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serious [3.7K]
4 years ago
8

The layers of planet Earth

Physics
1 answer:
lord [1]4 years ago
7 0

CRUST

   The crust is the thinnest layer. It is the layer of the earth that we stand and walk on. It is more than 25 miles thick on land and 3 miles under an ocean.  

MANTLE

   The mantle is the layer directly below the crust. It is the thickest layer. It is divided into two parts the lithosphere and the asthenosphere. The lithosphere is at the top of the mantle, the asthenosphere is at the bottom. Together they make up the mantle.

OUTER CORE

   The outer core is directly under the mantle and right above the inner core. the outer core is in a liquid state.

INNER CORE

   The inner core is hotter than the surface of the sun. It also is made of solid iron and nickel.


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The acceleration due to gravity on earth is 9.8 m/s2. what is the weight of a 75 kg person on earth
erastovalidia [21]
Weight = mass * gravity
W = 75 kg * 9.8
W = 735 N

hope this  helps :)
5 0
3 years ago
A 20-kg block is held at rest on the inclined slope by a peg. A 2-kg pendulum starts at rest in a horizontal position when it is
gregori [183]

Complete Question

The diagram of this question is shown on the first uploaded image

Answer:

The distance the block slides before stopping is d = 0.313 \ m

Explanation:

The free body diagram for the diagram in the question is shown

From the diagram the angle is \theta = 25 ^o

         sin \theta  = \frac{h}{d}

Where h = h_b - h_a

     So      d sin \theta  = h_b - h_a

From the question we are told that

      The mass of the block is  m = 20 \ kg

       The mass of the pendulum is  m_p = 2 \ kg

       The velocity of the pendulum at the bottom of swing is v_p = 15 m/s

        The coefficient of restitution is  e =0.7

         The coefficient of kinetic friction is  \mu _k = 0.5

The velocity of the block after the impact is mathematically represented as

            v_2 f = \frac{m_b - em_p}{m_b + m_p}  * v_2 i + \frac{[1 + e] m_1}{m_1 + m_2 } v_p

Where  v_2 i is the velocity of the block  before collision which is  0

                  = \frac{20 - (0.7 * 2)}{(2 + 20)} * 0 + \frac{(1 + 0.7) * 2 }{2 + 20}   * 15

Substituting value

                   v_2 f = 2.310\  m/s

According to conservation of energy principle

      The energy at point a  =  energy at point b

So    PE_A + KE _A = PE_B + KE_B  +  E_F

Where  

         PE_A is the potential energy at A which is mathematically represented as

          PE_A = m_b gh_a = 0 at the bottom

      KE _A is the kinetic energy at A  which is mathematically represented as

               K_A = \frac{1}{2} m_b * v_2f^2                  

         PE_B is the potential energy at B which is mathematically represented as  

            PE_B = m_b gh

From the diagram h = h_b -h_a

       PE_B = m_b g(h_b - h_a)

KE _B is the kinetic energy at B  which is 0 (at the top )

Where is E_F is the workdone against velocity  which from the diagram is

      \mu_k m_b g cos 25 *d

So

   \frac{1}{2} m_b v_2 f^2  = m_b g h_b + \mu_k m_b g cos \25 * d

Substituting values

   \frac{1}{2}  * 20 * 2.310^2 = 20 * 9.8 * d sin(25)  + 0.5* 20 * 9.8 * cos 25 * d    

So

       d = 0.313 \ m

       

   

6 0
4 years ago
To practice tactics box 13.1 hydrostatics. in problems about liquids in hydrostatic equilibrium, you often need to find the pres
OLEGan [10]

Answer:

A. The pressure denoted as Pa and Pb at the surfaces of A and B in the tube is

PA= Pgas

PB= Patmos

B. The second sketch

C. The gas pressure is

Pgas= Patmos+ rho.g(h2-h1)

= 1atm + rho.g (h2-h1)

Explanation:

5 0
3 years ago
What is the Ultraviolet Catastrophe?​
Sholpan [36]

The ultraviolet catastrophe was the prediction of late 19th century/early 20th century classical physics that an ideal black body (also blackbody) at thermal equilibrium will emit radiation in all frequency ranges, emitting more energy as the frequency increases.

7 0
3 years ago
You are performing a double slit experiment very similar to the one from DL by shining a laser on two nattow slits spaced 7.5 x
Blizzard [7]

Complete Question

You are performing a double slit experiment very similar to the one from DL by shining a laser on two nattow slits spaced 7.5 * 10^{-3} meters apart. However, by placing a piece of crystal in one of the slits, you are able to make it so that the rays of light that travel through the two slits are Ï out of phase with each other (that is to say, Ao,- ). If you observe that on a screen placed 4 meters from the two slits that the distance between the bright spot closest to center of the pattern is 1.5 cm, what is the wavelength of the laser?

Answer:

The  wavelength is  \lambda  =  56250 nm

Explanation:

From the question we are told that

   The  distance of slit separation is  d =  7.5 *10^{-3} \  m

   The  distance of the screen is  D =  4 \  m

    The  distance between the bright spot closest to the center of the interference  is  k   = 1.5 \ cm = 0.015 \  m

   

Generally the width of the central  maximum fringe produced is mathematically represented as

        y  =  2 *  k  = \frac{ D  *  \lambda}{d}

  =>    2 *  0.015 =  \frac{ \lambda  *  4}{ 7.5 *10^{-3}}

   =>   \lambda  =  56250 *10^{-9} \ m

=>      \lambda  =  56250 nm

7 0
4 years ago
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