Which of these statements best explains why the atmosphere of today was partly due to the interactions of spheres in the past?

Bacteria used nutrients in the soil for volcanic eruptions.
Chemicals released from water formed the atmosphere.
The molten Earth released hydrogen and helium into the atmosphere.
The increase in fertility of soil around the volcano helped produce rain.

Answer:

151.12 million miles i actually just did this question on a test

Explanation:

Mars orbits—or completes one revolution—around the Sun every 686.98 Earth days, or once every 1.88 Earth years.

The term "revolution" describes how an item moves in its orbit around another object. For instance, the 24-hour day is created as the Earth rotates on its own axis.

The 365-day year is made possible by the Sun's rotation of the Earth. A planet spins around a satellite.

While orbiting the Sun, Mars travels at an average speed of 53,979 miles per hour, which equates to 86,871 km per hour.

Thus, Mars orbits—or completes one revolution—around the Sun every 686.98 Earth days, or once every 1.88 Earth years.

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Answer:

[tex]1.199\ \mu\text{m}[/tex]

Explanation:

[tex]\lambda[/tex] = Wavelength = 600 nm

D = Distance of the light source from screen = 3 m

y = Distance of first order bright fringe from center = 4.84 mm

d = Distance between slits

m = Order = 1

We have the relation

[tex]y=\dfrac{D\lambda}{d}\\\Rightarrow d=\dfrac{D\lambda}{y}\\\Rightarrow d=\dfrac{3\times 600\times 10^{-9}}{4.84\times 10^{-3}}\\\Rightarrow d=0.0003719\ \text{m}[/tex]

From the question we have

[tex]y=\dfrac{\dfrac{1}{2}3\lambda}{d}\\\Rightarrow \lambda=\dfrac{2}{3}yd\\\Rightarrow \lambda=\dfrac{2}{3}\times 4.84\times 10^{-3}\times 0.0003719\\\Rightarrow \lambda=0.000001199\ \text{m}=1.199\ \mu\text{m}[/tex]

The required wavelength of light is [tex]1.199\ \mu\text{m}[/tex].

Answer: 0.8 g/cm

Explanation:

p= m/V

= 4 kg/ 5 liter

= 0.8

Answer:

.8

Explanation:

give them brainliest

Answer:

The right solution is "126 Psi".

Explanation:

The given values are:

P₁ = 130 psig

i.e.,

   = [tex]130\times 6.894[/tex]

   = [tex]896.22 \ Kpa[/tex]

or,

   = [tex]896.22\times 10^3 \ Pa[/tex]

Z₂ = 10ft

    = 3.05 m

[tex]\delta[/tex] = 1000 kg/m³

According to the question,

Z₁ = 0

V₁ = V₂

As we know,

⇒  [tex]\frac{P_1}{\delta_g} +\frac{V_1^2}{2g} +Z_1=\frac{P_2}{\delta_g} +\frac{V_2^2}{2g} +Z_2[/tex]

On substituting the values, we get

⇒  [tex]\frac{P_1}{\delta_g} +0+0=\frac{P_2}{\delta_g} +0+Z_2[/tex]

⇒  [tex]\frac{896.22\times 10^3}{1000\times 9.8} =\frac{P_2}{1000\times 9.8} +3.05[/tex]

⇒  [tex]P_2=866330 \ P_a[/tex]

i.e.,

⇒       [tex]=866330\times 0.000145[/tex]

⇒       [tex]=126 \ Psi[/tex]

Answer:  I'm not sure, but I think it would be a total lunar eclipse

When the moon passes through the penumbra of Earth’s shadow it is referred to as partial lunar eclipse. The correct option is C.

A partial lunar eclipse occurs when the moon is not completely immersed in the umbra of the earth's shadow.

During a partial solar eclipse, the Moon, Sun, and Earth do not align perfectly straight, and the Moon casts only the penumbra of its shadow on Earth. From our vantage point, it appears that the Moon has eaten the Sun.

A shadow's penumbra is the lighter outer edge. Partial solar eclipses are caused by the Moon's penumbra, while penumbral lunar eclipses are caused by the Earth's penumbra. The penumbra is a type of lighter shadow.

