2
Ca
+
O
2
→
2
CaO
What is the product of the reaction

Even before the advent of the telescope, ancient astronomers were able to observe the following:

1. Celestial Bodies: Ancient astronomers could observe celestial bodies such as the Sun, Moon, stars, and planets. They could track their movements across the sky and study their patterns and behaviors.

2. Solar and Lunar Eclipses: By carefully observing the positions of the Sun, Moon, and Earth, ancient astronomers could predict and witness solar and lunar eclipses. They noticed that during a solar eclipse, the Moon blocks the Sun's light, creating a temporary darkness on Earth, while during a lunar eclipse, the Earth casts a shadow on the Moon, causing it to appear reddish or darkened.

3. Stellar Positions: Ancient astronomers mapped and observed the positions of stars in the night sky. They recognized patterns and constellations, which helped them navigate and keep track of time.

4. Seasons and Celestial Movements: By observing the changing positions of the Sun and its daily and yearly motions, ancient astronomers could understand the changing seasons. They could determine solstices, equinoxes, and the length of days and nights.

5. Comet Appearances: Ancient astronomers were able to observe and document the appearance of comets in the night sky. They recognized these celestial objects as distinct from stars and noted their unusual and transient nature.

These observations formed the basis of ancient astronomy and laid the groundwork for the development of more advanced astronomical techniques and instruments, including the telescope.

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Given the mass of the cell is 1 * 10⁻¹⁴ kg, the number of cells in the 65 kg person is  6.5 * 10¹⁵  cells.

The number of cells in a human is calculated as follows:

The mass of an average cell is ten times the mass of a bacterium.

The mass of a bacterium = 1 * 10⁻¹⁵ kg

mass of an average cell = 10 *  1 * 10⁻¹⁵ kg =  1 * 10⁻¹⁴ kg

Number of cells = mass of person/mass of cell

Number of cells = 65 kg/1 * 10⁻¹⁴ kg

Number of cells = 6.5 * 10¹⁵  cells.

In conclusion, the number of cells is obtained by dividing the mass of the person by the mass of a average cell.

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Note that the complete question is given below:

Assuming the mass of an average cell is ten times the mass of a bacterium (which is 10⁻¹⁵ kg): Calculate the number of cells in a human assuming the mass of the person is 10² kg.

Answer:

heat

Explanation:

Answer:

Have a great day!

Explanation:

Barometers. A barometer is a scientific instrument used to measure atmospheric pressure, also called barometric pressure.

Answer:

atmospheric  pressure

Explanation:

Answer:

The thermal conductivity of the biomaterial is approximately 1.571 watts per meter-Celsius.

Explanation:

Let suppose that thermal conduction is uniform and one-dimensional, the conduction heat transfer (\(\dot Q\)), measured in watts, in the hollow cylinder is:

\(\dot Q = \frac{2\cdot k\cdot L}{\ln \left(\frac{D_{o}}{D_{i}} \right)}\cdot (T_{i}-T_{o})\)

Where:

\(k\) - Thermal conductivity, measured in watts per meter-Celsius.

\(L\) - Length of the cylinder, measured in meters.

\(D_{i}\) - Inner diameter, measured in meters.

\(D_{o}\) - Outer diameter, measured in meters.

\(T_{i}\) - Temperature at inner surface, measured in Celsius.

\(T_{o}\) - Temperature at outer surface, measured in Celsius.

Now we clear the thermal conductivity in the equation:

\(k = \frac{\dot Q}{2\cdot L\cdot (T_{i}-T_{o})}\cdot \ln\left(\frac{D_{o}}{D_{i}} \right)\)

If we know that \(\dot Q = 40.8\,W\), \(L = 0.6\,m\), \(T_{i} = 50\,^{\circ}C\), \(T_{o} = 20\,^{\circ}C\), \(D_{i} = 0.01\,m\) and \(D_{o} = 0.04\,m\), the thermal conductivity of the biomaterial is:

\(k = \left[\frac{40.8\,W}{2\cdot (0.6\,m)\cdot (50\,^{\circ}C-20\,^{\circ}C)}\right]\cdot \ln \left(\frac{0.04\,m}{0.01\,m} \right)\)

\(k \approx 1.571\,\frac{W}{m\cdot ^{\circ}C}\)

The thermal conductivity of the biomaterial is approximately 1.571 watts per meter-Celsius.

Answer:

The final speed of the second package is twice as much as the final speed of the first package.

