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3 years ago

What are earth's three different tidal patterns called?

Physics

1 answer:

mixer [17]3 years ago

8 0

I think those are it but I can't find the other one but I'll keep looking
2.tidal current
3.tidal force

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Question 8 (1 point)

sineoko [7]

Answer:

Explanation:

6 0

2 years ago

What percent of the energy used in the united states come from burning fossil fuels?

andreyandreev [35.5K]

Here is a link that should help you out
http://instituteforenergyresearch.org/topics/encyclopedia/fossil-fuels/

8 0

3 years ago

It is 2058 and you are taking your grandchildren to Mars. At an elevation of 34.7 km above the surface of Mars, your spacecraft

Paul [167]

Answer: $1.23\ m/s^2$

Explanation:

Given

At an elevation of $y=34.7\ km$ , spacecraft is dropping vertically at a speed of $u=293\ m/s$

Final velocity of the spacecraft is $v=0$

using equation of motion i.e. $v^2-u^2=2as$

Insert the values

$\Rightarrow 0-(293)^2=2\times a\times (34.7\times 10^3)\\\\\Rightarrow a=-\dfrac{293^2}{2\times 34.7\times 10^3}\\\\\Rightarrow a=-1.23\ m/s^2$

Therefore, magnitude of acceleration is $1.23\ m/s^2$ .

8 0

2 years ago

What pressing problem did the invention of the cotton gin solve?

AlexFokin [52]

Are you answering a question or asking? You have already seemed to get the answer, A is the correct answer :I

4 0

3 years ago

You hang a heavy ball with a mass of 10 kg from a gold wire 2.6 m long that is 1.6 mm in diameter. You measure the stretch of th

PolarNik [594]

<u>Answer:</u> The Young's modulus for the wire is $6.378\times 10^{10}N/m^2$

<u>Explanation:</u>

Young's Modulus is defined as the ratio of stress acting on a substance to the amount of strain produced.

The equation representing Young's Modulus is:

$Y=\frac{F/A}{\Delta l/l}=\frac{Fl}{A\Delta l}$

where,

Y = Young's Modulus

F = force exerted by the weight = $m\times g$

m = mass of the ball = 10 kg

g = acceleration due to gravity = $9.81m/s^2$

l = length of wire = 2.6 m

A = area of cross section = $\pi r^2$

r = radius of the wire = $\frac{d}{2}=\frac{1.6mm}{2}=0.8mm=8\times 10^{-4}m$ (Conversion factor: 1 m = 1000 mm)

$\Delta l$ = change in length = 1.99 mm = $1.99\times 10^{-3}m$

Putting values in above equation, we get:

$Y=\frac{10\times 9.81\times 2.6}{(3.14\times (8\times 10^{-4})^2)\times 1.99\times 10^{-3}}\\\\Y=6.378\times 10^{10}N/m^2$

Hence, the Young's modulus for the wire is $6.378\times 10^{10}N/m^2$

3 0

3 years ago