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Novosadov [1.4K]
3 years ago
13

Are these two correct ?

Physics
1 answer:
gulaghasi [49]3 years ago
5 0

Answer:

7. Your answer is correct dear, just add the unit

8. answer is 1.17m/s²

Explanation:

queation 7.

m = 3kg, F = 9N, a ?

F = ma

a = F/m = 9/3 = 3m/s²

Use the same approach for question 8

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What’s the difference between gravitational force & gravitational field strength?
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Gravitational field strength is the force experienced by a unit mass. Gravitational force is the amount of force acting on a body. It is the product of field strength times the mass under consideration. Gravitational pull is just a more colloquial name for gravitational force.

Explanation:

hope it helps u

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Cold syrup flows sluggishly because of??​
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Water is very different from honey, syrup, glycerine, or oil. It pours easily and is not thick and sticky like the others. The property that determines how easily a liquid pours is called VISCOSITY. Water has a low viscosity; syrup has a high viscosity. Liquids with a high viscosity are said to be viscous.

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A satellite is in a circular orbit around Mars, which has a mass M = 6.40 × 1023 kg and radius R = 3.40 ×106 m.
Pepsi [2]

Answer:

a) The orbital speed of a satellite with a orbital radius R (in meters) will have an orbital speed of approximately \displaystyle \sqrt\frac{4.27 \times 10^{13}}{R}\; \rm m \cdot s^{-1}.

b) Again, if the orbital radius R is in meters, the orbital period of the satellite would be approximately \displaystyle 9.62 \times 10^{-7}\, R^{3/2}\; \rm s.

c) The orbital radius required would be approximately \rm 2.04 \times 10^7\; m.

d) The escape velocity from the surface of that planet would be approximately \rm 5.01\times 10^3\; m \cdot s^{-1}.

Explanation:

<h3>a)</h3>

Since the orbit of this satellite is circular, it is undergoing a centripetal motion. The planet's gravitational attraction on the satellite would supply this centripetal force.

The magnitude of gravity between two point or spherical mass is equal to:

\displaystyle \frac{G \cdot M \cdot m}{r^{2}},

where

  • G is the constant of universal gravitation.
  • M is the mass of the first mass. (In this case, let M be the mass of the planet.)
  • m is the mass of the second mass. (In this case, let m be the mass of the satellite.)  
  • r is the distance between the center of mass of these two objects.

On the other hand, the net force on an object in a centripetal motion should be:

\displaystyle \frac{m \cdot v^{2}}{r},

where

  • m is the mass of the object (in this case, that's the mass of the satellite.)
  • v is the orbital speed of the satellite.
  • r is the radius of the circular orbit.

Assume that gravitational force is the only force on the satellite. The net force should be equal to the planet's gravitational attraction on the satellite. Equate the two expressions and solve for v:

\displaystyle \frac{G \cdot M \cdot m}{r^{2}} = \frac{m \cdot v^{2}}{r}.

\displaystyle v^2 = \frac{G \cdot M}{r}.

\displaystyle v = \sqrt{\frac{G \cdot M}{r}}.

Take G \approx 6.67 \times \rm 10^{-11} \; m^3 \cdot kg^{-1} \cdot s^{-2},  Simplify the expression v:

\begin{aligned} v &= \sqrt{\frac{G \cdot M}{r}} \cr &= \sqrt{\frac{6.67 \times \rm 10^{-11} \times 6.40 \times 10^{23}}{r}} \cr &\approx \sqrt{\frac{4.27 \times 10^{13}}{r}} \; \rm m \cdot s^{-1} \end{aligned}.

<h3>b)</h3>

Since the orbit is a circle of radius R, the distance traveled in one period would be equal to the circumference of that circle, 2 \pi R.

Divide distance with speed to find the time required.

\begin{aligned} t &= \frac{s}{v} \cr &= 2 \pi R}\left/\sqrt{\frac{G \cdot M}{R}} \; \rm m \cdot s^{-1}\right. \cr &= \frac{2\pi R^{3/2}}{\sqrt{G \cdot M}} \cr &\approx  9.62 \times 10^{-7}\, R^{3/2}\; \rm s\end{aligned}.

<h3>c)</h3>

Convert 24.6\; \rm \text{hours} to seconds:

24.6 \times 3600 = 88560\; \rm s

Solve the equation for R:

9.62 \times 10^{-7}\, R^{3/2}= 88560.

R \approx 2.04 \times 10^7\; \rm m.

<h3>d)</h3>

If an object is at its escape speed, its kinetic energy (KE) plus its gravitational potential energy (GPE) should be equal to zero.

\displaystyle \text{GPE} = -\frac{G \cdot M \cdot m}{r} (Note the minus sign in front of the fraction. GPE should always be negative or zero.)

\displaystyle \text{KE} = \frac{1}{2} \, m \cdot v^{2}.

Solve for v. The value of m shouldn't matter, for it would be eliminated from both sides of the equation.

\displaystyle -\frac{G \cdot M \cdot m}{r} + \frac{1}{2} \, m \cdot v^{2}= 0.

\displaystyle v = \sqrt{\frac{2\, G \cdot M}{R}} \approx 5.01\times 10^{3}\; \rm m\cdot s^{-1}.

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How would you explain force and movement to a friend who doesn’t know much about it?​
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I can explain it using the law of force

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