AveGali [126]

<em>The gravitational force between two objects is inversely proportional to the square of the distance between the two objects.</em>

The gravitational force between two objects is proportional to the product of the masses of the two objects.

The gravitational force between two objects is proportional to the square of the distance between the two objects.  <em> no</em>

The gravitational force between two objects is inversely proportional to the distance between the two objects.  <em> no</em>

The gravitational force between two objects is proportional to the distance between the two objects.  <em> no</em>

The gravitational force between two objects is inversely proportional to the product of the masses of the two objects. <em> no</em>

3 0
3 years ago
Read 2 more answers
Baseball player A bunts the ball by hitting it in such a way that it acquires an initial velocity of 1.3 m/s parallel to the gro
pashok25 [27]

Answer:

Explanation:

cSep 20, 2010

well, since player b is obviously inadequate at athletics, it shows that player b is a woman, and because of this, she would not be able to hit the ball. The magnitude of the initial velocity would therefore be zero.

Anonymous

Sep 20, 2010

First you need to solve for time by using

d=(1/2)(a)(t^2)+(vi)t

1m=(1/2)(9.8)t^2 vertical initial velocity is 0m/s

t=.45 sec

Then you find the horizontal distance traveled by using

v=d/t

1.3m/s=d/.54sec

d=.585m

Then you need to find the time of player B by using

d=(1/2)(a)(t^2)+(vi)t

1.8m=(1/2)(9.8)(t^2) vertical initial velocity is 0

t=.61 sec

Finally to find player Bs initial horizontal velocity you use the horizontal equation

v=d/t

v=.585m/.61 sec

so v=.959m/s

5 0
3 years ago
Read 2 more answers
A ball is spun around in circular motion such that it completes 50 rotations in 25 s.
Leokris [45]

Answer:

(A) The period of its rotation is 0.5 s (2) The frequency of its rotation is 2 Hz.

Explanation:

Given that,

a ball is spun around in circular motion such that it completes 50 rotations in 25 s.

(1). Let T be the period of its rotation. It can be calculated as follows :

T=\dfrac{25}{50}\\\\T=0.5\ s

(2). Let f be the frequency of its rotation. It can be defined as the number of rotations per unit time. So,

f=\dfrac{50}{25}\\\\f=2\ Hz

Hence, this is the required solution.

3 0
3 years ago
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