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3 years ago

Study the image of earths layer which statement correctly compares the thicknesses of earths layers

Physics

1 answer:

My name is Ann [436]3 years ago

4 0

This question is incomplete because the options are missing; here is the complete question:

Study the image of the Earth's layer which statement correctly compares the thicknesses of earths layers

A. Earth’s mantle is thinner than its oceanic crust.

B. Earth’s outer core is thicker than its mantle.

C. Earth’s continental crust is thicker than its lithosphere.

D. Earth’s lithosphere is thinner than its asthenosphere.

The answer to this question is D. Earth’s lithosphere is thinner than its asthenosphere.

Explanation:

The image shows the different layers that are part of Earth, as well as the thickness of each layer, in kilometers. In this, the thickest layer is the Mantle that is almost 2900 kilometers; this is followed in thickness by the outer and the inner core.

Additionally, other layers such as the continental/oceanic crust, the asthenosphere, and the lithosphere that are near the surface are thinner. About this, it can be concluded the lithosphere is thinner than the asthenosphere because the lithosphere has a thickness of 100 km, while the asthenosphere thickness is 660km. This makes option D the correct.

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Imagine two billiard balls on a pool table. Ball A has a mass of 7 kilograms and ball B has a mass of 2 kilograms. The initial v

wlad13 [49]

1) In a perfectly inelastic collision, the two balls stick together after the collision. In this type of collision, the total kinetic energy of the system is not conserved, while the total momentum is conserved.
If we call $v_f$ the final velocity of the two balls that stick together, the conservation of the total momentum before and after the collision can be written as
$m_a v_{Ai} + m_b v_{Bi} = (m_A+m_B)v_f$ (1)
where
$m_A=7 kg$ is the mass of ball A
$m_B=2 kg$ is the mass of ball B
$v_{Ai}=6 m/s$ is the initial velocity of ball A
$v_{Bi}=-12 m/s$ is the initial velocity of ball B (taken with a negative sign, since it goes in the opposite direction of ball A)

If we solve (1) to find $v_f$ , we find that the final velocity of the balls is
$v_f= \frac{m_Av_{Ai}+m_Bv_{Bi}}{m_A+m_B}= \frac{(7\cdot 6)+(2 \cdot (- 12))}{7+2}= \frac{18}{9}=2 m/s$
and the positive sign means the two balls are going to the right.

2) I assume here we are talking about an elastic collision. In this case, both total momentum and total kinetic energy are conserved:
$m_A v_{Ai}+m_B v_{Bi} = m_A v_{fA} + m_B v_{fB}$
$\frac{1}{2}m_A v_{Ai}^2+ \frac{1}{2}m_B v_{Bi}^2= \frac{1}{2}m_Av_{fA}^2+ \frac{1}{2}m_B v_{fB}^2$
where
$v_{fA}$ is the final velocity of ball A
$v_{fB}$ is the final velocity of ball B

If we solve simultaneously the two equations, we find:
$v_{fA}= \frac{v_{Ai}(m_A-m_B)+2m_Bv_{Bi}}{m_A+m_B} = \frac{(6)(7-2)+2(2)(-12)}{7+2}=-2 m/s$
$v_{fB}= \frac{v_{Bi}(m_B-m_A)+2m_Av_{Ai}}{m_A+m_B} = \frac{(-12)(2-7)+2(7)(6)}{7+2}= \frac{144}{9}=16 m/s$
So, after the collision, ball A moves to the left with velocity v=-2 m/s and ball B moves to the right with velocity v=16 m/s.

3) The total momentum before and after the collision is conserved.
In fact, the total momentum before the collision is:
$p_i = m_A v_{A} + m_B v_{fB} = (7\cdot 6)+(2 \cdot (-12))=42-24=18 m/s$
and the total momentum after the collision is:
$p_f = m_A v_{A} + m_B v_{fB} = (7\cdot (-2))+(2 \cdot 16)=-14+32=18 m/s$

3 0

3 years ago

True or False? 13. All living things are made of cells 14. All cells have DNA within their nucleus 15. The cell is the basic uni

Leni [432]

Answer:

1: True

2: False

3: True

4: False

5: True

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3 years ago

A man at point A directs his rowboat due north toward point B, straight across a river of width 100 m. The river current is due

prohojiy [21]

Answer:

1.35208 m/s

Explanation:

Speed of the boat = 0.75 m/s

Distance between the shores = 100 m

Time = Distance / Speed

$Time=\frac{100}{0.75}=133.33\ s$

Time taken by the boat to get across is 133.33 seconds

Point C is 150 m from B

Speed = Distance / Time

$Speed=\frac{150}{\frac{100}{0.75}}=1.125\ m/s$

Velocity of the water is 1.125 m/s

From Pythagoras theorem

$c=\sqrt{0.75^2+1.125^2}\\\Rightarrow c=1.35208\ m/s$

So, the man's velocity relative to the shore is 1.35208 m/s

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3 years ago

The strength of the electric field at a certain distance from a point charge is represented by E. What is the strength of the el

Nana76 [90]

Answer:

Explanation:

Using Coulomb's law equation

Force of the charge = k qQ /d²

and E = F/ q

substitute for F

E = ( K Qq/ d² ) / q

q cancel q

E = KQ / d²

so twice the distance of the from the point charge will lead to the E ( electric field ) decrease by a 4 = E/4. E is inversely proportional to d²

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3 years ago

In a certain time interval, natural gas with energy content of 19000 J was piped into a house during a winter day. In the same t

Bumek [7]

Explanation:

Since, it is mentioned the there occurs no change in the temperature. This also means that there will occur no change in thermal energy of the system.

Hence, $\Delta E$ = 0. And, as $\Delta E$ = 0 then there will be no work involved. This means that total energy added to the house will return to the outside air as heat.

Therefore,

Q = -(19000 J + 2000 J)

= -21000 J

or, |Q| = 21000 J

Thus, we can conclude that the magnitude of the energy transfer between the house and the outside air is 21000 J.

5 0

3 years ago