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Rama09 [41]
3 years ago
15

Scientists can estimate the age of a planetary surface by counting __________. scientists can estimate the age of a planetary su

rface by counting __________. volcanoes tectonic features impact craters erosion features
Physics
1 answer:
Viktor [21]3 years ago
4 0
The surface of the planets can explain the dynamics of planetary interiors (for example, how often volcanoes erupt or how often earthquakes occur). In order to see how much the surface has changed over time the age of the surface should be obtained. Scientists can estimate the age of a planetary surface by counting craters. The logic is that t<span>he largest bodies (the ones that would form the largest craters) were used up before the smaller ones, since there were fewer of the larger ones to start with. So the larger a crater is, the older the planet surface.</span>
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a pennyfarthing is a style of a bicycle with a very large front wheel and a small real wheel, the cyclist who sit high above and
s344n2d4d5 [400]

Answer:

In one rotation, the large wheel turns 4m.

Explanation:

The given values are:

Input distance,

= 0.64 m

Mechanical advantage,

= 0.16

As we know,

⇒ Out. \ Distance = \frac{Inp. \ distance}{Mechanical \ advantage}

On putting the values, we get

⇒                         =\frac{0.64}{0.16}

⇒                         =4 \ m

4 0
3 years ago
Which of the following is the correct name for CCl4
Ira Lisetskai [31]

Answer: Carbon tetrachloride Or Tetrachloromethane

Explanation: Carbon tetrachloride is an important nonpolar covalent compound. You determine its name based on the atoms present in the compound.

7 0
3 years ago
Read 2 more answers
The brainstem is to breathing and arousal as the limbic system is to memories and: (2 points)
riadik2000 [5.3K]

Answer:

The brainstem is to breathing and arousal as the limbic system is to memories and emotion. See the explanation below, please.

Explanation:

The limbic system develops certain responses to emotional stimuli (joy, anger, sadness, fear, pleasure). It interacts with the autonomic endocrine and nervous system. It is also related to sexuality, memory, attention, memories. It includes some areas such as hypothalamus, tonsil, hippocampus, among others.

8 0
3 years ago
A person swings a 0.546-kg tether ball tied to a 4.56-m rope in an approximately horizontal circle. If the maximum tension the r
Murrr4er [49]

Answer:

2.1 rad/s

Explanation:

Given that,

Mass of a tether ball, m = 0.546 kg

Length of a rope, l =  4.56 m

The maximum tension the rope can withstand before breaking is 11.0 N

We need to find the maximum angular speed of the ball. Let v is the linear velocity. The maximum tension is balanced by the centripetal force acting on it. It can be given by :

F=\dfrac{mv^2}{r}\\\\v=\sqrt{\dfrac{Fr}{m}} \\\\v=\sqrt{\dfrac{11\times 4.56}{0.546}} \\\\=9.584\ m/s

Let \omega is the angular speed of the ball. The relation between the angular speed and angular velocity is given by :

v=r\omega\\\\\omega=\dfrac{v}{r}\\\\=\dfrac{9.584}{4.56}\\\\=2.1\ rad/s

So, the maximum angular speed of the ball is 2.1 rad/s.

4 0
3 years ago
X rays of wavelength 0.0169 nm are directed in the positive direction of an x axis onto a target containing loosely bound electr
mamaluj [8]

Answer:

a) 4.04*10^-12m

b) 0.0209nm

c) 0.253MeV

Explanation:

The formula for Compton's scattering is given by:

\Delta \lambda=\lambda_f-\lambda_i=\frac{h}{m_oc}(1-cos\theta)

where h is the Planck's constant, m is the mass of the electron and c is the speed of light.

a) by replacing in the formula you obtain the Compton shift:

\Delta \lambda=\frac{6.62*10^{-34}Js}{(9.1*10^{-31}kg)(3*10^8m/s)}(1-cos132\°)=4.04*10^{-12}m

b) The change in photon energy is given by:

\Delta E=E_f-E_i=h\frac{c}{\lambda_f}-h\frac{c}{\lambda_i}=hc(\frac{1}{\lambda_f}-\frac{1}{\lambda_i})\\\\\lambda_f=4.04*10^{-12}m +\lambda_i=4.04*10^{-12}m+(0.0169*10^{-9}m)=2.09*10^{-11}m=0.0209nm

c) The electron Compton wavelength is 2.43 × 10-12 m. Hence you can use the Broglie's relation to compute the momentum of the electron and then the kinetic energy.

P=\frac{h}{\lambda_e}=\frac{6.62*10^{-34}Js}{2.43*10^{-12}m}=2.72*10^{-22}kgm\\

E_e=\frac{p^2}{2m_e}=\frac{(2.72*10^{-22}kgm)^2}{2(9.1*10^{-31}kg)}=4.06*10^{-14}J\\\\1J=6.242*10^{18}eV\\\\E_e=4.06*10^{-14}(6.242*10^{18}eV)=0.253MeV

5 0
3 years ago
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