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Liula [17]
3 years ago
14

Two spherical objects have masses of 3.1 x 10^5 kg and 6.5 x 10^3 kg. The gravitational attraction between them is 65 N. How far

apart are their centers? (answer to 2 digits)
Physics
1 answer:
nata0808 [166]3 years ago
3 0

Answer:

4.55 x 10⁹m

Explanation:

Given parameters:

Mass of object 1  = 3.1 x 10⁵kg

Mass of object 2 = 6.5 x 10³kg

Gravitational force  = 65N

Unknown:

Distance between them  = ?

Solution:

To solve this problem, we use the expression below from the universal gravitational law;

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

   G = 6.67 x 10⁻¹¹

        65  = \frac{6.67 x 10^{11} x 3.1 x 10^{5} x 6.5 x 10^{3}   }{distance^{2} }    

   Distance  = 4.55 x 10⁹m

         

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Answer:

t = 39.60 s

Explanation:

Let's take a careful look at this interesting exercise.

In the first case the two motors apply the force in the same direction

            F = m a₀          

           a₀ = F / m

with this acceleration it takes t = 28s to travel a distance, starting from rest

           x = v₀ t + ½ a t²

           x = ½ a₀ t²

           t² = 2x / a₀

           28² = 2x /a₀          (1)

in a second case the two motors apply perpendicular forces

we can analyze this situation as two independent movements, one in each direction

           

in the direction of axis a, there is a motor so its force is F/2

               

the acceleration on this axis is

          a = F/2m

          a = a₀ / 2

so if we use the distance equation

             x = v₀ t + ½ a t²

as part of rest v₀ = 0

             x = ½ (a₀ / 2) t²

             

let's clear the time

             t² = (2x / a₀)  2

we substitute the let of equation 1

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4 0
3 years ago
Formular for coefficient of friction<br>​
Bumek [7]

Answer:

hope you like it mark as brainliest

Explanation:

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3 years ago
Where c is a constant that depends on the initial gas pressure behind the projectile. The initial position of the projectile is
Ronch [10]

Complete Question

A gas gun uses high pressure gas tp accelerate projectile through the gun barrel.

If the acceleration of the projective is : a = c/s m/s​2

Where c is a constant that depends on the initial gas pressure behind the projectile. The initial position of the projectile is s= 1.5m and the projectile is initially at rest. The projectile accelerates until it reaches the end of the barrel at s=3m. What is the value of the constant c such that the projectile leaves the barrel with velocity of 200m/s?

Answer:

The value of the constant is  c = 28853.78 \ m^2 /s^2

Explanation:

From the question we are told that

         The acceleration is  a =  \frac{c}{s}\   m/s^2

         The  initial position of the projectile is s= 1.5m

         The final position of the projectile is s_f =  3 \ m

          The velocity is  v = 200 \ m/s

     Generally  time  =  \frac{ds}{dv}

   and  acceleration is a =  \frac{v}{time }

so

            a = v  \frac{dv}{ds}

 =>        vdv  =  a ds

             vdv  = \frac{c}{s}  ds

integrating both sides

           \int\limits^a_b  vdv  = \int\limits^c_d \frac{c}{s}  ds

Now for the limit

          a =  200 m/s

             b = 0 m/s  

         c = s= 3 m

          d =s_f= 1.5 m

So we have  

           \int\limits^{200}_{0}  vdv  = \int\limits^{3}_{1.5} \frac{c}{s}  ds

              [\frac{v^2}{2} ] \left | 200} \atop {0}} \right.  = c [ln s]\left | 3} \atop {1.5}} \right.

            \frac{200^2}{2}  =  c ln[\frac{3}{1.5} ]

=>           c = \frac{20000}{0.69315}

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andrey2020 [161]

Answer:

m1/6 ( c )

Explanation:

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After first collision

m1v = m1v1 + m2v2 --- ( 1 )

After second collision

m2v2 = m2v2 + m3v3   ---- ( 2 )

combining equations 1 and 2

m1v = m1v1 + m2v2 + m3v3  ----- ( 3 )

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solving with equation 9 and 10

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8 0
3 years ago
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