Why does one side of the Moon’s surface always remain much colder than the other side?
A) The colder side of the Moon has a cloud cover over it.
B) The colder side of the Moon is entirely covered with ice.
C) The colder side of the Moon always faces away from the Sun.
D) The colder side of the Moon has many high-altitude mountains.

The maximum compression in the spring is 0.06883 meters.

Given to us,

mass of bullet, [tex]m= 16.2\ grams[/tex]

velocity of bullet, [tex]v= 850\ meter/second[/tex]

mass of the block, [tex]m_2= 40\ kilograms[/tex]

velocity of block, [tex]v_2= 0\ meter/second[/tex]

spring constant, [tex]k = 1000\ Newton/meter[/tex]

Combined mass after collision,

[tex]\begin{aligned}M&= m_1+m_2\\&= 40.0162\ Kilograms\end{aligned}[/tex]

Using the momentum conservation equation, to find out final velocity of the system(bullet and block combined)

[tex]m_1v_1+m_2v_2= MV\\[/tex]

[tex]0.0162\times 850+40\times 0 = 40.0162\times V\\V= 0.3441 meter/sec[/tex]

Now using the Conservation of Energy (Kinetic energy of the system will be equal to potential energy in the spring), to find out the displacement of the spring [tex](x)[/tex]

[tex]\frac{1}{2}kx^2 = \frac{1}{2}MV^2\\\\\frac{1}{2}\times 1000\times x^2= \frac{1}{2} \times40.0162\times0.3441^2\\\\x^2= 0.004738\\\\x= 0.06883\ meters[/tex]

Hence, the maximum compression in the spring is 0.06883 meters.

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Answer:

Explanation:

Question 1 (1 point)

According to theory, which term best describes the universe before the big bang?

a small densely packed ball of matter

Question 2 (1 point)

Which example is used as evidence that the universe began with the big bang?

Question 2 options:

leftover energy in the form of background radiation in the universe from the event

Question 3 (1 point)

How do scientists use the Doppler effect to understand the universe?

to determine motion of objects

Question 4 (1 point)

The Doppler effect indicates that the universe is expanding because_________.

Which phrase best completes the sentence.

light from other galaxies is red shifted

Question 5 (1 point)

What does a blue shift in light from stars indicate?

The stars are moving closer.

The question is incomplete. The complete question is :

Consider a composite cube made of epoxy with fibers aligned along one axis of the cube (the fibers are parallel to four of the twelve cube edges). If the cube can only be loaded in axial tension such that the force is uniformly applied over - and is normal to - a cube face, what is the lowest possible positive length change the cube can experience under this tension? The applied tensile force is 102 KN. The unloaded cube edge length is 56 mm. The glass fibers have an elastic modulus of 200 GPa. The epoxy has an elastic modulus of 38 GPa. The cube is comprised of 18 vol% epoxy (the balancing vol % is glass fiber). Hint: The loading axis is intentionally unspecified. Answer Format: Lowest possible length increase (change of length) under tension.

Solution :

Given :

[tex]$E_{glass fibre}$[/tex] = 200 GPa

[tex]$V_{glass fibre} = 82\%$[/tex]

[tex]$E_{epoxy}$[/tex] = 38 GPa

[tex]$V_{epoxy} = 82\%$[/tex]

Edge length = 56 mm

Cube is loaded in axial tension such that the force is uniformly applied over a cube face.

[tex]$E_{\text{composite}}=\frac{E_{glass fibre} \times E_{epoxy}}{(E_{glass fibre .E_{epoxy}})+(E_{fibre}.V_{glass fibre})}$[/tex]

[tex]$E_{composite} = \frac{200 \times 38}{(200 \times 0.18)+(38\times 0.82)}$[/tex]

               [tex]$=113.16 $[/tex]  GPa

Applied stress [tex]$=\frac{\text{applied load}}{\text{area}}$[/tex]

                    [tex]$\sigma=\frac{102 \times 10^3 \ N}{56 \times 56 \times 10^{-6} \ m^2}$[/tex]

