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nikdorinn [45]
3 years ago
13

The planet Venus has a mass of 4.87 × 1024 kg and a radius of 6.05 × 106 km. What is the magnitude of the gravitational force th

at an 81-kg person would experience while standing on the surface of Venus?
Physics
1 answer:
Anastasy [175]3 years ago
6 0

Answer:

7.2 x 10⁻⁶N

Explanation:

Given parameters:

Mass of Venus  = 4.87 x 10²⁴kg

Radius  = 6.05 x 10⁶km = 6.05 x 10⁹m

Mass of person  = 81kg

Unknown:

Gravitational force  = ?

Solution:

To solve this problem, we use the Newtons law of universal gravitation:

  Fg  = \frac{G x mass of venus x mass of person}{radius^{2} }  

G is the universal gravitation constant  = 6.67 x 10⁻¹¹

 Fg  = \frac{6.67 x 10^{-11}  x 4.87 x 10^{22} x 81 }{(6.05 x 10^9)^{2} } }

 Fg  =  7.2 x 10⁻⁶N

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A ceiling fan draws a current of 0.625 A and has a voltage of 120 V. The resistance of the ceiling fan, to the nearest whole num
Fiesta28 [93]
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Can lack of sleep cause muscle injuries?
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3 0
1 year ago
A particular coaxial cable is comprised of inner and outer conductors having radii 1 mm and 3 mm respectively, separated by air.
noname [10]

Answer:

The value is  \rho_s  =  4.026 *10^{-6} \  C/m^2

Explanation:

From the question we are told that

   The radius of the inner conductor  is  r_1 = 1 \ mm =  0.001 \ m

    The radius of the outer conductor is  r_2 = 3 \ mm = 0.003 \  m

    The potential at the outer conductor is  V = 1.5 kV  =  1.5 *10^{3} \  V

Generally the capacitance per length of the capacitor like set up of the two conductors is

      C= \frac{2 * \pi * \epsilon_o }{ ln [\frac{r_2}{r_1} ]}

Here \epsilon_o is the permitivity of free space with value  \epsilon_o =  8.85*10^{-12} C/(V \cdot m)

=>   C= \frac{2 *  3.142  * 8.85*10^{-12}  }{ ln [\frac{0.003}{0.001} ]}

=>   C= 50.6 *10^{-12} \  F/m

Generally given that the potential  of the outer conductor with respect to the inner conductor is positive it then mean that the outer conductor is positively charge

Generally the line  charge density of the outer  conductor is mathematically represented as

      \rho_l  =  C *  V

=>   \rho_d  =  50.6*10^{-12} *  1.5*10^{3}

=>   \rho_d  =  7.59*10^{-8} \  C/m

Generally the surface charge density is mathematically represented as

        \rho_s  =  \frac{\rho_l }{2 \pi * r_2 }    here 2 \pi r = (circumference \ of \ outer \  conductor  )

=>    \rho_s  =  \frac{7.59 *10^{-8} }{2* 3.142 * 0.003 }

=>    \rho_s  =  4.026 *10^{-6} \  C/m^2

3 0
3 years ago
two vehicles have a head on collision. one vehicle has a mass of 3000 kg and moves at 25 m/s while the second vehicle has a mass
emmasim [6.3K]

Answer:

The speed of the combined vehicles is 6.82m/s

Explanation:

Using the law of conservation of momentum which stayed that the sum of momentum of bodies before collision is equal to their sum of momentum after collision. After collision, both object moves with the same velocity.

Momentum = mass×velocity

Before collision:

Momentum of vehicle or mass 3000kg moving with velocity 25m/s

= 3000×25

= 75000kgm/s

Pa = 75000kgm/s

Momentum of vehicle with mass 2500kg moving with velocity of -15m/s

= 2500×-15

= -37500kgm/s

After collision:

Momentum = (3000+2500)V

Where v is their common velocity

Momentum after collision = 5500V

Based on the law:

75000+(-37500) = 5500V

75000-37500 = 5500V

37500 = 5500V

V = 37500/5500

V = 6.82m/s

7 0
3 years ago
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