What is NOT an example of Refraction. Remember to not give me an example.

A melon is projected into the air with 100 j of kinetic energy in the presence of air resistance. when it returns to its initial

MAVERICK [17]

In the presence of air resistance, a watermelon is launched into the air with 100 j of kinetic energy.

Its kinetic energy is less than 100 J when it reaches its starting point. Its kinetic energy decreases as it encounters air resistance and returns to its starting point. In actuality, some of the energy has been lost because of air resistance. Since we use the ball's original height as a point of reference, there is no potential energy when the ball is in its initial state of motion, and K is its kinetic energy. This total energy is conserved if there is no air resistance, therefore when the ball returns to its starting position, its kinetic energy will remain at 100.

Learn more kinetic energy about here:

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8 0

1 year ago

Pressure is about 1000 hPa at sea level and about 500 hPa at an altitude of 5.5 km. Why doesn’t this vertical pressure gradient

saul85 [17]

Answer:

A. The upward pressure gradient force is balanced by gravity.

Explanation:

A. is correct because the pressure difference is actually generated by gravity. As in the following formula for the pressure at different points:

$p = p_0 + \rho g h$

where $p, p_0$ are the pressure at 2 points, ρ is the density of the fluid, g is the gravitational constant, and h is the height difference.

B is incorrect because friction in air is too small to make an effect.

C is incorrect because the Coriolis force is horizontal, not vertical.

D is incorrect because a difference of 500 hPa = 50000 Pa, this is half of the atmospheric pressure.

E is incorrect because temperature cannot generate force.

6 0

3 years ago

Which is an example of current electricity

Reil [10]

Answer:

A or B you choose

Explanation:

This is called current electricity or an electric current. A lightning bolt is one example of an electric current, although it does not last very long. Electric currents are also involved in powering all the electrical appliances that you use, from washing machines to flashlights and from telephones to MP3 players.

what is an electrical current, amp, ampere Current is the flow of electrons. When a circuit is closed then a current of electrons can flow and when a circuit is open then no current can flow. We can measure the flow of electrons just like you can measure the flow of water through a pipe.

4 0

3 years ago

Read 2 more answers

This is the process of water changing from a gas to a liquid. You will see clouds forming when this happens.

Goryan [66]

Its called condensation. condensation is <span>water that collects as droplets on a cold surface when humid air is in contact with it. aka. clouds</span>

4 0

3 years ago

A newly discovered planet is found to have a density if 2/3pe and a radius of 2RE, where PE and RED are the density and radius o

Liono4ka [1.6K]

Missing question in the text of the exercise. Found on internet:
"What is the acceleration due to gravity on the surface of the planet?"

Solution:
The gravitational acceleration at Earth's surface is given by:
$g= \frac{GM}{r^2}$
where
G is the gravitational constant
M is the Earth's mass
r is the Earth's radius

The Earth's mass can be rewritten also as the product between the Earth's density, $d$ , and its volume (the volume of a sphere of radius r):
$M=dV=d ( \frac{4}{3} \pi r^3)= \frac{4}{3} \pi d r^3$

Now let's call M' the mass of the new planet, r' its radius and d' its density. The acceleration due to gravity on the surface of the new planet is
$g' = \frac{GM'}{r'^2}$ (1)
so we need to find M' and r'.

The problem says the radius of the new planet is twice the Earth's radius:
$r'=2r$ (2)
and that its density is 2/3 of Earth's density:
$d'= \frac{2}{3} d$
so the mass M' of the new planet is, with respect to the Earth's mass:
$M' = d'V' = \frac{4}{3} \pi d' (r')^3 = \frac{4}{3} \pi ( \frac{2}{3}d) (2r)^3 = ( \frac{4}{3} \pi d r^3 )( \frac{16}{3}) = \frac{16}{3} M$ (3)

And if we substitute (2) and (3) into (1), we find the gravitational acceleration on the surface of the new planet:
$g'= \frac{G( \frac{16}{3}M) }{(2r)^2}= \frac{GM}{r^2} \frac{4}{3} = \frac{4}{3}g$
And since $g=9.81 m/s^2$ , we find
$g'= \frac{4}{3}(9.81 m/s^2)=13.1 m/s^2$

8 0

4 years ago