The Penumbra is a half-shadow region that occurs when an object only partially covers a light source.

Thus, the correct option is C.

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Answer: Both False

Explanation:

Our Milky Way Galaxy is a spiral galaxy. Some spiral galaxies are what we call "barred spirals" because the central bulge looks elongated

Irregualuar glaxyices are all over  the place

Answer:

Heat loss per unit length = 642.358 W/m

The heat loss per unit length on a breezy day during 8 m/s speed is = 1760.205 W/m

Explanation:

From the information given:

Diameter D [tex]= 100 mm = 0.1 m[/tex]

Surface emissivity ε = 0.8

Temperature of steam [tex]T_s[/tex] = 150° C = 423K

Atmospheric air temperature [tex]T_{\infty} = 20^0 \ C = 293 \ K[/tex]

Velocity of wind V = 8 m/s

To calculate average film temperature:

[tex]T_f = \dfrac{T_s+T_{\infty}}{2}[/tex]

[tex]T_f = \dfrac{423+293}{2}[/tex]

[tex]T_f = \dfrac{716}{2}[/tex]

[tex]T_f = 358 \ K[/tex]

To calculate volume expansion coefficient

[tex]\beta= \dfrac{1}{T_f} \\ \\ \beta= \dfrac{1}{358} \\ \\ \beta= 2.79 \times 10^{-3} \ K^{-1}[/tex]

From the table of "Thermophysical properties of gases at atmospheric pressure" relating to 358 K of average film temperature; the following data are obtained;

Kinematic viscosity (v) = 21.7984 × 10⁻⁶ m²/s

Thermal conductivity k = 30.608 × 10⁻³ W/m.K

Thermal diffusivity ∝ = 31.244 × 10⁻⁶ m²/s

Prandtl no. Pr = 0.698

Rayleigh No. for the steam line is determined as follows:

[tex]Ra_{D} = \dfrac{g \times \beta (T_s-T_{\infty}) \times D_b^3}{\alpha\times v}[/tex]

[tex]Ra_{D} = \dfrac{9.8 \times (2.79 *10^{-3})(150-20) \times (0.1)^3}{(31.244\times 10^{-6}) \times (21.7984\times 10^{-6})}[/tex]

[tex]Ra_{D} = 5.224 \times 10^6[/tex]

The average Nusselt number is:

[tex]Nu_D = \Big \{ 0.60 + \dfrac{0.387(Ra_D)^{1/6}}{[ 1+ (0.559/Pr)^{9/16}]^{8/27}} \Big \}^2[/tex]

[tex]Nu_D = \Big \{ 0.60 + \dfrac{0.387(5.224\times 10^6)^{1/6}}{[ 1+ (0.559/0.698)^{9/16}]^{8/27}} \Big \}^2[/tex]

[tex]Nu_D = \Big \{ 0.60 + \dfrac{5.0977}{[ 1.8826]^{8/27}}\Big \}^2[/tex]

[tex]Nu_D = \Big \{ 0.60 + 4.226 \Big \}^2[/tex]

[tex]Nu_D = 23.29[/tex]

However, for the heat transfer coefficient; we have:

[tex]h_D = \dfrac{Nu_D\times k}{D_b} \\ \\ h_D = \dfrac{(23.29) \times (30.608 \times 10^{-3} )}{0.1}[/tex]

[tex]h_D = 7.129 \ Wm^2 .K[/tex]

Hence, Stefan-Boltzmann constant [tex]\sigma = 5.67 \times 10^{-8} \ W/m^2.K^4[/tex]

Now;

To determine the heat loss using the formula:

[tex]q'_b = q'_{ev} + q'_{rad} \\ \\ q'_b = h_D (\pi D_o) (T_t-T_{\infty})+\varepsilon(\pi D_b)\sigma (T_t^4-T_{\infty }^4)[/tex]

[tex]q'_b = (7.129)(\pi*0.1) (423-293) + (0.8) (\pi*0.1) (5.67 *10^{-8}) (423^4-293^4) \\ \\ q'_b = 291.153 + 351.205 \\ \\ \mathbf{q'_b = 642.258 \ W/m}[/tex]

Now; here we need to determine the Reynold no and the average Nusselt number:

[tex]Re_D = \dfrac{VD_b}{v } \\ \\ Re_D = \dfrac{8 *0.1}{21.7984 \times 10^{-6}} \\ \\ Re_D = 3.6699 \times 10^4[/tex]

However, to determine the avg. Nusselt no by using Churchill-Bernstein correlation, we have;

[tex]Nu_D = 0.3 + \dfrac{0.62 \times Re_D^{1/2}* Pr^{1/3}}{[1+(0.4/Pr)^{2/3}]^{1/4}} [1+ (\dfrac{Re_D}{282000})^{5/8}]^{4/5}[/tex]

[tex]Nu_D = 0.3 + \dfrac{0.62 \times (3.6699*10^4)^{1/2}* (0.698)^{1/3}}{[1+(0.4/0.698)^{2/3}]^{1/4}} [1+ (\dfrac{3.669*10^4}{282000})^{5/8}]^{4/5}[/tex]

[tex]Nu_D = (0.3 +\dfrac{105.359}{1.140}\times 1.218) \\ \\ Nu_D = 112.86[/tex]

SO, the heat transfer coefficient for forced convection is determined as follows afterward:

[tex]h_D = \dfrac{Nu_{D}* k}{D_b} \\ \ h_D = \dfrac{112.86*30.608 *10^{-3}}{0.1} \\ \\ h_D = 34.5 \ W/m^2 .K[/tex]

Finally; The heat loss per unit length on a breezy day during 8 m/s speed is:

[tex]q'b = h_D (\pi D_b) (T_s-T_{\infty}) + \varepsilon (\pi D_b) \sigma (T_s^4-T_ {\infty}^4) \\ \\ q'b = (34.5) (\pi *0.1) (423-293) + (0.8) (\pi*0.1) (5.67*10^{-8}) (423^4 - 293^4) \\ \\ = 1409 +351.205 \\ \\ \mathbf{q'b = 1760.205 \ W/m}[/tex]

Answer:

An Oven

Explanation:

The heat is higher, so it moves faster. Shile in a freezer the particles are extremely slow!

Answer:

320 J

Explanation:

From the question,

KE = 1/2mv².................. Equation 1

Where KE = Kinetic Energy, m = mass of the rock, v = velocity of the rock

Given: m = 40 kg, v = 4m/s

Substitute these values into equation 1

KE = 1/2(40)(4²)

KE = 20×16

KE = 320 J

Hence the kinetic energy of the rock is 320 J

Answer:

electric force

Explanation:

its a contact and noncontact force

Answer:

Is where two or more inductors are “linked” so that voltage is induced in one coil proportional to the rate-of-change of current in another

Explanation:

रात में घूमने वाला arthaarat निशाचर

Answer:

Following are the solution to the given question:

Explanation:

The input linear polarisation was shown at an angle of [tex]2 \mu[/tex]. It's a very popular use of a half-wave plate. In particular, consider the case [tex]\mu = 45 \pm[/tex], at which the angle of rotation is [tex]90\pm[/tex]. HWP thereby provides a great way to turn, for instance, a linear polarised light that swings horizontally to polarise vertically. Illustration of action on event circularly polarized light of the half-wave platform. Customarily it is the slow axis of HWP that corresponds to either the rotation. Note that perhaps the vector of polarization is "double-headed," i.e., the electromagnetic current swinging back and forward in time. Therefore the turning angle could be referred to as the rapid axis to reach the same result. Please find the attached file.

Answer: 10.98 N

Explanation:

Given

mass of box is [tex]m=0.5\ slugs\approx 7.3\ kg[/tex]

The coefficient of kinetic friction is [tex]\mu= 0.4[/tex]

It is given that on the application of force box started moving with constant speed i.e. there is no net external force

[tex]\Rightarrow F+mg\sin 30^{\circ}=\mu mg\cos 30^{\circ}\\\Rightarrow F=mg(\mu \cos 30^{\circ}-\sin 30^{\circ})\\\Rightarrow F=7.3\times 9.8(0.866\times0 .4-0.5)=-10.98\ N\\[/tex]

The negative sign indicates direction of force is opposite i.e. it must be applied upwards