Explanation:

Free Fall Motion

If an object is dropped in the air, it starts a vertical movement with an acceleration equal to g=9.8 m/s^2. The speed of the object after a time t is:

\(v=gt\)

And the distance traveled downwards is:

\(\displaystyle y=\frac{gt^2}{2}\)

If we know the height at which the object was dropped, we can calculate the time it takes to reach the ground by solving the last equation for t:

\(\displaystyle t=\sqrt{\frac{2y}{g}}\)

Replacing into the first equation:

\(\displaystyle v=g\sqrt{\frac{2y}{g}}\)

Rationalizing:

\(\displaystyle v=\sqrt{2gy}\)

Let's call v1 the final speed of the package dropped from a height H. Thus:

\(\displaystyle v_1=\sqrt{2gH}\)

Let v2 be the final speed of the package dropped from a height 4H. Thus:

\(\displaystyle v_2=\sqrt{2g(4H)}\)

Taking out the square root of 4:

\(\displaystyle v_2=2\sqrt{2gH}\)

Dividing v2/v1 we can compare the final speeds:

\(\displaystyle v_2/v_1=\frac{2\sqrt{2gH}}{\sqrt{2gH}}\)

Simplifying:

\(\displaystyle v_2/v_1=2\)

The final speed of the second package is twice as much as the final speed of the first package.

Answer:

it is safe to stand at the end of the table

Explanation:

For this exercise we use the rotational equilibrium condition

         Στ = 0

         W x₁ - w x₂ - w_table x₃ = 0

         M x₁ - m x₂ - m_table x₃ = 0

where the mass of the large rock is M = 380 kg and its distance to the pivot point x₁ = 850 cm = 0.85m

the mass of the man is 62 kg and the distance

            x₂ = 4.5 - 0.85

            x₂ = 3.65 m

the mass of the table (m_table = 22 kg) is at its geometric center

            x_{cm} = L/2 = 2.25 m

            x₃ = 2.25 -0.85

            x₃ = 1.4 m

let's look for the maximum mass of man

            m_{maximum} = \(\frac{ M x_1 -m_{table} x_3}{ x_2}\)

let's calculate

             m_{maximum} = \(\frac{ 380 \ 0.85 - 22 \ 1.4}{3.65}\)(380 0.85 - 22 1.4) / 3.65

             m_{maximum} = 80 kg

we can see that the maximum mass that the board supports without turning is greater than the mass of man

             m_{maximum}> m

consequently it is safe to stand at the end of the table

Answer:

14.6 m/s

Explanation:

The total run time was 5 + 60 = 65 minute or 65(60) = 3900 s

At his average velocity, emu ran 15 m/s(3900 s) = 58,500 m

Which is a heck of a running distance for ANY emu.

In the first 5 minutes the emu traveled 20 m/s(5 min)(60 s/min) = 6000 m

So in the last hour (3600 s) the emu traveled 58,500 - 6000 = 52,500 m

at a speed of 52,500 m /3600 s = 14.583333333... m/s

The emu was running at a speed of 14.58 m/s when he was tired.

To solve this question, we'll begin by calculating the distance travelled in the first part of the trip. This can be obtained as follow:

Time (t₁) = 5 min = 5 × 60 = 300 s

Speed 1 (S₁) = 20 m/s

Speed = distance / time

S₁ = d₁ / t₁

20 = d₁ / 300

Cross multiply

d₁ = 20 × 300

Next, we shall determine  the total distance travelled by the emu.

Average speed = 15 m/s

Time 1 (t₁) = 300 s

Time 2 (t₂) = 1 h = 60 mins = 60 × 60 = 3600 s

Total time (T) = t₁ + t₂ = 300 + 3600 = 3900 s

Average speed = Total distance / total time

15 = D / 3900

Cross multiply

D = 15 × 3900

Next, we shall determine the distance travelled in the second part (i.e when he was tired) of the trip.

Total distance (D) = 58500 m

Distance 1 (d₁) = 6000 m

D = d₁ + d₂

58500 = 6000 + d₂

Collect like terms

58500 – 6000 = d₂

Finally, we shall determine the speed of the emu in the second part of the trip.

Distance 2 (d₂) = 52500 m

Time 2 (t₂) = 3600 s

Speed = distance / time

S₂ = 52500 / 3600

Therefore, the emu was running at a speed of 14.58 m/s when he was tired.

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The force acting on the particle at t = 1 s is 24.4 N.

Mass of the particle, m = 2 units

Distance of the curve, r(t) = (4t²- t³)i - 5tj + (t⁴- 2)k

The velocity of the particle,

v = d[r(t)]/dt = (8t - 3t²)i - 5j + 4t³k

The acceleration of the particle,

a = d²[r(t)]/dt² = (8 - 6t)i + 12t²k

Let the time for which the force is acting be 1 second.

