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3 years ago

How do think animals grow from babies to adults

Physics

1 answer:

Sladkaya [172]3 years ago

7 0

Answer:

As animals get older, their bodies begin to change as well as their instincts and priorities. They get bigger and might adapt or develop.

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Thick rubber gloves prevent electric shock because

Kaylis [27]

Answer:

Rubber is an insulator.

Explanation:

Rubber is an insulator. Electricity will always travel "the path of least resistance." Rubber has a very high resistance, so electricity will go somewhere else to find ground.

7 0

3 years ago

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The mean diameters of Mars and Earth are 6.9 ✕ 103 km and 1.3 ✕ 104 km, respectively. The mass of Mars is 0.11 times Earth's mas

Roman55 [17]

Answer:

(a) Ratio of mean density is 0.735

(b) Value of g on mars 0.920 $m,/sec^2$

Explanation:

We have given radius of mars $R_{mars}=6.9\times 10^3km=6.9\times 10^6m$ and radius of earth $R_{E}=1.3\times 10^4km=1.3\times 10^7m$

Mass of earth $M_E=5.972\times 10^{24}kg$

So mass of mars $M_m=5.972\times\times 0.11 \times 10^{24}=0.657\times 10^{24}kg$

Volume of mars $V=\frac{4}{3}\pi R^3=\frac{4}{3}\times 3.14\times (6.9\times 10^6)^3=1375.357\times 10^{18}m^3$

So density of mars $d_{mars}=\frac{mass}{volume}=\frac{0.657\times 10^{24}}{1375.357\times 10^{18}}=477.69kg/m^3$

Volume of earth $V=\frac{4}{3}\pi R^3=\frac{4}{3}\times 3.14\times (1.3\times 10^7)^3=9.198\times 10^{21}m^3$

So density of earth $d_{E}=\frac{mass}{volume}=\frac{5.972\times 10^{24}}{9.198\times 10^{21}}=649.271kg/m^3$

(A) So the ratio of mean density $\frac{d_{mars}}{d_E}=\frac{477.69}{649.27}=0.735$

(B) Value of g on mars

g is given by $g=\frac{GM}{R^2}=\frac{6.67\times 10^{-11}\times0.657\times 10^{24}}{(6.9\times 10^6)^2}=0.920m/sec^2$

$v=\sqrt{\frac{2GM}{R}}=\sqrt{\frac{2\times 6.67\times 10^{-11}\times 0.657\times 10^{24}}{6.9\times 10^6}}=3.563\times 10^4m/sec$

5 0

3 years ago

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Which is the correct definition for sea floor spreading

jok3333 [9.3K]

Seafloor spreading is a geologic process in which tectonic plates—large slabs of Earth's lithosphere—split apart from each other.

6 0

3 years ago

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The electric potential at the surface of a charged conductor _______.

frez [133]

Answer:

The electric potential at the surface of a charged conductor<u> is always such that the potential is zero at all points inside the conductor.</u>

Explanation:

Each point on the surface of a balanced charged conductor has the same electrical potential.

The surface on any charged conductor in electrostatic equilibrium is an equipotential surface. Since the electric field is equal to zero inside the conductor, the electric potential at any point inside and on the surface is equivalent to its value.

6 0

3 years ago

7. A beam of light converges at a point P. Now a lens is placed in the path of the convergent

Rudiy27

Answer:

Lens at a distance = 7.5 cm

Lens at a distance = 6.86 cm (Approx)

Explanation:

Given:

Object distance u = 12 cm

a) Focal length = 20 cm

b) Focal length = 16 cm

Computation:

a. 1/v = 1/u + 1/f

1/v = 1/20 + 1/12

v = 7.5 cm

Lens at a distance = 7.5 cm

b. 1/v = 1/u + 1/f

1/v = 1/16 + 1/12

v = 6.86 cm (Approx)

Lens at a distance = 6.86 cm (Approx)

5 0

3 years ago