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zimovet [89]
3 years ago
9

It is believed that during the late Cretaceous period sea levels rose drastically resulting in about a third of the Earth's pres

ent land masses being underwater. Which of the following types of evidence supports this theory?
A. minerals and oil found in underground deposits
B. corals and marine fossils found in the Great Plains
C. volcanoes and igneous rocks found on the ocean floor
D. dinosaur and bird fossils found in swamp areas
Physics
2 answers:
Naily [24]3 years ago
8 0

B corals and marine fossils found in the Great Plains!

forsale [732]3 years ago
3 0
The type of evidence which supports this theory is C. volcanoes and igneous rocks found on the ocean floor.
Normally, these types of ruptures and stones don't appear in water, however, after being submerged in oceans for such a long time, today it is a normal occurrence. 
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Answer:

0.098 N

Explanation:

From the question,

Spring scale reading = W-U............... Equation 1

Where W = weight of the cube, U = upthrust.

W = mg

Where m =  mass of the cube, g = acceleration due to gravity.

Given: m = 11 g = 0.011 kg, g = 9.8 m/s².

W = 0.011(9.8)

W = 0.1078 N.

From Archimedes principle,

Upthrust = weight of water displaced.

U = (Density of water×volume of metal cube)×acceleration due to gravity.

U = (D×V)g

Given: D = 1000 kg/m², V = 1 cm³ = (1/1000000) = 1×10⁻⁶ m³, g - 9.8 m/s²

U = 1000(9.8)(10⁻⁶)

U = 0.0098 N.

Substitute the value of W and U into equation 1

Reading of the spring scale = 0.1078-0.0098

Reading of the spring scale = 0.098 N

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A projectile is launched from ground level at angle u and speed v0 into a headwind that causes a constant horizontal acceleratio
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Answer:

Explanation:

Given

Launch angle =u

Initial Speed is v_0

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Total time of flight T=2t=\frac{2v_0\sin u}{g}

During this time horizontal range is

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For maximum range \frac{\mathrm{d} R}{\mathrm{d} u}=0

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\frac{\mathrm{d} R}{\mathrm{d} u}=\frac{2v_0^2}{g}\left [ \cos 2u-\frac{a}{g}\sin 2u\right ]=0

\tan 2u=\frac{g}{a}

u=\frac{1}{2}tan ^{-1}\frac{g}{a}

(b)If a =10% g

a=0.1g

thus u=\frac{1}{2}tan^{-1}\frac{g}{0.1g}

u=42.14^{\circ}

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