Therefore, acceleration at t = 1 is,

a₁ = 2i + 12k

Hence, the magnitude of acceleration,

|a₁| = √(2²+ 12²)

|a₁| = √148

|a₁| = 12.2 unit/s²

Therefore, the force acting on the particle at t = 1 s is,

F₁ = m x |a₁|

F₁ = 2 x 12.2

F₁ = 24.4 N

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The velocity of the car after 6.9 s if its acceleration is 1.5 m/s² due south would be  10.35 meters / seconds.

There are three equations of motion given by  Newton

v = u + at

S = ut + 1/2×a×t²

v² - u² = 2×a×s

Note that these equations are only valid for a uniform acceleration.

As given in the problem we have to find the velocity of the car we have to find the velocity of the car  after 6.9 s if its acceleration is 1.5 m/s² due south,

The acceleration of the car = 1.5 m/s²

The time taken by the car = 6.9 seconds

By using the first equation of the motion,

v = u + at

v = 0 + 1.5*6.9

v = 10.35 meters / seconds

Thus, the velocity of the car after 6.9 s, if its acceleration is 1.5 m/s² due south, would be  10.35 meters / seconds.

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The speed of light in material B is 1.6625 × 108 m/s.

The refractive index of a material is its optical density relative to that of a vacuum.

Material B has a refractive index of nB, and its speed of light is vB.

The speed of light in material A is given as 1.25 x 108 m/s.

The refractive indices of materials A and B have a ratio of nA/nB = 1.33.

We will use the formula:

nA/nB = vB/vA = nA/nB.

Therefore, nA/nB = vB/1.25 x 108 m/s.

This equation can be rearranged to give the speed of light in material B:

vB = nA/nB × 1.25 x 108 m/s.

Therefore, vB = 1.33 × 1.25 × 108 m/s.

We will perform this calculation:

vB = 1.6625 × 108 m/s.

Therefore, the speed of light in material B is 1.6625 × 108 m/s.

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Answer:

Difference between MeV and meV?

1 MV = 1000000 volts. 1 MeV stands for million electron volts. It is a unit of energy. 1 MeV = 1.60217653 x 10 13

Explanation:

hope this helps <3

Explanation:

Free Solutions to GR BATHLA & SONS BIOLOGY (HINGLISH) Class 12 book Chapters, Questions, Answers and Solutions 2022

Answer:

You get Day and Night

It takes 24 hour

Answer:

Explanation:

The Earth's orbit makes a circle around the sun. At the same time the Earth orbits around the sun, it also spins.Since the Earth orbits the sun and rotates on its axis at the same time we experience seasons, day and night, and changing shadows throughout the day.It only takes 23 hours, 56 minutes and 4.0916 seconds for the Earth to turn once on its axis.

Answer:

Attract

Explanation:

The balloons attract each other due to electrons.

Hope this helps :)

Answer:

Force = 50 N

Explanation:

Force = 50 N

Formula: Force = mass x acceleration

Answer: Force = 10kg x 5m/s²

Force = 50 N

In half life the number of the sample will reduce to 50% of therir initial number.

So after first half life, sample remained will be 50%.

After second half life 50% of the remaing sample will again disintegrate, so we are left with 25% of the sample.

After third half life the 50% of the remaining sample will again disintegrate so we are left with only,

So after 3 half life only 12.5% of the sample will be remaining.

a) The magnitude will be 0.838m

b) The displacement will be -17.35°

The path covered by an object from its initial point to final point.

Forces acting on the duck

x-axis:    0.13 + 0.16*cos(-56°) = 2.7 * ax  

ax = 0.0813 m/s^2

y-axis:    0.13*sin(-56°) = 2.7 * ay  

ay = -0.0491 m/s^2

The displacement on the x-axis

X = Vox * t + ax/2 * t²

X = 0.12* 3.2 + 0.0813/2*3.2²

X = 0.8

The displacement on the y-axis:

Y = Voy * t + ay/2 * t²

X = 0 - 0.0491/2*3.2²

Y=-0.25m

So, the magnitude and angle of this displacement [0.8,-0.25] is:

0.838m  at an angle of -17.35°

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Answer:

Explanation:

Let d be the distance to the center of mass from the front wheels

Sum moments about the front wheel contact point to zero

1240(9.8)[d] - 1240(9.8)(1 - 0.67)[3.2] = 0

1240(9.8)[d] = 1240(9.8)(1 - 0.67)[3.2]

                d = (1 - 0.67)[3.2]

                d = 1.056 m

Answer:

to show the arts and creativity of the person and to show also the culture of the place..