Answer:

a = 72 m/s^2

Explanation:

Since the object is moving in circle, the acceleration points TOWARDS the centre of the circle (it is centripetal acceleration)

a = v^2 / r

a = 60^2 / 50

a = 72 m/s^2

Explanation:

When hot water is poured on the can in a bucket of cold water, the can crushes off means it gets unshaped

Answer:

a) the plastic tube need to be 24.7 m long

b) the kilometer of copper wire required is 15.7

c) the resistance of this solenoid is 5538.8 2 ohms

Explanation:

Given the data in the question;

we determine the length of the plastic tube. assuming the solenoid is long.

the self inductance of a long solenoid is;

L = μ₀n²πr²l

μ₀ = 4π × 10⁻⁷ T-m/A

where

n = number of turns per unit length

r = radius of the solenoid = 8cm (as the diameter of the plastic hollow tube is 16 cm)

l = length of the solenoid or the length of the plastic tube

we find n = number of turns per unit length

given that, the copper wire to be wound around the solenoid is 0.79 mm in diameter

number of turns per meter = n = 1 / ( 0.79 × 10⁻³ m ) = 1265.8 turns/meter

So from our previous formula, we find l

L = μ₀n²πr²l

we substitute

1.0 H = (4π × 10⁻⁷ T-m/A)( 1265.8 )²(3.14)(0.08)² ( l)

1 = 0.04048 × l

l = 1 / 0.04048

l = 24.7 m

Therefore, the plastic tube need to be 24.7 m long

b)  

n = number of turns per unit length = 1265.8 turns/metre

so, the length of the plastic tube over which the copper wire is to be wound,

number of turns of copper wire required = n × l

= 1265.8 turns/meter × 24.7 m

= 31,265.26 turns

Now each turn of the copper wire is to be wound across the 18cm diameter of the plastic tube.

so for each turn length of copper wire required = 2π × r

= 2π × 0.08 m

= 0.5026548 m

So copper wire required for 31,265.26 turns will be;

⇒ 31,265.26 × 0.5026548 = 15715.63m = 15.7 km

Therefore, the kilometer of copper wire required is 15.7

c)

p = resistivity of copper = 1.68 × 10⁻⁸ ohm-m

Resistance = pl/a

where l is length of copper wire, a is cross sectional area;

diameter of copper wire is 0.79-mm

radius of copper wire is 0.79/2 = 0.395 mm = 0.000395 m

area of cross section of copper wire a = πr² = π( 0.00395)² = 4.9 × 10⁻⁷ m²

Resistance = pl/a

we substitute

Resistance = [(1.68 × 10⁻⁸ ohm-m)( 15715.63m )] / [ 4.9 × 10⁻⁷ m² ]

Resistance =  5538.8 2 ohms

Therefore,  the resistance of this solenoid is 5538.8 2 ohms

Answer:

they are right it is a new moon

Explanation:

took the test

Answer:

correct tysmm

Explanation:

Answer:

Explanation:

(A)

Force is said to be the rate of change of momentum.

At the instant, the parallel component of the change of momentum is zero. SInce centripetal force always acts towards the center, hence the direction upwards ( along the positive y-axis).

[tex]\dfrac{dp}{dt} = F(+\hat y)[/tex]

[tex]= \dfrac{mv^2}{r}(+\hat y) \ \\ \\ = \dfrac{28 \ kg \times (7 \ m/s)^2}{3.3 \ m }( + \hat y) \\ \\ = \mathbf{416 \ N ( + \hat y)}[/tex]

(B)

The net force acting on the child is:

[tex]F_{net} = \dfrac{mv^2}{r}( + \hat y) \\ \\[/tex]

[tex]= \dfrac{28 \ kg \times (7 \ m/s)^2}{3.3 \ m }( + \hat y) \\ \\ = \mathbf{416 \ N ( + \hat y)}[/tex]

