How long does it take for the moon to revolve once

Most geologists accept radiometric dating techniques as valid because

Rus_ich [418]

Most geologists accept radiometric dating techniques as valid because radioactive elements decay at a constant and measurable rate.

Answer: Option C

<u>Explanation:</u>

Scientists prefer radioactive dating to carbon dating because it is more accurate in measuring. The analysis depends upon the radioactive decay of radioactive isotopes of any matter in a given rock or soil.

The parent atoms and daughter atoms are compared while studying, and hence age can be calculated easily. Radioactive decay depends upon the given half-life of the atom, which is a constant and is known. So, it would be very easy to calculate the number of progeny atoms and parent atoms and find out their age.

7 0

3 years ago

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Because white dwarfs are small, as their name implies, they are hard to see. What is a way astronomers have to find white dwarfs

Bumek [7]

Astronomers find white dwarfs that distinguish them from main sequence stars because white dwarfs get really hot, we can search for their ultraviolet radiation.

<h3>What is a white dwarf?</h3>

A white dwarf is a very hot star that radiated much energy in the form of ultraviolet radiation.

This UV radiation is initially very bright and then these stars become red with time.

In conclusion, Astronomers find white dwarfs they can search for their ultraviolet radiation.

Learn more about white dwarfs here:

brainly.com/question/19602278

#SPJ1

3 0

2 years ago

Which of the following has no energy?

Readme [11.4K]

The correct answer is D: which is none of the above.
Hint: a wreckling ball contains pontential energy that acts like a pendulum

a pot of water contains pontential thermal energy

4 0

3 years ago

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A 125-kg astronaut (including space suit) acquires a speed of 2.50 m/s by pushing off with her legs from a 1900-kg space capsule

ryzh [129]

(a) 0.165 m/s

The total initial momentum of the astronaut+capsule system is zero (assuming they are both at rest, if we use the reference frame of the capsule):

$p_i = 0$

The final total momentum is instead:

$p_f = m_a v_a + m_c v_c$

where

$m_a = 125 kg$ is the mass of the astronaut

$v_a = 2.50 m/s$ is the velocity of the astronaut

$m_c = 1900 kg$ is the mass of the capsule

$v_c$ is the velocity of the capsule

Since the total momentum must be conserved, we have

$p_i = p_f = 0$

so

$m_a v_a + m_c v_c=0$

Solving the equation for $v_c$ , we find

$v_c = - \frac{m_a v_a}{m_c}=-\frac{(125 kg)(2.50 m/s)}{1900 kg}=-0.165 m/s$

(negative direction means opposite to the astronaut)

So, the change in speed of the capsule is 0.165 m/s.

(b) 520.8 N

We can calculate the average force exerted by the capsule on the man by using the impulse theorem, which states that the product between the average force and the time of the collision is equal to the change in momentum of the astronaut:

$F \Delta t = \Delta p$

The change in momentum of the astronaut is

$\Delta p= m\Delta v = (125 kg)(2.50 m/s)=312.5 kg m/s$

And the duration of the push is

$\Delta t = 0.600 s$

So re-arranging the equation we find the average force exerted by the capsule on the astronaut:

$F=\frac{\Delta p}{\Delta t}=\frac{312.5 kg m/s}{0.600 s}=520.8 N$

And according to Newton's third law, the astronaut exerts an equal and opposite force on the capsule.

(c) 25.9 J, 390.6 J

The kinetic energy of an object is given by:

$K=\frac{1}{2}mv^2$

where

m is the mass

v is the speed

For the astronaut, m = 125 kg and v = 2.50 m/s, so its kinetic energy is

$K=\frac{1}{2}(125 kg)(2.50 m/s)^2=390.6 J$

For the capsule, m = 1900 kg and v = 0.165 m/s, so its kinetic energy is

$K=\frac{1}{2}(1900 kg)(0.165 m/s)^2=25.9 J$

3 0

3 years ago

A small logo is embedded in a thick block of crown glass (n = 1.52), 4.70 cm beneath the top surface of the glass. The block is

harkovskaia [24]

The concept required to solve this problem is the optical relationship that exists between the apparent depth and actual or actual depth. This is mathematically expressed under the equations.

$d'w = d_w (\frac{n_{air}}{n_w})+d_g (\frac{n_{air}}{n_g})$

Where,

$d_g =$ Depth of glass

$n_w =$ Refraction index of water

$n_g =$ Refraction index of glass

$n_{air} =$ Refraction index of air

$d_w =$ Depth of water

I enclose a diagram for a better understanding of the problem, in this way we can determine that the apparent depth in the water of the logo would be subject to

$d'w = d_w (\frac{n_{air}}{n_w})+d_g (\frac{n_{air}}{n_g})$

$d'w = (1.7cm) (\frac{1}{1.33})+(4.2cm)(\frac{1}{1.52})$

$d'w = 4.041cm$

Therefore the distance below the upper surface of the water that appears to be the logo is 4.041cm

3 0

3 years ago