All categories

2 years ago

Choose one inner planet and one outer planet and compare their similarities and differences.

2 answers:

7 0

They both have gas in there cloudy atmosphere which creates the color of the planets. They both have a weird and off centered magnetic field which gives them strength

defon2 years ago

4 0

In our Solar System, astronomers often divide the planets into two groups — the inner planets and the outer planets. The inner planets are closer to the Sun and are smaller and rockier. The outer planets are further away, larger and made up mostly of gas.

You might be interested in

The minimum energy needed to eject an electron from a sodium atom is 4.41 x 10-19 j. what is the maximum wavelength of light, in

Minchanka [31]

The energy of an electron as it is ejected from the atom can be calculated from the product of the Planck's constant and the frequency of the light energy. We can calculate the wavelength from the frequency we can calculate. We do as follows:

E = hv
4.41 x 10-19 = 6.62607004 × 10<span>-34 (v)
v = 6.66x10^14 /s

wavelength = speed of light / frequency
</span>
wavelength = 3x10^8 / 6.66x10^14
wavelength = 4.51x10^-7 m = 450.75 nm

5 0

3 years ago

A Hooke's law spring is compressed 12.0 cm from equilibrium, and the potential energy stored is 72.0 J. What compression (as mea

Anuta_ua [19.1K]

Answer:14.14 cm

Explanation:

Given

Spring Compression $x=12 cm$

Potential energy Stored in spring $=72 J$

Suppose k is the spring constant of spring

Potential Energy of spring is given by $=\frac{kx^2}{2}$

$\frac{k(0.12)^2}{2}=72$

$k(0.12)^2=144$

$k=10,000 N/m$

$k=10 kN/m$

for 100 J energy

$\frac{k(x_0)^2}{2}=100$

$10\times 10^3\cdot (x_0)^2=200$

$(x_0)^2=2\times 10^{-2}$

$x_0=0.1414$

$x_0=14.14 cm$

6 0

2 years ago

What is endurance? a). the ability to run faster b). a combination of balance and coordination c). how much you can stretch d).

Alex787 [66]

Answer:

D. the ability to exercise for longer periods of time

Explanation:

For example, when someone does endurance training, they are stretching their body's ability to do a certain exercise for longer times as opposed to increasing strength.

8 0

3 years ago

A 2.30-kg cylindrical rod of length 2.00 m is suspended from a horizontal bar so it is free to swing about that end. a solid sph

Marina86 [1]

Solution:

initial sphere mvr = final sphere mvr + Iω
where I = mL²/3 = 2.3g * (2m)² / 3 = 3.07 kg·m²
0.25kg * (12.5 + 9.5)m/s * (4/5)2m = 3.07 kg·m² * ω
where: ω = 2.87 rad/s

So for the rod, initial E = KE = ½Iω² = ½ * 3.07kg·m² * (2.87rad/s)²
E = 12.64 J becomes PE = mgh, so
12.64 J = 2.3 kg * 9.8m/s² * h
h = 0.29 m

h = L(1 - cosΘ) → where here L is the distance to the CM
0.03m = 1m(1 - cosΘ) = 1m - 1m*cosΘ
Θ = arccos((1-0.29)/1) = 44.77 º

8 0

2 years ago

a 42.3 kg girl and a 7.93 kg sled are on the surface of a frozen lake, 15.0m apart and linked by a rope, but not moving yet. the

ycow [4]

Answer:

they meet from the girl's original position at: 2.37 (meters)

Explanation:

We need to use the Newton's law, exactly the second law that relate force, mass and acceleration as: $F=m*a$ with this we can get both accelerations; solving for acceleration $a=\frac{F}{m}$ . Now $a_{girl}=\frac{5.76}{42.3}=0.14 (m/s^{2})$ and $a_{sled}=\frac{5.76}{7.93}=0.73(m/s^{2})$ . Then knowing that they both travel at the same time and assuming that the distance among the girl and the sled is: 15.0-x, so, $x=\frac{1}{2}*a_{girl}*t^{2}$ and $15.0-x=\frac{1}{2}*a_{sled}*t^{2}$ , solving for the time we get: $t=\sqrt{\frac{2x}{a_{girl} } }$ and $t=\sqrt{\frac{2*(15.0-x)}{a_{sled} } }$ with this equations we solving for the x that is the distance between the girl and the sled after the apply the force and we get: $\sqrt{\frac{2x}{a_{girl}}} = \sqrt{\frac{2*(15.0-x)}{a_{sled} }$ . Finally we get: $\frac{x}{a_{girl} }=\frac{(15.0-x)}{a_{sled} }$ and replacing the values we have got: $\frac{x}{0.14} =\frac{(15.0-x)}{0.73}$ so $5.33*x=15-x$ so x=2.37 (meters).

5 0

3 years ago