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3 years ago

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Ancient species found in just one region, whose close relatives have all gone extinct, are designated __________.

1 answer:

Molodets [167]3 years ago

3 0

Ancient species living in earth with all its close relatives dead are called living fossils.

<h3><u>Explanation:</u></h3>

Living fossils are interesting as well important creations of nature. Let's take an example of a plant <em>Ginkgo biloba</em>. Its a plant of gymnosperm family along with pine and fir, but belongs to the order Ginkgoales. In this order, Gingko biloba is the only plant that's alive till now, all the other species are dead.

Living fossils give us important informations about the adaptation that used to prevail in earth that lead to survival of species in contemporary times. It also establishes relationships between the two or more phyla of organisms in a taxonomic way.

Animal living fossils are lungfish, horse shoe crab, coelacanth fish, Tuatara, lobe finned fishes etc.

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What is the difference between the velocity and the speed of an object?

timurjin [86]

Answer:

Velocity has a direction associated with it, while speed has no specific direction.

Explanation:

Velocity is a vector, while speed is a scalar.

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3 years ago

A charge of 9 pC is uniformly distributed throughout the volumebetween concentric spherical surfaces having radii of 1.6 cmand

balu736 [363]

Answer:

Electric Field = 3.369 x 10^4 N/C

Explanation:

Radius = r = (r1 + r2) / 2 = (1.6 + 3.6) /2 = 2.6 cm + 2.3 cm = 4.9 cm = 0.049 m

As we know, Electric field = E = kQ/r.r

= 8.98755 x 10^9 x 9 x 10^-9 / 0.049 x 0.049 = 33689.275 N/C

= 3.369 x 10^4 N/C

7 0

3 years ago

What would result from under-coverage? A. a lack of accurate result B. an incorrect independent variable C. an experiment D. acc

kherson [118]

A. a lack of accurate <span>result</span>

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3 years ago

Read 2 more answers

Radiation from the Sun reaching Earth (just outside the atmosphere) has an intensity of 1.39 kW/m2. (a) Assuming that Earth (and

Zigmanuir [339]

Answer:

$F=5.8\times 10^{8}\ N$

$F=35.57\times 10^{21}\ N$

Explanation:

Given that

Intensity I

$I= 1.39\ KW/m^2$

$Speed\ of \ light = 2.99\times 10^8\ m/s$

Radius of earth,R = 6370 Km

We know that surface area of earth, A

$A=\pi R^2$

$A=\pi (6370\times 10^3)^2\ m^2$

$A=1.27\times 10^{14}\ m^2$

As we know that pressure due to intensity given as

$P=\dfrac{I}{V}$

V =Velocity of light

$V=3\times 10^8\ m/s$

$P=\dfrac{1.39\times 1000}{=3\times 10^8}$

$P=4.6\times 10^{-6}\ Pa$

We know that force F

F = P .A

$F=4.6\times 10^{-6}\times 1.27\times 10^{14}\ N$

$F=5.8\times 10^{8}\ N$

b)Gravitational force F

$F=G\dfrac{m.M}{r^2}$

$M = mass\ of\ sun = 2\times 10^{30} kg\\m = mass\ of\ earth = 6\times 10^{24}kg$

$r =1.5\times 10^{11}\ m$

$G =6.67\times 10^{-11}Nm^2/kg^2$

So F

$F=6.67\times 10^{-11}\times \dfrac{2\times 10^{30}\times 6\times 10^{24}kg}{1.5\times 10^{11}}$

$F=35.57\times 10^{21}\ N$

7 0

3 years ago

Read 2 more answers

A 3.00-kg object undergoes an acceleration given by a = (2.00 i + 5.00 j) m/s^2. Find (a) the resultant force acting on the obje

kobusy [5.1K]

Answer:

(a): The resultant force acting on the object are F= (5.99 i + 14.98 j).

(b): The magnitude of the resultant force are F= 16.4 N < 68.19º .

Explanation:

m= 3kg

a= 2 i + 5 j = 5 .38 < 68.19 º

F= m * a

F= 3* ( 5.38 < 68.19º )

F= 16.4 N < 68.19º

Fx= F * cos(68.19º)

Fx= 5.99

Fy= F* sin(68.19º)

Fy= 14.98

3 0

3 years ago