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3 years ago

6

The increases in a planet's speed as it approaches the sun is described by kepler's second law what is the best explanation for

this empirical law

1 answer:

Keith_Richards [23]3 years ago

5 0

<span>Keplers second law, that gravitational potential energy is lost when planets get near the sun and it becomes kinetic energy, best describes the reason why a planets speed will increase as it becomes closer to the sun.</span>

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A sled that has a mass of 8 kg is pulled at a 50 degree angle with a force of 20 N. The force of friction acting on the sled is

ratelena [41]

Answer:

a = 1.3 m/s2; fn = 63.1 n :)

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2 years ago

Which of these is NOT a possible type of energy transformation?

Leona [35]

It would be A because Light energy can't be used to make a Nuclear Weapon.
Electricity turns on a light bulb so B is scientifically correct.
If potential energy can be turned to Kinetic, it can also be switched. so C is scientifically correct
Ever heard of Geothermal power plants? So D is scientifically correct.
Meanwhile, A is impossible. so it is the answer you want to choose. Because the question asked for the one which is impossible.

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3 years ago

What kind of education is required to be a ocean engineer?

ra1l [238]

To become an ocean engineer, the job would require a B<span>achelors degree in ocean engineering.

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4 0

3 years ago

Suppose that a 102.5 kg football player running at 8.5 m/s catches a 0.47 kg ball moving at a speed of 22.5 m/s with his feet of

nadya68 [22]

Answer:

a) $v=8.564\ m.s^{-1}$

b) $\Delta KE=45.76\ J$

c) $v=8.358\ m.s^{-1}$

d) $\Delta KE=225.24\ J$

Explanation:

Given:

mass of the player, $m_p=102.5\ kg$

mass of the ball, $m_b=0.47\ kg$

initial velocity of the player, $v_p=8.5\ m.s^{-1}$

initial velocity of the ball, $v_b=22.5\ m.s^{-1}$

a)

<u>Case:</u> When the player and the ball are moving in the same direction.

$m_t.v=m_p.v_p+m_b.v_b$

where:

$m_t=$ total mass after the player catches the ball

v = final velocity of the system

$v=\frac{102.5\times 8.5+0.47\times 22.5}{(102.5+0.47)}$

$v=8.564\ m.s^{-1}$

b)

Initial kinetic energy of the system:

$KE_i=\frac{1}{2} [m_p.v_p^2+m_b.v_b^2]$

$KE_i=\frac{1}{2} [102.5\times 8.5^2+0.47\times 22.5^2]$

$KE_i=3821.78\ J$

Final kinetic energy of the system:

$KE_f=\frac{1}{2} m_t.v^2$

$KE_f=\frac{1}{2}\times 102.97\times 8.564^2$

$KE_f=3776.02\ J$

∴Change in kinetic energy

$\Delta KE=KE_i-KE_f$

$\Delta KE=3821.78-3776.02$

$\Delta KE=45.76\ J$

c)

<u>Case:</u> When the player and the ball are moving in the opposite direction.

$m_t.v=m_p.v_p-m_b.v_b$

$v=\frac{102.5\times 8.5-0.47\times 22.5}{(102.5+0.47)}$

$v=8.358\ m.s^{-1}$

d)

Final kinetic energy in this case:

$KE_f=\frac{1}{2} m_t.v^2$

$KE_f=0.5\times 102.97\times 8.358^2$

$KE_f=3596.54\ J$

∴Change in kinetic energy:

$\Delta KE=KE_i-KE_f$

$\Delta KE=3821.78-3596.54$

$\Delta KE=225.24\ J$

3 0

3 years ago

the graph represent the relationship between the work done by a 40.8kg students running up a flight of 10 steps and time . each

diamong [38]

Since the graph is a straight line so the slope will remain same at all points

So here slope is given as

$slope = \frac{dy}{dx} = \frac{1000 - 0}{4 - 0}$

$slope = 250$

Now here we can say that slope will represent the rate of change in the physical quantity on Y axis with respect to the quantity on X axis

So here we will have

$Slope = \frac{dW}{dt}$

so it is rate of work done

So slope of this graph will same as power of the object

6 0

3 years ago