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svlad2 [7]
3 years ago
5

wildlife biologists are working to save some unique birds in the forest of north georgia.which of these questions must be answer

ed first to determine how the birds can be protected for future generations
Physics
1 answer:
Maru [420]3 years ago
7 0
Extinct<span> might be a word you associate with animals that lived long ago, like the dinosaurs, but did you know that over 18,000 species are classified as "threatened" (susceptible to extinction) today? Scientists involved in wildlife conservation have a tough job; they are in charge of determining what needs to be done to prevent a species from becoming extinct. Habitat, food supply, and impacts of local human populations are just a few of the factors these scientists take into account. It is a lot to keep track of for a single location, but the job becomes even harder when it is a migratory animal. In this science project, you will get a firsthand look at their job. You will access </span>real<span> data about migratory birds and use satellite images to analyze their habitats, then come up with a conservation plan to protect the species from extinction.</span>
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A 3 kg object is moving along a horizontal surface. The kinetic energy of the object is increasing at a constant rate of 6 J/m;
Whitepunk [10]

To solve this problem we will apply the concepts of energy conservation and Newton's second law that defines force as the product of the object's mass with its acceleration. Additionally we will apply concepts related to the kinematics equations of linear motion.

For conservation of energy we have that work is equal to kinetic energy therefore,

W = KE

Fd = \frac{1}{2} mv^2

Here,

F = Force

d = Displacement

m = Mass

v= Velocity

At the same time we have the relation of

F = \frac{W}{d}

Therefore the value of the force can be interpreted as the rate of increase in energy per unit of distance, which makes it equivalent to

F = \frac{W}{d} = 6J/m

Applying Newton's Second Law

F = ma

6J/m = (3kg)a

a = 2m/s^2

In 4 seconds final velocity of the object becomes

v = at

v= 2*4

v= 8m/s

Then the work done is equal to,

W = KE

W = \frac{1}{2} mv^2

W = \frac{1}{2} (3)(82)

W = 96J

Then the displacement is,

W = F*d

d = \frac{W}{F}

d = \frac{96}{6}

d = 16 m

Therefore the distance moved is 16m

7 0
4 years ago
A child is swinging back and forth with a constant period and amplitude. Somewhere in front of the child, a stationary horn is e
Amanda [17]

Answer:

Explanation:

  We shall apply concept of Doppler's effect of apparent frequency to this problem . Here observer is moving sometimes towards and sometimes away from the source . When observer moves towards the source , apparent frequency is more than real frequency and when the observer moves away from the source , apparent frequency is less than real frequency . The apparent frequency depends upon velocity of observer . The formula for apparent frequency when observer is going away is as follows .

f = f₀ ( V - v₀ ) / V , f is apparent , f₀ is real frequency , V is velocity of sound and v is velocity of observer .

f will be lowest when v₀ is highest .

velocity of observer is highest when he is at the equilibrium position or at middle point .

So apparent frequency is lowest when observer is at the middle point and going away from the source  while swinging to and from before the source of sound .

3 0
3 years ago
A 2400-kg satellite is in a circular orbitaround a planet. The
Fofino [41]

Answer

given,

mass of satellite = 2400 Kg

speed of the satellite =  6.67 x 10³ m/s

acceleration of satellite = ?

gravitational force of the satellite will be equal to the centripetal force

F = \dfrac{mv^2}{r}

F = \dfrac{2400\times (6.67\times 10^3)^2}{r}

Assuming the radius of circular orbit = 8.92 x 10⁶ m

now,

F = \dfrac{2400\times (6.67\times 10^3)^2}{8.92\times 10^6}

F = 11970.11 N

acceleration,

a = \dfrac{F}{m}

a = \dfrac{11970.11}{2400}

  a = 4.98 m/s²

4 0
3 years ago
An electron is accelerated from rest by a potential difference of (24.5 A) V for a distance of (4.50 B) cm. Determine the de Bro
DiKsa [7]

Answer:

206 pm

Explanation:

We are given that

Potential difference,\Delta V=24.5+A V

Distance,d=4.5+B cm

We have to determine the de Brogile wavelength of the electron.

A=11 and B=5

\Delta V=24.5+11=35.5 V

d=4.5+5=9.5 cm

Charge on electron,q =1.6\times 10^{-19} C

Mass of electron=m=9.1\times 10^{-31} kg

Speed of electron,v=\sqrt{\frac{2q\Delta V}{m}}

Using the formula

v=\sqrt{\frac{2\times 1.6\times 10^{-19}\times 35.5}{9.1\times 10^{-31}}

v=3.53\times 10^6 m/s

de Brogile wavelength, \lambda=\frac{h}{mv}

Where h=6.626\times 10^{-34}

\lambda=\frac{6.626\times 10^{-34}}{9.1\times 10^{-31}\times 3.53\times 10^6}=2.06\times 10^{-10}=206\times 10^{-12} m

1 pm=10^{-12} m

\lambda=206 pm

7 0
3 years ago
Read 2 more answers
If the distance between the source of sound and observer is reduced to one half, then
Dmitriy789 [7]

Answer:

Let the power delivered by the sound wave be 'P'

Intensity by definition is the power delivered per unit area.

i.e. I =

dA

dP

;

In spherical polar coordinate system dA = r

2

Ω

I =

r

2

Ω

dP

;

I ∝

r

2

1

So if distance increases 3 fold. The Intensity becomes

9

1

times the initial

5 0
3 years ago
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