                       = 32.5 MPa

By Hooke's law

[tex]$\sigma = E . \epsilon$[/tex]

[tex]$\sigma = E. \frac{\Delta l}{l}$[/tex]

[tex]$\Delta l = \frac{\sigma}{E}\times l$[/tex]

Length change, [tex]$\Delta l =\frac{32.5 \times 10^6 \ Pa}{113.16 \times 10^9 \ Pa}\times 56 \times 10^{-2} \ m$[/tex]

[tex]$\Delta l = \frac{32.5 \times 56}{113.16} \times 10^{-3} \ mm$[/tex]

   = 0.016 mm

Answer:

d

Explanation:

The sphere is positively charged so that when it comes in contact with the rod, its positive charges will repel to the other end. Hence, Hold the sphere near each end of the block, because that will give experimental data about both ends of the block.

Polarization is the separation of charges into two poles due electrostatic repulsion. Two similar charges will repel and opposite charges attracts each other.

When a charged or polar substance get in contact with a non-polar substance, the nonpolar one will deform in such a way that the electrons from the polar one will repel the electrons of nonpolar and there occurs a temporary charge separation in non polar substance creating an induced polarity.

Here, when the positive sphere comes in contact with the charged rod, the positive charges will align to the opposite pole of the rod away from the sphere and the negative charges will align near the sphere.

Thus there occurs an attractive force at one end and repulsion at the other end. Thus, holding the sphere at each end will data about charge distribution.

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Answer:

I think its D, Does the wave travel In a Vaccum?

Answer:

It's C. Does the wave transfer energy?

Answer:

The angular width is "[tex]1.575\times 10^{-3} \ rad[/tex]" and the linear width is "[tex]2.835\times 10^{-3} \ m[/tex]".

Explanation:

Given:

Wavelength,

[tex]\lambda = 630 \ nm[/tex]

or,

  [tex]=630\times 10^{-9} \ m[/tex]

Slit width,

[tex]a = 0.8 \ mm[/tex]

or,

  [tex]=0.8\times 10^{-3} \ m[/tex]

Distance between slit and screen,

[tex]D = 1.8 \ m[/tex]

As we know,

The central bright fringe's angular height,

⇒ [tex]\theta = \frac{2 \lambda}{a}[/tex]

       [tex]= \frac{2\times 630\times 10^{-9}}{0.8\times 10^{-3}}[/tex]

       [tex]=1.575\times 10^{-3} \ rad[/tex]

and,

The linear width will be:

⇒ [tex]x_o=\frac{2D \lambda}{a}[/tex]

By substituting the values, we get

        [tex]=\frac{3\times 1.8\times 630\times 10^{-9}}{0.8\times 10^{-3}}[/tex]

        [tex]=2.835\times 10^{-3} \ m[/tex]

Each electron has a charge of :

[tex]1.6 \times 10 {}^{ - 19} \: C[/tex]

For producing 1 Cuolomb charge we need :

Now, for producing 0.009 C of charge, the number of electrons required is :

_____________________________

So, Number of electrons passing through the cross section in 3.6 seconds is :

[tex] \mathrm{5.616 \times 10 {}^{16} \: \: electrons} [/tex]

Number of electrons passing through it in 1 Second is :

Now, in 10 seconds the number of electrons passing through it is :

_____________________________

[tex]\mathrm{ \#TeeNForeveR}[/tex]

Answer:

The correct answer is "53.15 days".

Explanation:

Given that:

Half life of [tex]131_{I}[/tex],

[tex]T_{\frac{1}{2} }= 8 \ days[/tex]

To find t when R will be "1%" of [tex]R_o[/tex], then

⇒ [tex]R=\frac{1}{100}R_o[/tex]

As we know,

⇒ [tex]R=R_o e^{-\lambda t}[/tex]

or,

∴ [tex]e^{\lambda t}=\frac{R_o}{R}[/tex]

By putting the values, we get

        [tex]=\frac{R_o}{\frac{R}{100} }[/tex]

        [tex]=100[/tex]

