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xz_007 [3.2K]
3 years ago
12

If Mars were 10 times closer to the Sun, then the Sun would attract Mars with

Physics
2 answers:
tamaranim1 [39]3 years ago
7 0

The velocity does not change greatly between Mars and Venus —explained Dong— but Venus's closer proximity to the sun boosts the density by almost a factor of 4.5. This would mean that atmosphere on Mars would be lost even more rapidly than at its current position.

Goshia [24]3 years ago
4 0

Answer:

more

Explanation:

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A 2.0 m × 4.0 m flat carpet acquires a uniformly distributed charge of −10 μC after you and your friends walk across it several
mamaluj [8]

Answer:

The  charge on the dust particle is  q_d  = 6.94 *10^{-13} \  C

Explanation:

From the question we are told that

    The length is  l = 2.0 \ m

    The width is  w = 4.0 \ m

   The charge is  q =  -10\mu C= -10*10^{-6} \ C

    The mass suspended in mid-air is m_a =  5.0 \mu g =  5.0 *10^{-6} \ g =  5.0 *10^{-9} \  kg

   

Generally the electric field on the carpet is mathematically represented as

           E =  \frac{q}{ 2 *  A  *  \epsilon _o}

Where \epsilon _o is the permittivity of free space with value \epsilon_o = 8.85*10^{-12}  \ \  m^{-3} \cdot kg^{-1}\cdot  s^4 \cdot A^2

substituting values

           E =  \frac{-10*10^{-6}}{ 2 *  (2 * 4 )  *  8.85*10^{-12}}

           E = -70621.5 \  N/C

Generally the electric force keeping the dust particle on the air  equal to the force of gravity acting on the particles

        F__{E}} =  F__{G}}

=>     q_d *  E  =  m * g

=>      q_d  =  \frac{m * g}{E}

=>      q_d  =  \frac{5.0 *10^{-9} * 9.8}{70621.5}

=>     q_d  = 6.94 *10^{-13} \  C

4 0
3 years ago
What happens when the sun emits more energy than normal
xxTIMURxx [149]

Answer:

When the Sun emits more amount of energy than normal, "Solar flares and sunspots" occur, increasing temperature of Earth. Explanation: The Earth's temperature is governed by many factors. One of these factors is 'Solar flare'.

4 0
3 years ago
An airplane travels 2100 km at a speed of 1000 km/h. It then encounters a headwind which slows it to 800 km/h for the next 1300
Sidana [21]
I don’t get it I need help
5 0
3 years ago
An electron enters a region of uniform electric field with an initial velocity of 64 km/s in the same direction as the electric
8090 [49]

Answer:

1.) 11 km/s

2.) 9.03 × 10^-5 metres

Explanation:

Given that an electron enters a region of uniform electric field with an initial velocity of 64 km/s in the same direction as the electric field, which has magnitude E = 48 N/C.

Electron q = 1.6×10^-19 C

Electron mass = 9.11×10^-31 Kg

(a) What is the speed of the electron 1.3 ns after entering this region?

E = F/q

F = Eq

Ma = Eq

M × V/t = Eq

Substitute all the parameters into the formula

9.11×10^-31 × V/1.3×10^-9 = 48 × 1.6×10^-19

V = 7.68×10^-18 /7.0×10^-22

V = 10971.43 m/s

V = 11 Km/s approximately

(b) How far does the electron travel during the 1.3 ns interval?

The initial velocity U = 64 km/s

S = ut + 1/2at^2

S = 64000×1.3×10^-6 + 1/2 × 8.4×10^12 × ( 1.3×10^-9)^2

S =8.32×10^-5 + 7.13×10^-6

S = 9.03 × 10^-5 metres

3 0
2 years ago
A 5 μF capacitor is connected to a 12 V battery. The charge on each plate of the capacitor is:
ziro4ka [17]
1 farad = 1 coulomb/volt

5 μF = 5 x 10⁻⁶ coulomb/volt

        = 60 x 10⁻⁶ coulomb / 12 volts

The charge is 60 x 10⁻⁶ coulombs = 6 x 10⁻⁵  (choice-A) 


3 0
3 years ago
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