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Which of the following is an example of the concepts of growth and development functioning together?

DedPeter [7]

Hello there.

Which of the following is an example of the concepts of growth and development functioning together? 

 C. Children learning to print their names

7 0

3 years ago

Light travels approximately 982,080,000 ft/s, and one year has approximately 32,000,000 seconds. A light year is the distance li

diamong [38]

So to solve this problem, we must convert units and then cancel units simultaneously:

$(\frac{982,080,000ft}{s})*(\frac{32,000,000s}{1})*(\frac{12inches}{1ft})=3.77x10^{17} inches$

Therefore we know that the distance that light travels in one year is $3.77x10^{17}$ inches.

7 0

4 years ago

Pwease hlp me (>O<) Darkness falls during daytime hours. Which event might have occurred?

Tcecarenko [31]

Answer:

D. Solar Eclipse

Explanation:

This is quite a phenomenal occurrence. Please, allow me to explain why D is correct.

A. A New Moon: This is the moon's first lunar phase. This is such a phase during which the sun and earth are located at nearly precise opposite ends of the moon, an occurrence that renders the moon nearly (if not) entirely invisible by the naked eye from an earthly perspective. This is not an occurrence that renders darkness during daylight.

B & C: (Partial & Total) Lunar Eclipse: A Lunar Eclipse is a beautiful occurrence that is visible only during nightfall, but can be witnessed from any location on earth that is currently facing away from the sun (a.k.a. nighttime hours). This occurs when the earth, sun, and moon are nearly perfectly aligned, with the sun and moon at opposite ends of the earth. The earth blocks the light of the sun, and though the side facing the sun won't experience any difference, the side facing away from the sun will be able to see the light from the sun beam around the earth, hitting the moon and reflecting back on us. The shadow of the earth entirely covers the moon during this event. The phases determines exactly what we see and to what degree/extent. This eclipse itself can happen numerous times a year, though only in Partial or Penumbral phases.

D. A Solar Eclipse: A solar eclipse is a breathtaking and oftentimes somewhat terrifying occurrence during which the moon passes perfectly in between the earth and the sun. Unlike a New Moon, however, this occurrence is perceived during the day, so for the side of the earth FACING the sun. When the moon is making its journey in between the sun and earth (which takes roughly 3-5m), the massive silhouette of the moon can be seen blotting out the sun, effectively causing a brief moment of darkness during daytime hours. This occurence is considered remarkably rare because DESPITE it occuring once per year, it can ONLY be viewed from specific locations on earth.

7 0

3 years ago

Read 2 more answers

A 0.0550-kg ice cube at −30.0°C is placed in 0.400 kg of 35.0°C water in a very well-insulated container. What is the final temp

KatRina [158]

Answer:

19.34°C

Explanation:

When the ice cube is placed in the water, heat will be transferred from the hot water to it such that the heat gained (Q₁) by the ice is equal to the heat lost(Q₂) by the hot water and a final equilibrium temperature is reached between the melted ice and the cooling/cooled hot water. i.e

Q₁ = -Q₂ ----------------------(i)

{A} Q₁ is the heat gained by the ice and it is given by the sum of ;

(i) the heat required to raise the temperature of the ice from -30°C to 0°C. This is given by [m₁ x c₁ x ΔT]

Where;

m₁ = mass of ice = 0.0550kg

c₁ = a constant called specific heat capacity of ice = 2108J/kg°C

ΔT₁ = change in the temperature of ice as it melts from -30°C to 0°C = [0 - (-30)]°C = [0 + 30]°C = 30°C

(ii) and the heat required to melt the ice completely - This is called the heat of fusion. This is given by [m₁ x L₁]

Where;

m₁ = mass of ice = 0.0550kg

L₁ = a constant called latent heat of fusion of ice = 334 x 10³J/kg

Therefore,

Q₁ = [m₁ x c₁ x ΔT₁] + [m₁ x L₁] ------------------(ii)

Substitute the values of m₁, c₁, ΔT₁ and L₁ into equation (ii) as follows;

Q₁ = [0.0550 x 2108 x 30] + [0.0550 x 334 x 10³]

Q₁ = [3478.2] + [18370]