We know that,

Decay constant, [tex]\lambda = \frac{ln2}{T_{\frac{1}{2} }}[/tex]

hence,

⇒ [tex]\lambda t=ln100[/tex]

     [tex]t=\frac{ln100}{\lambda}[/tex]

        [tex]=\frac{ln100}{\frac{ln2}{8} }[/tex]

        [tex]=53.15 \ days[/tex]  

Answer:

A. The medium through which it travels

Answer:

the ball lose kentic energy and gains potential energy rolling upward

Answer:

c

Explanation:

when a ball is rolling down it loses potential and gains kinetic, but in this case since the ball is going upwards it is losing kinetic energy and gaining potentail like a roller coaser

Stopping a 1000 kg car moving at a speed of 174 km/h will be more difficult than stopping an identical car moving at 100 km/h.

This can be explained by using Newton's second law of motion which says that the rate of change of momentum of a body is equal to the force applied.

Therefore, the car moving at 174 km/h has higher momentum and from Newton's second law of motion, a higher force would be required to stop this car.

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Answer:

The theory of evolution by natural selection, first formulated in Darwin's book "On the Origin of Species" in 1859, is the process by which organisms change over time as a result of changes in heritable physical or behavioral traits

The best demonstration that applies to Newton's Third Law of motion would be D) When you walk your foot pushes down on the ground while the ground pushes back on your foot.

------------------

Newton's Third Law states that for every action, there is an equal and opposite reaction. This is actually explains that forces come in pairs and forces are an interaction between two objects. As per the correct option given in the question explains Newton’s Third Law.  

------------------

When you walk your foot (say object A) pushes down on the ground while the ground (say object Q) pushes back on your foot with the same force but in the opposite direction.

Answer:

Isotopic dating of meteor fragments

Explanation:

Earth is so old that its age can only be determined using isotopic dating. The oldest measurement of Earth's age was determined using isotopic dating of meteor fragments.

Answer: Isotopic dating of meteor fragments

Explanation:

This question is incomplete, the complete question is;

A ladder 25 feet long is leaning against the wall of a house. The base of the ladder is pulled away from the wall at a rate of 2 feet per second. (a) What is the velocity of the top of the ladder when the base is given below

7 feet away from the wall

15 feet away from the wall

20 feet away from the wall

Answer:

- When x is 7 feet away from the wall, [tex]\frac{dr}{dt}[/tex] = -7/12 ft/s

- When x is 15 feet away from the wall, [tex]\frac{dr}{dt}[/tex] = -3/2 ft/s

- When x is 20 feet away from the wall, [tex]\frac{dr}{dt}[/tex] = -8/3 ft/s

Explanation:

Given the data in the question;

Let [tex]r[/tex]ft and [tex]x[/tex]ft represent the distance of the top of ladder from the ground and the distance of the bottom of ladder from the wall respectively.

Now using the Pythagoras Theorem ( right angled triangle ).

(hypotenuse)² = ( adjacent )² + ( opposite )²

so

r² = ( 25 )² - x²

r² = 625 - x² ------- let this be equation 1

now, we differentiate with respect to t

2r [tex]\frac{dr}{dt}[/tex] = 0 - 2x [tex]\frac{dx}{dt}[/tex]

r [tex]\frac{dr}{dt}[/tex] = -x [tex]\frac{dx}{dt}[/tex]

[tex]\frac{dr}{dt}[/tex] = -x [tex]\frac{dx}{dt}[/tex] × 1/r

[tex]\frac{dr}{dt}[/tex] = -x/r [tex]\frac{dx}{dt}[/tex]

Now, from equation, r² = 625 - x²

r = √( 625 - x² )

Then

[tex]\frac{dr}{dt}[/tex] = -x / √( 625 - x² ) [tex]\frac{dx}{dt}[/tex]

given that, The base of the ladder is pulled away from the wall at a rate of 2 feet per second i.e [tex]\frac{dx}{dt}[/tex] = 2 ft/s

so;

