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2 years ago

What are the reasons for the establishment of UGA?

Physics

1 answer:

3241004551 [841]2 years ago

8 0

Answer:

In February 1784, just after the close of the Revolutionary War, the General Assembly of Georgia earmarked 40,000 acres of land to endow "a college or seminary of learning." The following year, Abraham Baldwin, a lawyer and minister educated at Yale University in New Haven, Connecticut, who had settled in Georgia

Explanation:

please mark this answer as brainliest

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An engineer wants to design a circular racetrack of radius R such that cars of mass m can go around the track at speed without t

gtnhenbr [62]

1. $tan \theta = \frac{v^2}{Rg}$

For the first part, we just need to write the equation of the forces along two perpendicular directions.

We have actually only two forces acting on the car, if we want it to go around the track without friction:

- The weight of the car, mg, downward

- The normal reaction of the track on the car, N, which is perpendicular to the track itself (see free-body diagram attached)

By resolving the normal reaction along the horizontal and vertical direction, we find the following equations:

$N cos \theta = mg$ (1)

$N sin \theta = m \frac{v^2}{R}$ (2)

where in the second equation, the term $m\frac{v^2}{R}$ represents the centripetal force, with v being the speed of the car and R the radius of the track.

Dividing eq.(2) by eq.(1), we get the following expression:

$tan \theta = \frac{v^2}{Rg}$

2. $F=\frac{m}{R}(w^2-v^2)$

In this second situation, the cars moves around the track at a speed

$w>v$

This means that the centripetal force term

$m\frac{v^2}{R}$

is now larger than before, and therefore, the horizontal component of the normal reaction, $N sin \theta$ , is no longer enough to keep the car in circular motion.

This means, therefore, that an additional radial force F is required to keep the car round the track in circular motion, and therefore the equation becomes

$N sin \theta + F = m\frac{w^2}{R}$

And re-arranging for F,

$F=m\frac{w^2}{R}-N sin \theta$ (3)

But from eq.(2) in the previous part we know that

$N sin \theta = m \frac{v^2}{R}$

So, susbtituting into eq.(3),

$F=m\frac{w^2}{R}-m\frac{v^2}{R}=\frac{m}{R}(w^2-v^2)$

4 0

2 years ago

Students use a stretched elastic band to launch carts of known mass horizontally on a track. The elastic bands exert a force F,

rodikova [14]

Answer:

the correct answer is E

A graph of the cart's maximum speed squared as a function of x^3

Explanation:

For this exercise let's use Newton's second law

F = m a

force has the form

F = k x²

and acceleration is related to velocity

a = dv / dt

Let's use the chain rule or L'Hospital

a = dv /dx dx/dt

a = dv /dx v

let's substitute

k x² = m v dv / dx

k /m x² dx = v dv

we integrate

k /m x³ /3 = v² / 2

v² = (2k /3m) x³

This is the expression for the variation of the speed as a function of the position, to make a linear graph realism the changes of variable

y = v²

x´ = x³

y = (2k/3m) x´

if we graph y vs x 'we have a linear graph whose slope is

m = 2k / 3m

By reviewing the different answers, the correct answer is E

4 0

2 years ago

Please re-explain the following phrases in terms of momentum

zheka24 [161]

A. An object in motion will remain in motion unless acted upon by an outside force : The momentum of an object is constant unless an outside force acts on the object.

B. Force is defined as mass times acceleration : the rate of change of the momentum of a particle is proportional to the force F acting on it, hence the force is equal to <span>mass times acceleration.

C. </span>For every action there is an equal and opposite reaction : <span>to every action force there is an equal and opposite reaction force. </span>

8 0

2 years ago

Which statement describes one way in which global winds affect weather and climate? A. Polar easterlies move warm air to the mid

Genrish500 [490]

The answer your looking for is "D".

4 0

2 years ago

Read 2 more answers

13. A diver swims to a depth of 3.2 m in a freshwater lake. What is the increase in the force pushing in on her eardrum, compare

Gnoma [55]

Answer:

1.88 N

Explanation:

h = 3.2 m, A = 0.6 cm^2 = 0.6 x 10^-4 m^2

density of water, d = 1000 kg/m^3, g = 9.8 m/s^2

Pressure at depth h, P = h x d x g = 3.2 x 1000 x 9.8 = 31.36 x 10^3 Pa

Force = pressure x Area = 31.36 x 10^3 x 0.6 x 10^-4 = 1.88 N

Thus, the force on ear drum is 1.88 N.

6 0

2 years ago