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Dimas [21]
2 years ago
10

What is the gravitational force of attraction between two asteroids in space, each with a mass of 50,000 kg, separated

Physics
1 answer:
Fynjy0 [20]2 years ago
5 0

Answer:

1.16 x 10⁻⁸N

Explanation:

Given parameters:

Mass of the two asteroids  = 50000kg

Distance between the asteroids  = 3800m

Unknown:

Gravitational force between the two asteroids = ?

Solution:

To solve this problem, we apply the newton's law of universal gravitation:

  Fg  = \frac{G mass 1 x mass 2}{r^{2} }  

 G is the universal gravitation constant

 r is the distance between them

So;

       Fg  = \frac{6.67 x 10^{-11}  x 50000 x 50000}{3800^{2} }   = 1.16 x 10⁻⁸N

You might be interested in
An object of height 2.7 cm is placed 29 cm in front of a diverging lens of focal length 16 cm. Behind the diverging lens, and 12
kati45 [8]

Answer:

A)  q = -8.488 cm ,  B)  m = 0.29

Explanation:

A) For this exercise in geometric optics, we will use the equation of the constructor

          \frac{1}{f} = \frac{1}{p} +  \frac{1}{q}

where p and q are the distance to the object and image, respectively and f is the focal length

in our case the distance the object is p = 29 cm the focal length of a diverging lens is negative and indicates that it is f = - 12 cm

         \frac{1}{q} = \frac{1}{f} - \frac{1}{p}

     

we calculate

          \frac{1}{q} = - \frac{1}{12} - \frac{1}{29}

          \frac{1}{q} = - 0.1178

          q = -8.488 cm

the negative sign indicates that the image is virtual

B) the magnification is given

          m = \frac{h'}{h} = - \frac{q}{p}

       

we substitute

          m = - \frac{-8.488}{29}

          m = 0.29

the positive sign indicates that the image is right

4 0
3 years ago
How do satellites stay in orbit?
Vinil7 [7]

Answer:A satellite maintains its orbit by balancing two factors its velocity the speed it takes to travel in a straight line and the gravitational pull that Earth has on it. A satellite orbiting closer to the Earth requires more velocity to resist the stronger gravitational pull.

Explanation:

6 0
3 years ago
Read 2 more answers
The question states: two large, parallel conducting plates are 12cm
AnnZ [28]

Answer:

1. 24375 N/C

2. 2925 V

Explanation:

d = 12 cm = 0.12 m

F = 3.9 x 10^-15 N

q = 1.6 x 10^-19 C

1. The relation between the electric field and the charge is given by

F = q E

So, E=\frac{F}{q}

E=\frac{3.9 \times 10^{-15}}{1.6 \times 10^{-19}}

E = 24375 N/C

2. The potential difference and the electric field is related by the given relation.

V = E x d

where, V be the potential difference, E be the electric field strength and d be the distance between the electrodes.

By substituting the values, we get

V = 24375 x 0.12 = 2925 Volt

6 0
3 years ago
What are the average velocity and average acceleration of the tip of a 2.4 cm long hour hand of a clock?
ra1l [238]
A circle has a revolution of 360°. Since there are 12 hour markings, each hour interval has an angle of 30°. In radians, that would be equal to π/6 radians. So, in every 1 hour that passes, it covers π/6 of an angle. So, the angular velocity denoted as ω is π/6 ÷ 1 hour = π/6 rad/h. We can compute the average linear velocity, v, from the relationship:

v = rω, where r is the radius of the circle which is the length of the hour hand
v = (2.4 cm)(π/6 rad/h)
v = 1.257 cm/hour

Therefore, the average velocity is 1.257 cm per hour.

For the average acceleration, it is equal to zero. The hands of the clock move at a constant velocity. Since acceleration is the change of velocity per unit time, there is no change of velocity because it's constant. That's why it is zero.

8 0
3 years ago
Starting from rest, a basketball rolls from the top to the bottom of a hill, reaching a translational speed of 6.1 m/s. Ignore f
tatiyna

Answer:

a) h=3.16 m, b)  v_{cm }^ = 6.43 m / s

Explanation:

a) For this exercise we can use the conservation of mechanical energy

Starting point. Highest on the hill

           Em₀ = U = mg h

final point. Lowest point

           Em_{f} = K

Scientific energy has two parts, one of translation of center of mass (center of the sphere) and one of stationery, the sphere

           K = ½ m v_{cm }^{2} + ½ I_{cm} w²

angular and linear speed are related

           v = w r

           w = v / r

            K = ½ m v_{cm }^{2} + ½ I_{cm} v_{cm }^{2} / r²

            Em_{f} = ½ v_{cm }^{2} (m + I_{cm} / r2)

as there are no friction losses, mechanical energy is conserved

             Em₀ = Em_{f}

             mg h = ½ v_{cm }^{2} (m + I_{cm} / r²)         (1)

             h = ½ v_{cm }^{2} / g (1 + I_{cm} / mr²)

for the moment of inertia of a basketball we can approximate it to a spherical shell

             I_{cm} = ⅔ m r²

we substitute

            h = ½ v_{cm }^{2} / g (1 + ⅔ mr² / mr²)

            h = ½ v_{cm }^{2}/g    5/3

             h = 5/6 v_{cm }^{2} / g

           

let's calculate

           h = 5/6 6.1 2 / 9.8

           h = 3.16 m

b) this part of the exercise we solve the speed of equation 1

          v_{cm }^{2} = 2m gh / (1 + I_{cm} / r²)

in this case the object is a frozen juice container, which we can simulate a solid cylinder with moment of inertia

              I_{cm} = ½ m r²

we substitute

             v_{cm } = √ [2gh / (1 + ½)]

             v_{cm } = √(4/3 gh)

let's calculate

             v_{cm } = √ (4/3 9.8 3.16)

             v_{cm }^ = 6.43 m / s

4 0
3 years ago
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