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3 years ago

6

4. Look at the graph above. What information is on the graph and what does it mean? Please help!!!

1 answer:

Aliun [14]3 years ago

6 0

Answer:

CO2 increases in the winter

Explanation:

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What is it called when there’s an atom that has fewer neutrons than protons and more electrons than protons??l

HACTEHA [7]

Answer: The answer is an atom.

Explanation: This is because an atom has fewer neutrons than protons and more electrons than protons

6 0

3 years ago

6) Find the speed a spherical raindrop would attain by falling from 4.00 km. Do this:a) In the absence of air dragb) In the pres

sleet_krkn [62]

We are asked to determine the velocity of a rain drop if it falls from 4 km.

To do that we will use the following formula:

$2ah=v_f^2-v_0^2$

Where:

$\begin{gathered} a=\text{ acceleration} \\ h=\text{ height} \\ v_f,v_0=\text{ final and initial velocity} \end{gathered}$

If we assume the initial velocity to be 0 we get:

$2ah=v_f^2$

The acceleration is the acceleration due to gravity:

$2gh=v_f^2$

Now, we take the square root to both sides:

$\sqrt{2gh}=v_f$

Now, we substitute the values:

$\sqrt{2(9.8\frac{m}{s^2})(4000m)}=v_f$

solving the operations:

$280\frac{m}{s}=v$

Therefore, the velocity without air drag is 280 m/s.

Part B. we are asked to determine the velocity if there is air drag. To do that we will use the following formula:

$F_d=\frac{1}{2}C\rho_{air}Av^2$

Where:

$\begin{gathered} F_d=drag\text{ force} \\ C=\text{ constant} \\ \rho_{air}=\text{ density of air} \\ A=\text{ area} \\ v=\text{ velocity} \end{gathered}$

We need to determine the drag force. To do that we will use the following free-body diagram:

Since the velocity that the raindrop reaches is the terminal velocity and its a constant velocity this means that the acceleration is zero and therefore the forces are balanced:

$F_d=mg$

Now, we determine the mass of the raindrop using the following formula:

$m=\rho_{water}V$

Where:

$\begin{gathered} \rho_{water}=\text{ density of water} \\ V=\text{ volume} \end{gathered}$

The volume is the volume of a sphere, therefore:

$m=\rho_{water}(\frac{4}{3}\pi r^3)$

Since the diameter of the raindrop is 3 millimeters, the radius is 1.5 mm or 0.0015 meters. Substituting we get:

$m=(0.98\times10^3\frac{kg}{m^3})(\frac{4}{3}\pi(0.0015m)^3)$

Solving the operations:

$m=1.39\times10^{-5}kg$

Now, we substitute the values in the formula for the drag force:

$F_d=(1.39\times10^{-5}kg)(9.8\frac{m}{s^2})$

Solving the operations:

$F_d=1.36\times10^{-4}N$

Now, we substitute in the formula:

$1.36\times10^{-4}N=\frac{1}{2}C\rho_{air}Av^2$

Now, we solve for the velocity:

$\frac{1.36\times10^{-4}N}{\frac{1}{2}C\rho_{air}A}=v^2$

Now, we substitute the values. We will use the area of a circle:

$\frac{1.36\times10^{-4}N}{\frac{1}{2}(0.45)(1.21\frac{kg}{m^3})(\pi r^2)}=v^2$

Substituting the radius:

$\frac{1.36\cdot10^{-4}N}{\frac{1}{2}(0.45)(1.21\frac{kg}{m^{3}})(\pi(0.0015m)^2)}=v^2$

Solving the operations:

$70.67\frac{m^2}{s^2}=v^2$

Now, we take the square root to both sides:

$\begin{gathered} \sqrt{70.67\frac{m^2}{s^2}}=v \\ \\ 8.4\frac{m}{s}=v \\ \end{gathered}$

Therefore, the velocity is 8.4 m/s

7 0

10 months ago

Select the correct answer. A golfer hits a golf ball with a velocity of 36.0 meters/second at an angle of28.0°. If the hang time

lions [1.4K]

Its 83 im pretty sure :)

5 0

3 years ago

When two hydrogen atoms bond, the positive nucleus of one atom attracts the a. negative nucleus of the other atom. c. negative e

Aleksandr-060686 [28]

Answer:

c

Explanation:

4 0

3 years ago

A reaction taking place in a container with a piston-cylinder assembly at constant temperature produces a gas, and the volume in

djverab [1.8K]

Answer:

60.4 J

Explanation:

The work done by the gas is given by:

$W=p(V_f-V_i)$

where

p is the gas pressure

$V_f$ is the final volume of the gas

$V_i$ is the initial volume

We must convert all the quantities into SI units:

$p=860 torr \cdot \frac{1.013\cdot 10^5 pa}{760 torr}=1.146\cdot 10^5 Pa$

$V_i = 127 mL = 0.127 L = 0.127 dm^3 = 0.127 \cdot 10^{-3}m^2$

$V_f = 654 mL = 0.654 L = 0.654 dm^3 = 0.654 \cdot 10^{-3}m^2$

So the work done is

$W=(1.146\cdot 10^5 Pa)(0.654\cdot 10^{-3} m^3-0.127\cdot 10^{-3} m^3)=60.4 J$

4 0

2 years ago