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pishuonlain [190]
3 years ago
13

Explain why Earth and other planets were not solid when they formed during the beginning of the Precambrian, approximately 4,600

million years ago (MYA). What was this particular time period called, and why is this time period not really considered a part of Earth’s geological history?
Physics
1 answer:
alexandr402 [8]3 years ago
5 0

earth still is not solid ... core is molten ... volcanoes, sea is liquid

at v v v beginning was gas ????

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How we will measure centimeters
Bingel [31]

Answer:

  • .

Explanation:

——»To measure centimeters, we can use ruler.

  • Use a ruler with the side marked either cm or mm. Align the edge of the object with the first centimeter line on the ruler, then find the length in whole centimeters, or the larger numbers on the ruler.
7 0
1 year ago
please solve this for me ,A garden roller is pulled with a force of 200N acting at an angle of 50 degree with the ground level.f
Bingel [31]

Answer:

The force pulling the roller along the ground is 128.55 N

Explanation:

A force of 200 N acting at an angle of 50° with the ground level

This force is pulled a garden roller

We need to find the force pulling the roller along the ground

The force that pulling the roller along the ground is the horizontal

component of the force acting

→ The force acting is 200 N at direction 50° with ground (horizontal)

→ The horizontal component = F cosФ

→ F = 200 N , Ф = 50

→ The horizontal component = 200 cos(50) = 128.55 N

128.55 N is the horizontal component of the force that pulling the

roller along the ground

<em>The force pulling the roller along the ground is 128.55 N</em>

8 0
2 years ago
An aluminum bar 600mm long, with diameter 40mm, has a hole drilled in the center of the bar. The hole is 40mm in diameter and 10
s2008m [1.1K]

Answer:

<em>1.228 x </em>10^{-6}<em> mm </em>

<em></em>

Explanation:

diameter of aluminium bar D = 40 mm  

diameter of hole d = 30 mm

compressive Load F = 180 kN = 180 x 10^{3} N

modulus of elasticity E = 85 GN/m^2  = 85 x 10^{9} Pa

length of bar L = 600 mm

length of hole = 100 mm

true length of bar = 600 - 100 = 500 mm

area of the bar A = \frac{\pi D^{2} }{4} =  \frac{3.142* 40^{2} }{4} = 1256.8 mm^2

area of hole a = \frac{\pi(D^{2} - d^{2}) }{4} = \frac{3.142*(40^{2} - 30^{2})}{4} = 549.85 mm^2

Total contraction of the bar = \frac{F*L}{AE} + \frac{Fl}{aE}

total contraction = \frac{F}{E} * (\frac{L}{A} +\frac{l}{a})

==> \frac{180*10^{3}}{85*10^{9}} *( \frac{500}{1256.8} + \frac{100}{549.85}) = <em>1.228 x </em>10^{-6}<em> mm </em>

6 0
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