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4 years ago

How does the temperature of the Earth’s mantle compare to the other layers of the Earth?

Physics

2 answers:

wlad13 [49]4 years ago

8 0

The anwser is d the others don't make sense

marta [7]4 years ago

4 0

Its not a A or B cause the core is the hottest. Its not D either the answer is D

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The terminal velocity of a person falling in air depends upon the weight and the area of the person facing the fluid. Find the t

DENIUS [597]

Answer:

terminal velocity is;

v = 117.54 m/s

v = 423.144 km/hr

Explanation:

Given the data in the question;

we know that, the force on a body due to gravity is;

$F_g$ = mg

where m is mass and g is acceleration due to gravity

Force of drag is;

$F_d$ = $\frac{1}{2}$ pCAv²

where p is the density of fluid, C is the drag coefficient, A is the area and v is the terminal velocity.

Terminal velocity is reach when the force of gravity is equal to the force of drag.

$F_g = F_d$

mg = $\frac{1}{2}$ pCAv²

we solve for v

v = √( 2mg / pCA )

so we substitute in our values

v = √( [2×(86 kg)×9.8 m/s² ] / [ 1.21 kg/m³ × 0.7 × 0.145 m²] )

v = √( 1685.6 / 0.122015 )

v = √( 13814.6949 )

v = 117.54 m/s

v = ( 117.54 m/s × 3.6 ) = 423.144 km/hr

Therefore terminal velocity is;

v = 117.54 m/s

v = 423.144 km/hr

5 0

3 years ago

Noah stands 170 meters away from a steep canyon wall. He shouts and hears the echo of his

bezimeni [28]

340 ms

I got it right and I hope you do as well

6 0

3 years ago

An inductor has an inductance of 0.12 mH. Calculate the number of turns per unit length of this inductor if it has a total of 50

SSSSS [86.1K]

Answer:

Explanation:

5 0

3 years ago

Two particles with charges +6e and -6e are initially very far apart (effectively an infinite distance apart). They are then fixe

JulijaS [17]

Answer:

$\rm EPE_{final}-EPE_{initial}=-1.478\times 10^{-15}\ J.$

Explanation:

Given charges are:

$\rm q_1 = +6e.\\q_2 = -6e.$

The electric potential energy of a charge due to the electric field of another charge is given by

$\rm EPE=\dfrac{kq_1q_2}{r}.$

where,

k = Coulomb's constant, having value = $\rm 9\times 10^9\ Nm^2/C^2.$
r = distance between the charges.

When the charges are infinite distance apart, $\rm r = \infty$ ,

$\rm EPE_{initial} = \dfrac{kq_1q_2}{r}=0\ J.$

When the charges are $\rm 5.61\times 10^{-12}\ m$ apart, $\rm r=5.61\times 10^{-12}\ m$ ,

$\rm EPE_{final}=\dfrac{kq_1q_2}{r}\\=\dfrac{(9\times 10^9)\times (+6e)\times (-6e)}{5.61\times 10^{-12}}\\=-5.775\ e^2\times 10^{22}.$

Here, e is the charge on one electron, such that, $\rm e = -1.6\times 10^{-19}\ C$ .

Therefore,

$\rm EPE_{final}=-5.775\times (-1.6\times 10^{-19})^2\times 10^{22} = -1.478\times 10^{-15}\ J.$

Thus,

$\rm EPE_{final}-EPE_{initial}=-1.478\times 10^{-15}-0=-1.478\times 10^{-15}\ J.$

4 0

3 years ago

If we could see our own galaxy from 2 million light-years away, it would appear _________.

yaroslaw [1]

Answer:

It would appear as a flattened disk with a central bulge and spiral arms, spanning a few degrees across the sky.

Explanation:

I hope this helps! Have a great day!

Anygays-

5 0

2 years ago