Explanation:

Based on health or personal choices, some foods that I avoid include:

Food comprises nutrients, which are things required for the development, upkeep, and repair of bodily tissues as well as the control of vital functions. The energy that nutrients give our bodies allows them to function.

Processed foods are food that has been modified industrially by the addition or removal of certain compounds or chemicals.

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Answer:

20.8 ft³

Explanation:

The following data were obtained from the question:

Initial pressure (P1) = 80 psi

Initial volume (V1) = 13 ft³

Final pressure (P2) = 50 psi

Final volume (V2) =?

The new volume of the gas can be obtained by using the Boyle's law equation as shown below:

P1V1 = P2V2

80 × 13 = 50 × V2

1040 = 50 × V2

Divide both side by 50

V2 = 1040 / 50

V2 = 20.8 ft³

Thus, the volume of the gas at a pressure of 50 psi is 20.8 ft³

Answer:

P = 147,75 W

Explanation:

A man whose mass is 59.1kg climbs up 30 steps of a stair each step is 25 cm high

Height at 30 steps , h=30×2.5=  7.5 m

Change in potential energy , =mgh=59.1×10×7.5 = 4432.5 J

So, Work done by the man , W= 4432.5J

Power used , P= \(\frac{W}{T}\)

P = 4432.5 /30

P = 147,75 W

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Answer:

C = (7 + 8) i + (5 - 1) j     adding vectors

C = 15 i + 4 j

theta = arctan (4 / 15) = 14.9 deg

Note that this is the same as adding x and y components

Each of the five friends needs to pay £12 to cover the cost of their own meals and contribute towards the birthday person's meal. Using mean allows us to distribute the cost equally among the friends, ensuring a fair division of expenses for the meal.

To determine how much each of the five friends needs to pay, we can use the concept of averages (mean) and divide the total cost by the number of people paying.

In this scenario, the total cost of the meal for 6 people is £12 per person. Since the other 5 friends have decided to pay for the birthday person's meal, they will collectively cover the cost of their own meals plus the birthday person's meal.

To calculate the total cost covered by the five friends, we can subtract the cost of one person's meal (since the birthday person's meal is being paid by the group) from the total cost. The cost of one person's meal is £12.

Total cost covered by the five friends = Total cost - Cost of one person's meal

= (£12 x 6) - £12

= £72 - £12

= £60

Now, to find out how much each of the five friends needs to pay, we divide the total cost covered by the five friends (£60) by the number of friends (5).

Amount each friend needs to pay = Total cost covered by the five friends / Number of friends

= £60 / 5

= £12

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Answer:

A

Explanation:

Any sort of honk can frighten the animal, drive slow enough that the person would know you're coming/you're there and make sure you are taking a lot of care.

The thing that should be done when driving is A. Use reasonable care, and don't make any loud noises when approaching them, to avoid frightening the animal.

It should be noted that the person is driving on a rural road and see a person on the shoulder ahead, riding or leading an animal.

In this case, the person should reasonable care, and don't make any loud noises when approaching them, to avoid frightening the animal.

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Answer:

Refer to the step-by-step Explanation.

Step-by-step Explanation:

Simplify the equation with given substitutions,

Given Equation:

\(mgh+(1/2)mv^2+(1/2)I \omega^2=(1/2)mv_{_{0}}^2+(1/2)I \omega_{_{0}}^2\)

Given Substitutions:

\(\omega=v/R\\\\ \omega_{_{0}}=v_{_{0}}/R\\\\\ I=(2/5)mR^2\)\(\hrulefill\)

Start by substituting in the appropriate values: \(mgh+(1/2)mv^2+(1/2)I \omega^2=(1/2)mv_{_{0}}^2+(1/2)I \omega_{_{0}}^2 \\\\\\\\\Longrightarrow mgh+(1/2)mv^2+(1/2)\bold{[(2/5)mR^2]} \bold{[v/R]}^2=(1/2)mv_{_{0}}^2+(1/2)\bold{[(2/5)mR^2]}\bold{[v_{_{0}}/R]}^2\)

Adjusting the equation so it easier to work with.\(\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{2} \Big[\dfrac{2}{5} mR^2\Big]\Big[\dfrac{v}{R} \Big]^2=\dfrac12mv_{_{0}}^2+\dfrac12\Big[\dfrac25mR^2\Big]\Big[\dfrac{v_{_{0}}}{R}\Big]^2\)

\(\hrulefill\)

Simplifying the left-hand side of the equation:

\(mgh+\dfrac{1}{2} mv^2+\dfrac{1}{2} \Big[\dfrac{2}{5} mR^2\Big]\Big[\dfrac{v}{R} \Big]^2\)

Simplifying the third term.