Q₁ = 21848.2 J

{B} Q₂ is the heat lost by the hot water and is given by

Q₂ = m₂ x c₂ x ΔT₂ -----------------(iii)

Where;

m₂ = mass of water = 0.400kg

c₂ = a constant called specific heat capacity of water = 4200J/Kg°C

ΔT₂ = change in the temperature of water as it cools from 35°C to the final temperature of the hot water (T) = (T - 35)°C

Substitute these values into equation (iii) as follows;

Q₂ = 0.400 x 4200 x (T - 35)

Q₂ = 1680 x (T-35) J

{C} Now to get the final temperature, substitute the values of Q₁ and Q₂ into equation (i) as follows;

Q₁ = -Q₂

=> 21848.2 = - 1680 x (T-35)

=> 35 - T = 21848.2 / 1680

=> 35 - T = 13

=> T = 35 - 13

=> T = 22

Therefore the final temperature of the hot water is 22°C.

Now let's find the final temperature of the mixture.

The mixture contains hot water at 22°C and melted ice at 0°C

At this temperature, the heat ( $Q_{W}$ ) due to the hot water will be equal to the negative of the one ( $Q_{I}$ ) due to the melted ice.

i.e

$Q_{W}$ = - $Q_{I}$ -----------------(a)

Where;

$Q_{I}$ = $m_{I}$ x $c_{I}$ x Δ $T_{I}$ [ $m_{I}$ = mass of ice, $c_{I}$ = specific heat capacity of melted ice which is now water and Δ $T_{I}$ = change in temperature of the melted ice]

and

$Q_{W}$ = $m_{W}$ x $c_{W}$ x Δ $T_{W}$

[ $m_{W}$ = mass of water, $c_{W}$ = specific heat capacity of water and Δ $T_{W}$ = change in temperature of the water]

Substitute the values of $Q_{W}$ and $Q_{I}$ into equation (a) as follows

$m_{W}$ x $c_{W}$ x Δ $T_{W}$ = - $m_{I}$ x $c_{I}$ x Δ $T_{I}$

Note that $c_{W}$ and $c_{I}$ are the same since they are both specific heat capacities of water. Therefore, the equation above becomes;

$m_{W}$ x Δ $T_{W}$ = - $m_{I}$ x Δ $T_{I}$ -----------------------(b)

Now, let's analyse Δ $T_{W}$ and Δ $T_{I}$ . The final temperature ( $T_{F}$ ) of the two kinds of water(melted ice and cooled water) are now the same.

=> Δ $T_{W}$ = change in temperature of water = final temperature of water( $T_{F}$ ) - initial temperature of water( $T_{IW}$ )

Δ $T_{W}$ = $T_{F}$ - $T_{IW}$

Where;

$T_{IW}$ = 22°C [which is the final temperature of water before mixture]

=> Δ $T_{I}$ = change in temperature of melted ice = final temperature of water( $T_{F}$ ) - initial temperature of melted ice ( $T_{II}$ )

Δ $T_{I}$ = $T_{F}$ - $T_{II}$

$T_{II}$ = 0°C (Initial temperature of the melted ice)

Substitute these values into equation (b) as follows;

$m_{W}$ x Δ $T_{W}$ = - $m_{I}$ x Δ $T_{I}$

0.400 x ( $T_{F}$ - $T_{IW}$ ) = -0.0550 x ( $T_{F}$ - $T_{II}$ )

0.400 x ( $T_{F}$ - 22) = -0.0550 x ( $T_{F}$ - 0)

0.400 x ( $T_{F}$ - 22) = -0.0550 x ( $T_{F}$ )

0.400 $T_{F}$ - 8.8 = -0.0550 $T_{F}$

0.400 $T_{F}$ + 0.0550 $T_{F}$ = 8.8

0.455 $T_{F}$ = 8.8

$T_{F}$ = 19.34°C

Therefore, the final temperature of the mixture is 19.34°C

8 0

3 years ago

A ufo was detected on the radar flying 7400 miles in 3 min, what is the estimated speed per hour?

Ghella [55]

S=7400mi
t=3 min= 0.05h
v=7400mi/0.05h
v=148000mph

8 0

4 years ago