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oksano4ka [1.4K]
3 years ago
7

What describes Newton’s law of universal gravitation

Physics
1 answer:
VikaD [51]3 years ago
8 0

Newton's law of universal gravitation gives the gravitational force between two objects:

F = GMm/r²

F = gravitational force, G = gravitational constant, M & m are the masses of the two objects, r = distance between the objects

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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}

              c = 28853.78 \ m^2 /s^2

     

5 0
3 years ago
A 3 kg toy car with a speed of 7 m/s collides head-on with a 2 kg car traveling in the opposite direction with a speed of 5 m/s.
Mademuasel [1]

Answers:

kinetic energy lost = 86.4J

Explanation:

let Kf be the kinetic energy after the collision and Ki be the kinetic energy before the collision. let the 3kg car be 1 and 2kg car be 2.

Kf = K1(f) + K2(f)

Ki = K1(i) + k2(i)

loss in kinetic energy = Kf - Ki

                                 = 1/2(3)(2.20)^2 + 1/2(2)(2.20)^2 - 1/2(3)(7)^2 - 1/2(2)(-5)^2

                                 = 12.1 - 98.5

                                 = -86.4 J

therefore, the kinetic energy lost in the collision is 86.4 J.

6 0
2 years ago
on the surface of planet x a body with a mass of 10 kilograms weighs 40 newtons. The magnitude of the acceleration due to gravit
tamaranim1 [39]
Based on the Newton's second law of motion, the value of the net force acting on the object is equal to the product of the mass and the acceleration due to gravity. If we let a be the acceleration due to gravity, the equation that would allow us to calculate it's value is,
      W = m x a
where W is weight, m is mass, and a is acceleration. Substituting the known values,
    40 kg m/s² = (10 kg) x a
Calculating for the value of a from the equation will give us an answer equal to 4. 
ANSWER: 4 m/s². 

6 0
3 years ago
Read 2 more answers
The value 10.00 has____ significant figures
natka813 [3]

Answer:

4

Explanation:

You assume that when they were writing two zeroes after the decimal point, they wanted to show that these are significant in determining the precision of this number.

8 0
3 years ago
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Define the Earth's atmosphere
Alex Ar [27]
<span> The blanket of gas on the surface of a planet or satellite. Note: The </span>atmosphere<span> of the </span>Earth<span> is roughly eighty percent nitrogen and twenty percent oxygen, with traces of other gases. (See ionosphere, stratosphere, and troposphere.)

</span>
7 0
3 years ago
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