[tex]\frac{dr}{dt}[/tex] = ( -x / √( 625 - x² ) ) × 2

[tex]\frac{dr}{dt}[/tex] = -2x / √( 625 - x² )

∴

When x is 7 feet away from the wall;

[tex]\frac{dr}{dt}[/tex] = -2(7) / √( 625 - (7)² )

[tex]\frac{dr}{dt}[/tex] = -14 / √( 625 - 49 )

[tex]\frac{dr}{dt}[/tex] = -14 / √576

[tex]\frac{dr}{dt}[/tex] = -14 / 24

we simplify

[tex]\frac{dr}{dt}[/tex] = -7/12 ft/s

Therefore, When x is 7 feet away from the wall, [tex]\frac{dr}{dt}[/tex] = -7/12 ft/s

When x is 15 feet away from the wall

[tex]\frac{dr}{dt}[/tex] = -2(15) / √( 625 - (15)² )

[tex]\frac{dr}{dt}[/tex] = -30 / √( 625 - 225 )

[tex]\frac{dr}{dt}[/tex] = -30 / √400

[tex]\frac{dr}{dt}[/tex] = -30 / 20

we simplify

[tex]\frac{dr}{dt}[/tex] = -3/2 ft/s

Therefore, When x is 15 feet away from the wall, [tex]\frac{dr}{dt}[/tex] = -3/2 ft/s

When x is 20 feet away from the wall

[tex]\frac{dr}{dt}[/tex] = -2(20) / √( 625 - (20)² )

[tex]\frac{dr}{dt}[/tex] = -40 / √( 625 - 400 )

[tex]\frac{dr}{dt}[/tex] = -40 / √225

[tex]\frac{dr}{dt}[/tex] = -40 / 15

we simplify

[tex]\frac{dr}{dt}[/tex] = -8/3 ft/s

Therefore, When x is 20 feet away from the wall, [tex]\frac{dr}{dt}[/tex] = -8/3 ft/s

Answer:

parallel circuit is a kind of circuits in which all components are connected between the same two sets of electrically common points, creating multiple paths for the current to flow from one end of the battery to the other:

Answer:

Gadhe hote hai log and usme sabse pehele aati hai Jonain jaise log waah waah xD

Brainliest krdena warna chappal padenge -,-

Answer:

length

Explanation:

Legth is derived physical quantity

Answer:

The fifth harmonic frequency is 2.9 KHz.

Explanation:

For a closed tube, the fifth harmonic frequency is also the second overtone. So that;

fifth harmonic frequency = [tex]\frac{5V}{4L}[/tex]

where V is the velocity of the sound in air, and L is the length of the tube.

But L = 0.15 m and V = 343 m/s, then:

fifth harmonic frequency = [tex]\frac{5*343}{4*0.15}[/tex]

                                     = [tex]\frac{1715}{0.6}[/tex]

                                     = 2858.33 Hertz

fifth harmonic frequency = 2.9 KHz

The fifth harmonic frequency of the soda bottle is approximately 2.9 KHz.

Answer:

2828.33

Explanation:

answer for acellus

Combined energy of the two sleds will be conserved when they collides.

Energy is nothing but the ability to do work. there are different energies in different form which are thermal energy, mechanical energy, electric energy and sound energy etc.

According to first law of thermodynamic, Energy neither be created nor be destroyed. it can only be transferred from one form into another form. Energy is expressed in joule (J). its dimensions are [M¹ L² T⁻²].

Energy is conserved throughout the motion,

according to conservation law of energy, initial energy is equal to final energy.

When two sleds coming in opposite direction, it is having mass as well as velocity. hence it has kinetic energy, when they get collide with each other some of total energy gets converted into sound energy as collision cause sound(boom). some of the energy will use to break the material which made the sled and remaining energy will convert in kinetic energy of broken sleds it can be moved in opposite direction or in same direction depending initial energies.

in this collision combined kinetic energy can be converted into sound energy, mechanical energy and again kinetic energy but totally energy is conserved.

Hence combined energy will be conserved.

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Answer:

it's very easy and simple answer u can't do it

The electric field intensity at a point of 15cm from a charge of 10 μc will be 3.9 ×10⁶ N/C.

