Which best supports the idea that the surface of the moon has changed very little?

1 answer:

8 0

The correct answer is<span> c) Because the moon has no atmosphere, it is not possible for geologic events to occur on the moon.

Atmosphere is related to geological change because of things such as wind and water erosion that over time change the landscape, and there's neither on the moon. It's just a floating rock.</span>

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A fully loaded cart with a mass of 2200 kg starts from the top of a 12-meter hill on a roller coaster.

Salsk061 [2.6K]

Answer:

A. potential energy is 258720 Joule

Explanation:

A.Gravitational potential energy is: PE = m × g × h

velocity = 15.33 m/s when the car reaches the bottom of the hill.

where, m = mass

g = acceleration due to gravity

h = height from the bottom of hill.

The potential energy is : m×g×h

=(2200×9.8×12)

=258720 Joule

B. at the bottom of the hill, the potential energy is converted into kinetic energy so PE at top = KE at bottom

kinetic energy= $\frac{1}{2}$ ( $m*v^{2}$ )

where v = velocity

m= mass

therefore, v= $\sqrt\frac{2*K.E}{m} {}$

or, v= $\sqrt{\frac{2*258720}{2200} }$

or, v=15.33 m/s

7 0

3 years ago

An isolated conducting sphere has a 17 cm radius. One wire carries a current of 1.0000020 A into it. Another wire carries a curr

notsponge [240]

14 ms is required to reach the potential of 1500 V.

<u>Explanation:</u>

The current is measured as the amount of charge traveling per unit time. So the charge of electrons required for each current is determined as the product of current with time.

$Charge = Current \times Time$

As two different current is passing at two different times, the net charge will be the different in current. So,

$\text { Charge }=(1.0000020-1.0000000) \times t=2 \times 10^{-6} \times t$

The electric voltage on the surface of cylinder can be obtained as the ratio of charge to the radius of the cylinder.

$V=\frac{k q}{R}$

Here $k = 9 * 10^9$ , q is the charge and R is the radius. As $q=2 \times 10^{-6} \times t$ and R =17 cm = 0.17 m, then the voltage will be

$V=\frac{9 \times 10^{9} \times 2 \times 10^{-6} \times t}{0.17}$

The time is required to find to reach the voltage of 1500 V, so

$1500 =\frac{9 \times 10^{9} \times 2 \times 10^{-6} \times t}{0.17}$

$\begin{aligned}&t=\frac{1500 \times 0.17}{\left(9 \times 10^{9} \times 2 \times 10^{-6}\right)}\\&t=14.1666 \times 10^{-3} s=14\ \mathrm{ms}\end{aligned}$