\(\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{2} \Big[\dfrac{2}{5} mR^2\Big]\Big[\dfrac{v}{R} \Big]^2\\\\\\\\\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{2}\cdot \dfrac{2}{5} \Big[mR^2\Big]\Big[\dfrac{v}{R} \Big]^2\\\\\\\\\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{5} \Big[mR^2\Big]\Big[\dfrac{v}{R} \Big]^2\)

\(\\ \boxed{\left\begin{array}{ccc}\text{\Underline{Power of a Fraction Rule:}}\\\\\Big(\dfrac{a}{b}\Big)^2=\dfrac{a^2}{b^2} \end{array}\right }\)

\(\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{5} \Big[mR^2\Big]\Big[\dfrac{v^2}{R^2} \Big]\\\\\\\\\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{5} \Big[mR^2 \cdot\dfrac{v^2}{R^2} \Big]\)

"R²'s" cancel, we are left with:

\(\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{5} \Big[mR^2\Big]\Big[\dfrac{v^2}{R^2} \Big]\\\\\\\\\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{5}mv^2\)

We have like terms, combine them.

\(\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{5} \Big[mR^2\Big]\Big[\dfrac{v^2}{R^2} \Big]\\\\\\\\\Longrightarrow mgh+\dfrac{7}{10} mv^2\)

Each term has an "m" in common, factor it out.

\(\Longrightarrow m(gh+\dfrac{7}{10}v^2)\)

Now we have the following equation:

\(\Longrightarrow m(gh+\dfrac{7}{10}v^2)=\dfrac12mv_{_{0}}^2+\dfrac12\Big[\dfrac25mR^2\Big]\Big[\dfrac{v_{_{0}}}{R}\Big]^2\)

\(\hrulefill\)

Simplifying the right-hand side of the equation:

\(\Longrightarrow \dfrac12mv_{_{0}}^2+\dfrac12\cdot\dfrac25\Big[mR^2\Big]\Big[\dfrac{v_{_{0}}}{R}\Big]^2\\\\\\\\\Longrightarrow \dfrac12mv_{_{0}}^2+\dfrac15\Big[mR^2\Big]\Big[\dfrac{v_{_{0}}}{R}\Big]^2\\\\\\\\\Longrightarrow \dfrac12mv_{_{0}}^2+\dfrac15\Big[mR^2\Big]\Big[\dfrac{v_{_{0}}^2}{R^2}\Big]\\\\\\\\\Longrightarrow \dfrac12mv_{_{0}}^2+\dfrac15\Big[mR^2\cdot\dfrac{v_{_{0}}^2}{R^2}\Big]\\\\\\\\\Longrightarrow \dfrac12mv_{_{0}}^2+\dfrac15mv_{_{0}}^2\Big\\\\\\\\\)

\(\Longrightarrow \dfrac{7}{10}mv_{_{0}}^2\)

Now we have the equation:

\(\Longrightarrow m(gh+\dfrac{7}{10}v^2)=\dfrac{7}{10}mv_{_{0}}^2\)

\(\hrulefill\)

Now solving the equation for the variable "v":

\(m(gh+\dfrac{7}{10}v^2)=\dfrac{7}{10}mv_{_{0}}^2\)

Dividing each side by "m," this will cancel the "m" variable on each side.

\(\Longrightarrow gh+\dfrac{7}{10}v^2=\dfrac{7}{10}v_{_{0}}^2\)

Subtract the term "gh" from either side of the equation.

\(\Longrightarrow \dfrac{7}{10}v^2=\dfrac{7}{10}v_{_{0}}^2-gh\)

Multiply each side of the equation by "10/7."

\(\Longrightarrow v^2=\dfrac{10}{7}\cdot\dfrac{7}{10}v_{_{0}}^2-\dfrac{10}{7}gh\\\\\\\\\Longrightarrow v^2=v_{_{0}}^2-\dfrac{10}{7}gh\)

Now squaring both sides.

\(\Longrightarrow \boxed{\boxed{v=\sqrt{v_{_{0}}^2-\dfrac{10}{7}gh}}}\)

Thus, the simplified equation above matches the simplified equation that was given.