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DIA [1.3K]
4 years ago
3

Which theory is supported most by the finding that the moon is similar in composition to the outer portions of Earth?

Physics
2 answers:
nadezda [96]4 years ago
5 0
<h3><u>Answer;</u></h3>

Co-formation theory

<h3><u>Explanation;</u></h3>
  • <em><u>Co-formation theory</u></em> is the theory that is supported most by the finding that the moon is similar in composition to the outer portions of the Earth.
  • <em><u>Co-formation theory is therefore a theory that states that moon developed at the same time as the Earth from the solar nebula. This theory explains the origin of the moon as a body that was formed from the solar nebula at the same time and roughly at a similar place as the Earth, and also goes on to explain the location of the moon</u></em>
  • During the formation of the moon and the earth from the primitive solar nebula , the nuclei of the moon drew materials obtained materials from the cloud of gas and dust that surrounded them. The nebular material from which the moon and the earth were formed ought to be very similar.
Monica [59]4 years ago
3 0

Answer:

Co-formation theory

Explanation:

Co-formation theory is therefore a theory that states that moon developed at the same time as the Earth from the solar nebula. This theory explains the origin of the moon as a body that was formed from the solar nebula at the same time and roughly at a similar place as the Earth, and also goes on to explain the location of the moon

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Derive an expression for the block's centripetal acceleration ac in terms of m , θ , and physical constants, as appropriate
maxonik [38]

The expression for the block's centripetal acceleration is derived as ω²r or v²/r.

<h3>What is centripetal acceleration?</h3>

The centripetal acceleration of an object is the inward or radial acceleration of an object moving in a circular path.

The expression for the block's centripetal acceleration is derived as follows;

ω = dθ/dt

where;

  • ω is the angular speed
  • θ is the angular displacement
  • t is the time of motion

ac = ω²r

where;

  • r is the radius of the circular path

Also, ω = v/r

ac = (v/r)²r

ac = v²/r

Thus, the expression for the block's centripetal acceleration is derived as ω²r or v²/r.

Learn more about centripetal acceleration here: brainly.com/question/79801

7 0
2 years ago
Which of the diagrams below show forces that would result in a movement of the block to the left?
xenn [34]

Diagram A will result in the movement of the block to the left as a result of the forces.

<h3>What is Force?</h3>

This is referred to an influence which is capable of changing the motion of an object.

Diagram A has equal upward and downward force and left side which is 60N is higher than the right side which has 20N. The block will therefore move to the left.

Read more about Force here brainly.com/question/25239010

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4 0
2 years ago
A plane is landing. It started at 400m/s and ended up at 50m/s after 30 seconds. What is it's acceleration?
quester [9]

Give u = start velocity

        v = end velocity

v = u + at

50 = 400 + a*30

30a = -350

a = -116.67 m/s^{2}

**Why the accecleration is negative number**

Because displacement, velocity, and acceleration are VECTOR QUANTITIES.

Vector Quantity must have direction.

4 0
4 years ago
A bullet is fired with a muzzle velocity of 1178 ft/sec from a gun aimed at an angle of 26° above the horizontal. Find the horiz
dalvyx [7]

Answer:

1058.78 ft/sec

Explanation:

Horizontal Component of Velocity; This is the velocity of a body that act on the horizontal axis. I.e Velocity along x-axis

The horizontal velocity of a body can be calculated as shown below.\

Vh = Vcos∅.......................... Equation 1

Where Vh = horizontal component of the velocity, V = The velocity acting between the horizontal and the vertical axis, ∅ = Angle the velocity make with the horizontal.

Given: V = 1178 ft/sec, ∅ = 26°

Substitute into equation 1

Vh = 1178cos26

Vh = 1178(0.8988)

Vh = 1058.78 ft/sec

Hence the horizontal component of the velocity = 1058.78 ft/sec

8 0
3 years ago
The work done by an external force to move a -6.70 μc charge from point a to point b is 1.20×10−3 j .
ASHA 777 [7]

Answer:

108.7 V

Explanation:

Two forces are acting on the particle:

- The external force, whose work is W=1.20 \cdot 10^{-3}J

- The force of the electric field, whose work is equal to the change in electric potential energy of the charge: W_e=q\Delta V

where

q is the charge

\Delta V is the potential difference

The variation of kinetic energy of the charge is equal to the sum of the work done by the two forces:

K_f - K_i = W + W_e = W+q\Delta V

and since the charge starts from rest, K_i = 0, so the formula becomes

K_f = W+q\Delta V

In this problem, we have

W=1.20 \cdot 10^{-3}J is the work done by the external force

q=-6.70 \mu C=-6.7\cdot 10^{-6}C is the charge

K_f = 4.72\cdot 10^{-4}J is the final kinetic energy

Solving the formula for \Delta V, we find

\Delta V=\frac{K_f-W}{q}=\frac{4.72\cdot 10^{-4}J-1.2\cdot 10^{-3} J}{-6.7\cdot 10^{-6}C}=108.7 V

4 0
3 years ago
Read 2 more answers
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