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4 years ago

Koala bears can eat only certain kinds of Australian eucalyptus leaves. Koalas are considered _____.

2 answers:

3 0

They are marsupials that are considered "specialist eaters" since they only eat a certain type of leaf. While generalist feeders aren't as picky as the koala bears. I hope this helps!

defon4 years ago

3 0

Answer:

Herbivore Marsupials

Explanation:

Koalas are the herbivore marsupials that are found to be dwelling on the trees, and are dominantly found in the continent of Australia. They are very selective in terms of food, and are able to eat only certain types of plant leaves but not all. They obtain relatively less nutritious food, thereby receives less amount of energy. Due to this, the Koalas are mostly lazy and spend most of their time by sleeping and sitting.

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Explain the differences between horticulture and agriculture.

sammy [17]

<h2><u>Hello </u><u>There</u></h2>

$\tt{Difference \: Between \: \underline\red {Horticulture} \: And \: \underline\green{Agriculture}}$

$\text{\underline\red {Horticulture}}$

The Cultivation of Fruits, Vegetables and Flowers for domestic and international markets are called Horticulture.

$\text{\underline\green{Agriculture}}$

It is science, art and occupation of cultivating the soil, producing crops and livestock. It is systematic and controlled use of living organisms and the environment to improve the human condition.

<h3>Hope This Helps</h3>

6 0

2 years ago

A police car chases a speeder along a straight road towards a cliff both vehicles move at 160km/h the siren on the police car pr

natta225 [31]

Answer:

f ’= 97.0 Hz

Explanation:

This is an exercise of the doppler effect use the frequency change due to the relative movement of the fort and the observer

in this case the source is the police cases that go to vs = 160 km / h

and the observer is vo = 120 km / h

the relationship of the doppler effect is

f ’= f₀ (v + v₀ / v- $v_{s}$ )

let's reduce the magnitude to the SI system

v_{s} = 160 km / h (1000 m / 1km) (1h / 3600s) = 44.44 m / s

v₀ = 120 km / h (1000m / 1km) (1h / 3600s) = 33.33 m / s

we substitute in the equation of the Doppler effect

f ‘= 100 (330+ 33.33 / 330-44.44)

f ’= 97.0 Hz

4 0

3 years ago

6. A child tugs on a rope attached to a 0.62-kg toy with a horizontal force

ser-zykov [4K]

Answer:

Explanation:

Force Mass * acceleration

F = ma

a = F/m

Sum of force = 16.3N - 15.8N = 0.5N

Mass = 0.62kg

Substitute

a = 0.5/ * 0.62

a = 0.81m/s²

The acceleration of the toy is 0.81m/s²

8 0

3 years ago

Assignments

sladkih [1.3K]

Hello there!

I hope you and your family are staying safe and healthy during this unprecendented time.

A) What is the work done?

Answer: We need to use the formula

$w=-F_f(d)$

$w=-(35)(20)$

$w=-700J$

B) What is the work done on the cart by the gravitational force?

Alright, we know that the gravitional force is perpendicular to the diplacement. Therefore, we gonna use the following formula:

$w=Fdcos90$

$w=0$

C) What is the work done on the cart by the shopper?

This is the easier part, since we already know that the work done by the shopper is the same as the work done by the friction force

$W shopper + W friction = 0\\W shopper = W-friction \\W shopper = 700J$

D) Find the force the shopper exerts, using energy considerations.

$F_f+Fcos25=0\\-35+Fcos25=0\\F=38.6N$

E) What is the total work done?

You just need to add them:

$w=wshopper+wfriction\\w=0$

4 0

3 years ago

In the nuclear fission process mass is converted into energy. Determine the total mass that must be converted to energy in one y

brilliants [131]

Answer:

1) Mass that needs to be converted at 100% efficiency is 0.3504 kg

2) Mass that needs to be converted at 30% efficiency is 1.168 kg

Explanation:

By the principle of mass energy equivalence we have

$E=mc^{2}$

where,

'E' is the energy produced

'm' is the mass consumed

'c' is the velocity of light in free space

Now the energy produced by the reactor in 1 year equals

$Energy=Power\times time\\\\\therefore Energy=1\times 10^{9}\times 365\times 24\times 3600\\\\Energy=31.536\times 10^{15}Joules$

Thus the mass that is covertred at 100% efficiency is

$mass=\frac{Energy}{c^{2}}\\\\mass=\frac{31.536\times 10^{15}}{(3\times 10^{8})^{2}}\\\\mass=\frac{31.536\times 10^{15}}{9\times 10^{16}}\\\\\therefore mass=0.3504kg$

Part 2)

At 30% efficiency the mass converted equals

$mass|_{30}=\frac{mass|_{100}}{0.3}\\\\mass|_{30}=\frac{0.3504}{0.3}\\\\mass|_{30}=1.168kg$

8 0

4 years ago