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LUCKY_DIMON [66]

3 years ago

5

PLZ help !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

1 answer:

vlada-n [284]3 years ago

6 0

Answer:

The leftmost red square.

Explanation:

As the earth surface we are looking at is moving toward the right, the impact point of the wind would be left of the original appearing target.

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Question 1 of 4 Attempt 4 The acceleration due to gravity, ???? , is constant at sea level on the Earth's surface. However, the

Evgen [1.6K]

Answer:

g(h) = g ( 1 - 2(h/R) )

<em>*At first order on h/R*</em>

Explanation:

Hi!

We can derive this expression for distances h small compared to the earth's radius R.

In order to do this, we must expand the newton's law of universal gravitation around r=R

Remember that this law is:

$F = G \frac{m_1m_2}{r^2}$

In the present case m1 will be the mass of the earth.

Additionally, if we remember Newton's second law for the mass m2 (with m2 constant):

$F = m_2a$

Therefore, we can see that

$a(r) = G \frac{m_1}{r^2}$

With a the acceleration due to the earth's mass.

Now, the taylor series is going to be (at first order in h/R):

$a(R+h) \approx a(R) + h \frac{da(r)}{dr}_{r=R}$

a(R) is actually the constant acceleration at sea level

and

$a(R) =G \frac{m_1}{R^2} \\ \frac{da(r)}{dr}_{r=R} = -2 G\frac{m_1}{R^3}$

Therefore:

$a(R+h) \approx G\frac{m_1}{R^2} -2G\frac{m_1}{R^2} \frac{h}{R} = g(1-2\frac{h}{R})$

Consider that the error in this expresion is quadratic in (h/R), and to consider quadratic correctiosn you must expand the taylor series to the next power:

$a(R+h) \approx a(R) + h \frac{da(r)}{dr}_{r=R} + \frac{h^2}{2!} \frac{d^2a(r)}{dr^2}_{r=R}$

6 0

4 years ago

Read 2 more answers

A heavy boy and a lightweight girl are balanced on a massless seesaw. If they both

Len [333]

Answer:

D. Nothing will happen; the seesaw will still be balanced.

Explanation:

D. Nothing will happen; the seesaw will still be balanced. Since both toruqes or momentums respect to the center have changed in the same amount (one-half their original distance) the seesaw will remain balanced, if the children change distance in a different amount then it will be out of balance

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3 years ago

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plzzz PLZZZ PLZZZ HELP WILL GIVE BRAINLIST What is one thing you have learned about the platform of one of the Third Parties? Be

ICE Princess25 [194]

Answer:

Third parties may also help voter turnout by bringing more people to the polls. Third party candidates at the top of the ticket can help to draw attention to other party candidates down the ballot, helping them to win local or state office.

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3 years ago

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a fluid has a density of 921 kg/m^3. at a depth of 1.22 m, the fluid pressure is 122,000 Pa. what is the pressure at the top of

Zielflug [23.3K]

Answer:

Patm or the presion at the top is equal to 110977.28[Pa]

Explanation:

First we must define the absolute pressure value, which is equal to the sum of the atmospheric pressure plus the manometrial pressure.

$P_{abs} = P_{atm}+P_{man}\\where:\\P_{abs}=122000[Pa]$

Now we can find the manometric pressure.

$P_{man} = density*g*h\\where:\\density = 921[kg/m^3]\\g = gravity = 9.81[m/s^2]\\h = depth = 1.22[m]$

$P_{man} = 921*9.81*1.22\\ P_{man}= 11022.71[Pa]$

Therefore:

$P_{atm} = P_{abs} - P_{man} \\P_{atm} = 122000 - 11022.71\\P_{atm}= 110977.28[Pa]$

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3 years ago

Most people in south and south east Asia live in

Dimas [21]

Answer small rual villages

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3 years ago