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kvv77 [185]
2 years ago
7

Suppose the entire population of the world gathers in ONE spot and everyone jumps at the sound of a prearranged signal. While ev

eryone is in the air, does the Earth gain momentum in the opposite direction?a) No, the inertial mass of Earth is so large that the planet's change in motion is imperceptibleb) Yes, however the change in momentum of Earth is much less than that of all the jumping people because of Earth's large inertial massc) Yes, Earth recoils like that of a rifle firing a bullet with a change in momentum equal to and opposite of peopled) It depends
Physics
1 answer:
puteri [66]2 years ago
6 0

Answer:

(b) Yes, the earth gains momentum but the change in momentum of the earth is much lesser compared to that of everyone in the air. The resistance to motion (inertia of the earth), which is a function of its mass is so great that the earth's acceleration is small in the given time frame.

Explanation:

From Newton's second law which can be stated mathematically as

F = m(v-u)/t = ma.

By Newton's law of gravitation, there is a force between the earth and everyone in the air. This force is responsible for the change in momentum of everyone in the air and this force gives them an acceleration equal to g = 9.80m/s². By Newton's law of gravitation and Newton's third law of motion, this force is also equal to the force exerted by everyone on the earth.

For this to be true,

F = M (everyone) ×a (everyone) = M(earth) × a (earth).

And

a (earth) = {M (everyone) ×a (everyone) }/M (earth)

Then

a (earth) must be lesser than a (everyone) since M(earth) >> M(everyone).

a = change in momentum/ time

Therefore the earth will have a much lesser change in momentum which is the reason we won't notice the earth's movement.

Thank you for reading.

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How many electrons are in the outer energy level of an atom of carbon
xz_007 [3.2K]

Answer:

4 electrons

Explanation:

Carbon, is the group 14 element, with four electrons in its outer shell. Carbon always shares electrons to reach a complete valence shell, making bonds with other atoms.

8 0
3 years ago
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
During heavy rain, a section of a mountainside measuring 2.5 km horizontally, 0.80 km up along the slope, and 2.0 m deep slips i
Crank

Answer:

The mass of the mud is 3040000 kg.

Explanation:

Given that,

length = 2.5 km

Width = 0.80 km

Height = 2.0 m

Length of valley = 0.40 km

Width of valley = 0.40 km

Density = 1900 Kg/m³

Area = 4.0 m²

We need to calculate the mass of the mud

Using formula of density

\rho=\dfrac{m}{V}

m=\rho\times V

Where, V = volume of mud

\rho = density of mud

Put the value into the formula

m=1900\times4.0\times0.40\times10^{3}

m =3040000\ kg

Hence, The mass of the mud is 3040000 kg.

4 0
3 years ago
Review:
Tresset [83]

Answer:

1. 1. A quantity is completely described by magnitude alone. A quantity Is completely described by a magnitude with a direction.

[a]. scalar, vector

b. vector, scalar

2.2. Speed is a velocity is a quantity and quantity.

a. scalar, vector

[b]. vector, scalar

3 0
2 years ago
Part A
geniusboy [140]

Answer:

a) b = -5

b) slope = 3/2

Explanation:

a) The equation of a line is given as y = mx + b, where m is the slope of the line and b is the intercept on the y axis.

Given that y = 3x + b and it passes through the point (2, 1). Hence when x = 2, y = 1. Therefore, substituting for x and y:

1 = 3(2) + b

1 = 6 + b

b = 1 - 6

b = -5

b) The equation of a line passing through two points (x_1,y_1) and x_2,y_2 is given by:

y-y_1=\frac{y_2-y_1}{x_2-x_1}(x-x_1)

The equation of the line passing through the two points (0,3) and (4,9) is:

y-3=\frac{9-3}{4-0}(x-0)\\ \\y-3=\frac{3}{2}x\\ \\y = \frac{3}{2}x+3

Comparing y = (3/2)x + 3 with y = mx + b, the slope (m) is 3/2

4 0
2 years ago
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