The electric field is defined as the area over an electric charge where its impact may be felt. The force experienced by a unit positive charge deposited at a given position is known as the electric field intensity.

The formula for the electric field intensity is given as;

[tex]\rm E= \frac{q}{4 \pi \epsilon d^2 } \\\\ \rm E= \frac{10 \times 10^{-6}}{4 \times 3.14 \times 8.85 \times 10^{-12}(15 \times 10^{-2}^2 } \\\\ E=3.9 \times 10^6 \ NC^{-1}[/tex]

Hence the electric field intensity at a point of 15cm from a charge of 10 μc will be 3.9 ×10⁶ N/C.

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Answer:

Fc = 123 Newton

Explanation:

Net force can be defined as the vector sum of all the forces acting on a body or an object i.e the sum of all forces acting simultaneously on a body or an object.

Mathematically, net force is given by the formula;

[tex] Fnet = Fapp + Fg[/tex]

Where;

Fnet is the net force.

Fapp is the applied force.

Fg is the force due to gravitation.

Given the following data;

Normal force = 25N

Mass = 10kg

To find the centripetal force;

From the net force, we have the following formula;

Fc = N + mg

Where;

Fc is the centripetal force.

N is the normal force.

mg is the the weight of the object.

Substituting into the formula, we have;

Fc = 25 + 10(9.8)

Fc = 25 + 98

Fc = 123 Newton

Answer:

4.16×103 m/s

Explanation:

Answer:

A practical activity about the study of the structure of the natural and physical world that can be shown thorough experiments and study's.

Explanation:

Explanation:

hope it's helpful

have a nice day!!!! ;)

Answer:

Hi how are you doing today Jasmine

Answer:

The answer is B

Explanation:

I did the USA test prep

Answer:

In addition to the sun's angle, atmospheric conditions can affect radiation levels. Cloud cover, air pollution and the hole in the ozone layer all alter the amount of solar radiation that can reach the surface. These factors all cause typical radiation levels to differ.

hope it helps

To solve this, we must be knowing each and every concept behind solar radiation. Therefore, cloud cover, air pollution, and the ozone hole all reduce the quantity of solar energy that reaches the earth's surface.

The energy generated by the sun as just a consequence of vast internal processes is known as solar radiation. In a word, the sun's capacity to generate intense nuclear fusion within and around the core enables it to radiate such vast amounts of energy inside the form of both heat and light. The entire process begins at the core of the sun.

Aside from the sun's angle, meteorological variables can influence radiation levels. Cloud cover, air pollution, and the ozone hole all reduce the quantity of solar energy that reaches the earth's surface. All of these variables contribute to variations in normal radiation levels.

Therefore, cloud cover, air pollution, and the ozone hole all reduce the quantity of solar energy that reaches the earth's surface.

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Answer:

a) 627840000 W

b) 0.9233

Explanation:

The rate at which energy flows through the dam for the generation of electricity is proportional to the volulme flow rate times the potential energy per unit volume.

Mathematically,

P = Q(ρgh)

Where, ρ is water density, g is accelaration due to gravity, h is height through which water falls over the turbine and Q is the discharge.

a) The power can be calculated as

P = 640(1000×9.81×221)

P = 627840000 W

b) The ratio = Calculated power / avarage power

= 62784×10^4/680×10^6

= 0.9233

Answer:

Ming treasure voyages

Explanation:

Answer:

The speed of sound in the argon is 340 m/s.

Explanation:

Given;

fundamental frequency, f₀ = 200 Hz

length of the pipe, L = 42.5 cm = 0.425 m

A pipe that is open at one end and closed at another end is known as a closed pipe.

The wavelength for the first harmonic is calculated as;

L = Node -------> Antinode

L = λ/4

λ = 4L

The speed of the sound is calculated as;

v = fλ

where;

v is the speed of the sound

v = f x 4L

v = 200 x (4 x 0.425)

v = 340 m/s

Therefore, the speed of sound in the argon is 340 m/s.