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marshall27 [118]

3 years ago

6

Well into the twentieth century, many scientists believed the ocean floor was

1 answer:

marusya05 [52]3 years ago

4 0

<span>The correct answer is B. flat. It was believed that if one could reach the bottom of the ocean that they would find a flat floor. That is because they didn't have the technology to dive so deep to explore it, but eventually things changed and it was found out that it's not just a flat panel and that it's full of craters and similar things.</span>

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A car travels 50 km in 25 km /h and then travels 60km with a velocity 20 km/h and then travels 60km with a velocity 20 km/h in t

Butoxors [25]

Answer:

v = 21.25 km/h

The average velocity is 21.25km/h

Explanation:

Average velocity = total displacement/time taken

v = d/t

Given;

A car travels 50 km in 25 km /h

d1 = 50km

v1 = 25km/h

time taken = distance/velocity

t1 = d1/v1

t1 = 50/25 = 2 hours

and then travels 60km with a velocity 20 km/h

d2 = 60km

v2 = 20km/h

t2 = d2/v2 = 60/20

t2 = 3 hours

and then travels 60km with a velocity 20 km/h in the same direction

d3 = 60km

v3 = 20km/h

t3 = d3/v3 = 60/20

t3 = 3 hours

Average velocity = total displacement/total time taken

v = (d1+d2+d3)/(t1+t2+t3)

v = (50+60+60)/(2+3+3)

v = 170/8

v = 21.25 km/h

The average velocity is 21.25km/h

3 0

3 years ago

A long thin rod of mass M and length L is situated along the y axis with one end at the origin. A small spherical mass m1 is pla

Yuri [45]

Answer:

$= - 5.65\times 10^{-2} J$

Explanation:

Given data:

L =2.00 *10^4 m

d = 18*10^4 m

M = 18 *10^6 kg

m_1 = 8*10^6 kg

Gravitational energy is given as

$U =- G \frac{m_1 m_2}{r}$

mass per unit length is given as

$\sigma = \frac{M}{L} = \frac{18 \times 10^6}{2\times 10^4 m} = 900 kg/m$

calculating potential energy

$dU ==-G\int_{16\times 10^4}^{18\times 10^4} \frac{m_1 *dm}{r}$

$=-G\int_{16\times 10^4}^{18\times 10^4} \frac{m_1 *\sigma dr}{r}$

$=-G*m_1*\sigma\int_{16\times 10^4}^{18\times 10^4} \frac{dr}{r}$

$=-G*m_1*\sigma \left | ln r \right |_{16\times 10^4}^{18\times 10^4}$

$=-G*m_1*\sigma ln(1.125)$

$=-6.673 \times 10^{-11}*8*10^6*900*ln(1.125)$

$= - 5.65\times 10^{-2} J$

5 0

3 years ago

The number of depletion layers in a transistor is. *

Tema [17]

Here is your answer

Answer :- two(2)

3 0

3 years ago

\Which are consumers within a spruce-fir forest ecosystem?

lara [203]

The answer is mushroom.

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yarga [219]

Answer:

Please reframe your questions

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3 years ago