Ask question

Ask question

All categories

3 years ago

13

Mercury, Mars, and Earth’s moon all show evidence of heavy cratering from meteor impacts while Earth shows much less evidence of

meteor impacts. The most reasonable conclusion to make from this data is that the early solar system had a much higher density of rocky material, but–
a) Earth has been resurfaced by ongoing processes.
b) the rocky material was less likely to reach Earth.
c) Earth formed much later, when the solar system had fewer meteors.
d) the protection from impact provided by Earth’s atmosphere is decreasing.

1 answer:

algol [13]3 years ago

4 0

Answer:

D)

Explanation:

You might be interested in

A car drives 100 km north, 50 km east, then 10 km south. What is the car's displavement?

Over [174]

Answer:

$\sqrt{ ({40}^{2} } + {50}^{2} ) = \sqrt{(1600 + 2500)} = \sqrt{4100} = 64.03$

6 0

3 years ago

In the bohr model, what is the ratio of its kinetic energy to its potential energy?

stich3 [128]

The centrifugal force C = mv^2/r = kq^2/r^2 = P centripetal force. m is the electron mass, q are the proton and electron charges (opposites), and r is the Bohr radius.

Thus 1/2 mv^2/r = 1/2 kq^2/r^2 and KE = 1/2 mv^2 = 1/2 kq^2/r = 1/2 PE

Therefore KE/PE = 1/2, no matter what state the electron is in.

8 0

3 years ago

A glider with mass m = 0.230 kg sits on a frictionless horizontal air track, connected to a spring with force constant k = 4.50

loris [4]

Answer

given,

mass of glider = 0.23 Kg

spring constant = k = 4.50 N/m

spring stretched to 0.130 m

The springs potential energy =

$U = \dfrac{1}{2}kx^2$

$U = \dfrac{1}{2}\times 4.5 \times 0.13^2$

U = 0.038 J

at x = 0,the only energy will be kinetic .

$\dfrac{1}{2}mv^2=0.038$

$\dfrac{1}{2}\times 0.23 \times v^2=0.038$

v² = 0.3304

v = 0.575 m/s

displacement of the glider

using conservation of energy

$\dfrac{1}{2}mv^2=\dfrac{1}{2}kx^2$

$x =v\sqrt{\dfrac{m}{k}}$

$x =3\times \sqrt{\dfrac{0.23}{4.5}}$

x = 0.678 m

8 0

3 years ago

Star A and Star B have different apparent brightnesses but identical luminosities. Star A is 10 light years away from earth and

STALIN [3.7K]

intensity of a star is inversely depends on the square of the distance from the star

we can say it is given as

$\frac{I_1}{I_2} = \frac{r_2^2}{r_1^2}$

here we know that

$\frac{I_1}{I_2} = 36 times$

also we know that

$r_1 = 10 Ly$

now we will have

$\frac{I_1}{I_2} = \frac{r_2^2}{r_1^2}$

$36 = \frac{r_2^2}{10^2}$

$r_2 = 60 Ly$

so other star is at distance 60 Light years

8 0

3 years ago

A spherical capacitor is formed from two concentric sphericalconducting shells separated by vacuum. The inner sphere has radius1

zubka84 [21]

Explanation:

(1). Formula to calculate the potential difference is as follows.

$\Delta V = -\int E dr$

= $-\int \frac{kq}{r^{2}} dr$

= $\frac{kq}{r_{f}} - \frac{kq}{r_{i}}$

= $\frac{kq(r_{f} - r_{i})}{r_{f}r_{i}}$

= $\frac{9 \times 10^{9} \times 3.30 \times 10^{-9}(0.1 - 0.015)}{0.1 \times 0.015}$

= 38.7 volts

Therefore, magnitude of the potential difference between the two spheres is 38.7 volts.

(2). Now, formula to calculate the energy stored in the capacitor is as follows.

E = $\frac{1}{2}QV$

= $\frac{1}{2} \times 3.30 \times 10^{-9} \times 3.87 V$

= $6.39 \times 10^{-8} J$

Thus, the electric-field energy stored in the capacitor is $6.39 \times 10^{-8} J$ .

7 0

3 years ago

Read 2 more answers