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mel-nik [20]
3 years ago
15

The moon travels in a nearly circular path around earth. If somehow gravitation between earth and moon disappeared, how would th

e moon's path differ?

Physics
2 answers:
erma4kov [3.2K]3 years ago
7 0
The gravitational force is a balance between centripetal and centrifugal force that keeps the planets in their orbits when they are in motion. The gravitational force is a factor of the masses of the Earth and moon and the distance between them. If there were no gravitational force, the moon would stay out of the circular path and follow the direction as shown in the picture.

m_a_m_a [10]3 years ago
4 0

Answer:

moon must follow a movement tangential to the circle at the point where the force disappears

Explanation:

The gravitational force is the attraction due to the mass of the bodies, if this force disappears the centripetal acceleration between the moon and the Earth would disappear and using the first Newton law, every body remains in uniform motion until an external force changes this movement

 

Consequently, the moon must follow a movement tangential to the circle at the point where the force disappears

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PLS HELP!!! A 1200-kg whale swims horizontally to the right at a speed of 6.0 m/s. It suddenly collides directly with a stationa
Fittoniya [83]

Answer:

7200 kg.m/s

Explanation:

According the law of conservation of linear momentum, the sum of momentum before and after collision are equal.

Using this principle, the sum of initial momentum will be given as p=mv where p is momentum, m is mass and v is velocity

Initial momentum

Mass of whale*initial velocity of whale + mass of seal*initial seal velocity

Since the seal is initially stationary, its velocity is zero. By substitution and taking right direction as positive

Initial momentum will be

1200*6+(280*0)=7200 kg.m/s

Since both initial and final momentum should be equal, hence the final momentum will also be 7200 kg.m/s

7 0
3 years ago
BRAINLIEST WELL BE AWARDED.<br>what is an alloy?​
DedPeter [7]

Answer:

Alloy, metallic substance composed of two or more elements, as either a compound or a solution. The components of alloys are ordinarily themselves metals, though carbon, a nonmetal, is an essential constituent of steel.

Explanation:

Alloys are usually produced by melting the mixture of ingredients. The value of alloys was discovered in very ancient times; brass (copper and zinc) and bronze (copper and tin) were especially important. Today, the most important are the alloy steels, broadly defined as steels containing significant amounts of elements other than iron and carbon. The principal alloying elements for steel are chromium, nickel, manganese, molybdenum, silicon, tungsten, vanadium, and boron have a wide range of special properties, such as hardness, toughness, corrosion resistance, magnetizability, and ductility. Nonferrous alloys, mainly copper–nickel, bronze, and aluminum alloys, are much used in coinage. The distinction between an alloying metal and an impurity is sometimes subtle; in aluminum, for example, silicon may be considered an impurity or a valuable component, depending on the application, because silicon adds strength though it reduces corrosion resistance.

8 0
3 years ago
Please help me with this
mezya [45]

Answer:

.067 so C

Explanation:

I asked my sister who is in 2nd grade and she said it was right so you are good! =). have a great day!

5 0
3 years ago
1. What is the pull that all objects exert on each other?
Andrei [34K]

Answer:

Im pretty sure 1 is gravity, 2 is force

Explanation:

8 0
3 years ago
The train passes point A with a speed of 30 m/s and begins to decrease its speed at a constant rate of at = - 0.25 m/s^2. Determ
prisoha [69]

Explanation:

At point B, the velocity speed of the train is as follows.

          \nu^{2}_{B} = \nu^{2}_{A} + 2a_{t} (s_{B} - s_{A})

                           = (30)^{2} + 2(-0.25(412 - 0))

                           = 26.34 m/s

Now, we will calculate the first derivative of the equation of train.

          y = 200 e^{\frac{x}{1000}}

      \frac{dy}{dx} = 0.2 e^{\frac{x}{1000}}

Now, second derivative of the train is calculated as follows.

         \frac{dy}{dx} = 0.2 e^{\frac{x}{1000}}      

       \frac{d^{2}y}{dx^{2}} = 0.2 (10^{-3}) e^{\frac{x}{1000}}    

Radius of curvature of the train is as follows.  

   \rho = \frac{[1 + (\frac{dy}{dx})^{2}]^{\frac{3}{2}}}{\frac{d^{2}y}{dx^{2}}}

               = \frac{[1 + 0.2e^{\frac{400}{1000}}^{2}]^{\frac{3}{2}}}{0.2(10^{-3})e^{\frac{400}{1000}}}

              = 3808.96 m

Now, we will calculate the normal component of the train as follows.

            a_{n} = \frac{\nu^{2}_{B}}{\rho}

                        = \frac{(26.34)^{2}}{3808.96}

                        = 0.1822 m/s^{2}

The magnitude of acceleration of train is calculated as follows.

            a = \sqrt{(a_{t})^{2} + (a_{n})^{2}}

               = \sqrt{(-0.25)^{2} + (0.1822)^{2}}

              = 0.309 m/s^{2}

Thus, we can conclude that magnitude of the acceleration of the train when it reaches point B, where sAB = 412 m is 0.309 m/s^{2}.

6 0
3 years ago
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