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3 years ago

Describe what happens when the rock on the top of a hill is pushed.

Physics

1 answer:

Zigmanuir [339]3 years ago

6 0

Answer:

The precise point is that gravitational energy is potential energy unless it makes something move, and then the energy is converted to kinetic energy, no work can be done. So a big rock at the top of a hill has no kinetic energy. Its only when it rolls down that work is done, and this can be converted to useful energy.

Explanation:

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Io and Europa are two of Jupiter's many moons. The mean distance of Europa from Jupiter is about twice as far as that for Io and

Anastaziya [24]

The universal gravitation law and Newton's second law allow us to find that the answer for the relation of the rotation periods of the satellites is:

$\frac{T_{Eu}}{T_{Io}}$ = 2.83

The universal gravitation law states that the force between two bodies is proportional to their masses and inversely proportional to their distance squared

$F = G \frac{Mm}{r^2}$

Where G is the universal gravitational constant (G = 6.67 10⁻¹¹ $\frac{N m^2 }{kg^2}$ ), F the force, m and m the masses of the bodies and r the distance between them

Newton's second law states that force is proportional to the mass and acceleration of bodies

F = m a

Where F is the force, m the mass and the acceleration

In this case the body is the satellites of Jupiter and the planet,

$G \frac{Mm}{r^2} = m a$

Suppose the motion of the satellites is circular, then the acceleration is centripetal

a = $\frac{v^2}{r}$ r

Where v is the speed of the satellite and r the distance to the center of the planet

we substitute

$G \frac{Mm}{r^2} = m \frac{v^2}{r} \\G \frac{M}{r} = v^2$

Since the speed is constant, we can use the uniform motion ratio

v = $\frac{\Delta x}{t}$

In the case of a complete orbit, the time is called the period.

The distance traveled is the length of the orbit circle

Δx = 2π r

We substitute

$G \frac{M}{r} = (\frac{2 \pi r}{T} )^2 \\T^2 = (\frac{4 \pi ^2}{GM}) \ r^3$

Let's write this expression for each satellite

Io satellite

Let's call the distance from Jupiter is

r = $r_{Io}$

$T_{Io}^2 = (\frac{4 \pi ^2}{GM} ) \ r_{Io}^3$ TIo² = (4pi² / GM) rIo³

Europe satellite

Distance from Jupiter is

$r_{Eu} = 2 \ r_{Io}$

We calculate

$T_{Eu} = ( \frac{4\pi ^2 }{GM} (2 \ r_{Io})^3\\T_{Eu} = ( \frac{4 \pi ^2 }{GM}) r_{Io} \ 8$

$T_{Eu}^2 = 8 T_{Io}^2$

$\frac{ T_{Eu}}{T_{Io}} = \sqrt{8} = 2.83$

In conclusion, using the universal gravitation law and Newton's second law, we find that the answer for the relationship of the relation periods of the satellites is:

$\frac{T_{Eu}}{T_{Io}} = 2.83$

Learn more about universal gravitation law and Newton's second law here:

brainly.com/question/10693965

6 0

3 years ago

A commonly used unit of mass in the English system is the pound-mass, abbreviated lbm, where 1 lbm = 0.454 kg. What is the densi

Gemiola [76]

Answer: The density of glycerine will be $0.028lbm/foot^3$

Explanation:

Density is defined as the mass contained per unit volume.

$Density=\frac{mass}{Volume}$

Given:

Density of glycerine= $1.26\times 10^3kg/m^3$

1 lbm = 0.454 kg

1 kg = $\frac{1}{0.454}=2.2lbm$

Thus $0.454kg=\frac{2.2}{1}\times 0.454=0.99$

Also $1m^3=35.3147foot^3$

Putting in the values we get:

$Density=\frac{0.99lbm}{35.3147}=0.028lbm/foot^3$

Thus density of glycerine will be $0.028lbm/foot^3$

3 0

3 years ago

An object moves along the x-axis. Its position is given by the equation y( X = 4t^2 - 41t+ 78 \). Find the position of the

prohojiy [21]

The object changes direction when its velocity changes sign. You can get the velocity function by differentiating the position function with respect to time t :

x(t) = 4t ² - 41t + 78

→ v(t) = dx(t)/dt = 8t - 41

Solve v(t) = 0:

8t - 41 = 0

8t = 41

t = 41/8 = 5.125

Just to confirm that the velocity indeed changes sign:

• Pick any time before this one to check the sign of v :

v (0) = 8•0 - 41 = -41 < 0

• Pick any time after and check the sign again:

v (6) = 8•6 - 41 = 7 > 0

Now just find the position at this time:

x (5.125) = -433/16 = -27.0625

which means the object is 27.0625 units on the negative x-axis.

You can also do this without calculus by completing the square in the position function:

4t ² - 41t + 78 = 4 (t ² - 41/4 t ) + 78

… = 4 (t ² - 2• 41/8 t + (41/8)² - (41/8)²) + 78

… = 4 (t ² - 2• 41/8 t + (41/8)²) - 4•1681/64 + 78

… = 4 (t - 41/8)² - 433/16

which describes a parabola that opens upward. When t = 41/8 = 5.125, the quadratic term vanishes and the turning point of the parabola occurs at a position of -433/16 units.

7 0

3 years ago

Shock best describes the current model of the atom ?

const2013 [10]

Answer:

IDK?//

Explanation:

8 0

3 years ago

An endangered species is a species that has died out and no individuals are left. True or false

valentinak56 [21]

Answer:

the answer is false... ....

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3 years ago

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