Answer:

hydrogen, helium, lithium, beryllium, boron, carbon, nitrogen, oxygen, fluorine, neon, sodium, magnesium, aluminum, silicon, phosphorus, sulfur, chlorine, argon, potassium, calcium, scandium, titanium, vanadium, chromium, maganese, iron, cobalt, nickel, copper, zinnc, gallium, germanium, aresnic, selenium, bromine, krypton, rubidium, strongtium, yttrium, zirconium, niobium, molybdenum, technetium, ruthenium, rhodium, palladium, silver, cadmium, indium, tin, atimony, tellurium, iodine, xenon, cesium, barium, lanthanum, cerium, praseodyumium, neodymim, promethium, samarium, europium, gadolinium, terbium, dysprosium, holmium, erbium, thulium, ytterbium, luteium, hafnium, tantalum, tungsten, rhenium, osmium, irdium, platinum, gold, mercury, thallium, lead bismuth, polnium, astatine, radon, rancium, radium, actinum, thorium, protactinium

Explanation:

Answer:

20 J/s

Explanation:

Given data

Kinetic energy=100 J

Time= 5 seconds

Hence the rate at which the kinetic friction work is

=100/5

=20 J/s

Therefore the answer is 20 J/s

Answer:

The correct answer is A) It is upright and 1.5m behind the mirror

Explanation:

Your reflection must be upright, meaning vertical/erect, and the distance will be the exact same. Also, the reflected ray appears as if it had traveled from an object located behind the mirror.

Answer:

distance = 112 miles

Explanation:

its 12 miles every 0.6 in a hour

Answer:

The frequency of the oscillation is 2.45 Hz.

Explanation:

Given;

mass of the spring, m = 0.5 kg

total mechanical energy of the spring, E = 12 J

Determine the spring constant, k as follows;

E = ¹/₂kA²

kA² = 2E

k = (2E) / (A²)

k = (2 x 12) / (0.45²)

k = 118.519 N/m

Determine the angular frequency, ω;

[tex]\omega = \sqrt{\frac{k}{m} } \\\\\omega = \sqrt{\frac{118.519}{0.5} } \\\\\omega = 15.396 \ rad/s[/tex]

Determine the frequency of the oscillation;

ω = 2πf

f = (ω) / (2π)

f = (15.396) / (2π)

f = 2.45 Hz

Therefore, the frequency of the oscillation is 2.45 Hz.

Answer:

5.0/s

Explanation:

Answer:

b and a it is this that abewsr

Answer:

32.25 s

Explanation:

From the question,

P = W/t.............. Equation 1

Where P = Power, W = work done, t = time.

But

W = F×d................. Equation 2

Where F = force and d = distance

Substitute equation 2 into equation 1

P = F×d/t............... Equation 3

make t the subject of euqation 3

t = (F×d)/P............. Equation 4

Givn: F = 150 N, d = 4.3 m, P = 20 watts.

Substitute these values into equation 4

t = (150×4.3)/20

t = 32.25 s

Answer:

The respiratory system provides oxygen for cells, while the circulatory system transports oxygen to cells.

Explanation:

so the answer is D

Answer:

[tex]3.52\times 10^{25}\ \text{J}[/tex]

Explanation:

G = Gravitational constant = [tex]6.674\times 10^{-11}\ \text{Nm}^2/\text{kg}^2[/tex]

M = Mass of Sun = [tex]1.989\times 10^{30}\ \text{kg}[/tex]

m = Mass of Earth = [tex]5.972\times 10^{24}\ \text{kg}[/tex]

[tex]r_i[/tex] = Initial radius of orbit = [tex]1.5\times 10^{11}\ \text{m}[/tex]

[tex]r_f[/tex] = Final radius of orbit = [tex]((1.5\times 10^{11})+2\times 10^3)\ \text{m}[/tex]

Energy required is given by

[tex]E=\dfrac{1}{2}\Delta U\\\Rightarrow E=\dfrac{GMm}{2}(\dfrac{1}{r_i}-\dfrac{1}{r_f})\\\Rightarrow E=\dfrac{6.674\times 10^{-11}\times 1.989\times 10^{30}\times 5.972\times 10^{24}}{2}(\dfrac{1}{1.5\times 10^{11}}-\dfrac{1}{(1.5\times 10^{11})+2\times 10^3})\\\Rightarrow E=3.52\times 10^{25}\ \text{J}[/tex]

The energy required would be [tex]3.52\times 10^{25}\ \text{J}[/tex].