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3 years ago

8

Your're sailing a boat from spain west to florida. which winds would you use?

1 answer:

timurjin [86]3 years ago

6 0

The wind you would use to travel from Spain west to Florida are The Trade Winds

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What could be the plot of a story about planets/ astronomy?

Finger [1]

Answer:

Some stars in space can come to life causing humans to be amazed but at thesame time in great/critical danger.

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2 years ago

Read 2 more answers

A person lowers a bucket into a well by turning the hand crank, as the drawing illustrates. The crank handle moves with a consta

dimulka [17.4K]

Answer:

0.453 m/s

Explanation:

Assuming the handle has diameter of 0.4 m while inner part diameter is 0.1 m then the circumference of outer part is $\pi d_h$ where d is diameter and subscript h denote handle. By substituting 0.4 for the handle's diameter then cirxumference of outer part is $\pi\times 0.4\approx 1.256 m$

The rate of rotation will then be 1.81/1.256=1.441 rev/s

Similarly, circumference of inner part will be $\pi d_i$ where subscript i represent inner. Substituting 0.1 for inner diameter then

$\pi\times 0.1\approx 0.3142 m$

The rate of rotation found for outer handle applies for inner hence speed will be 0.3142*1.441=0.453 m/s

7 0

2 years ago

during the course of songraphic exam, you notice lateral splaying of echoes in the far field. what can you do to improve the ima

horrorfan [7]

Answer:

lateral splaying of echoes in the far field can be improved by Increasing the maximum number of transmit focal zones and optimize their location.

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2 years ago

The Hubble Space Telescope has a mass of 1.16*10^ 4 kg and orbits the Earth at an altitude of 5.68 * 10 ^ 5 above Earth's surfac

andrezito [222]

Answer:

$E=8.13\times 10^{12}\ J$

Explanation:

Given that,

The mass of a Hubble Space Telescope, $m_1=1.16\times 10^4\ kg$

It orbits the Earth at an altitude of $5.68\times 10^5\ m$

We need to find the potential energy the telescope at this location. The formula for potential energy is given by :

$E=\dfrac{Gm_1m_e}{r}$

Where

$m_e$ is the mass of Earth

Put all the values,

$E=\dfrac{6.67\times 10^{-11}\times 1.16\times 10^4\times 5.97\times 10^{24}}{5.68\times 10^5}\\\\E=8.13\times 10^{12}\ J$

So, the potential energy of the telescope is $8.13\times 10^{12}\ J$ .

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2 years ago

A Porsche challenges a Honda to a 500 m race

zlopas [31]

A Porsche will always win no matter what

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3 years ago