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3 years ago

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Although the class has no water, the students want to know if the cube will float or sink in water. Explain, in detail, the step

s that must be taken to determine if the cube will float or sink. Be sure to use the following key terms: triple beam balance, metric ruler ,and density.

1 answer:

Nutka1998 [239]3 years ago

3 0

Okay, so the density of water is 1g/cm3. In order for the cube to float, it has to be less than 1, and it will sink if it is more than 1 g/cm3. Use a triple beam balance to weigh the cube, looking at the metric ruler on the balance. Then, if the cube's density is more than 1, then you know it will float. If the density is less than 1, you know it will sink.
hope this helps, and I didn't know how to use the word "metric ruler"

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Using any data you can find in the ALEKS Data resource, calculate the equilibrium constant K at 25.0 °C for the following reacti

Yakvenalex [24]

The equilibrium constant of the reaction at 25⁰c will be 426827.5.

<u />

<u>Theory-</u>

<u>Equilibrium constant</u> :The equilibrium constant comes from the chemical equilibrium law. For the chemical equilibrium state, at a fixed constant temperature, the ratio of the product of the reaction's multiplication to the concentration of its reactants' multiplication, and each is raised to the power to the corresponding coefficients of the elements in the reaction.

The chemical equilibrium is given by for a general chemical reaction.

a. A+ b. B ⇌ c. C+ d. D,.

Kc =[C]c [D]d/[A]a [B]b.

<u>Gibb's free energy</u> :The second law of thermodynamics can be arranged in such a way that it gives a new expression when a chemical reaction happens at a constant temperature and constant pressure.

G=H-TS

<u>Calculations</u>:-

T=25⁰c

G=51.4 x 10³J

$\\\\k=GR+\frac{nRT}{Z} \\$

k= equilibrium constant ,G=Gibbs free energy ,n= no. of moles ,R=Gas constant ,T=temperature ,Z=compressibility

$Ideal.Situation=\left \{ {{Z=1} \atop {n=1}} \right.$

$\\\\\\k=GR+RT$

k=51.4 x 10³ x 8.3 + 8.3 x 25

k=426827.5

To learn equilibrium constant-

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1 year ago

Two bicyclist, originally separated by a distance of 20 miles, are each traveling at a uniform speed of 10 miles per hour toward

Radda [10]

Answer:

D = 25 miles

Explanation:

To solve this problem, we just need to know how much time it took both bicyclists to collide and that will be the same amount of time that the bee flew at 25miles per hour. With those values we could calculate the distance it traveled.

Since both bicyclists collide, we know that Xa=Xb, so:

Xa = V*t = 10*t and Xb = 20 - V*t = 20 - 10*t

10*t = 20 - 10*t Solving for t:

t = 1 hour Now we can calculate the distance for the bee:

D = Vbee * t = 25 * 1 = 25 miles

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4 years ago

Calculate the current in a circuit when the voltage of the circuit is 5 V and the circuit has a total resistance of 5 Ω.

seropon [69]

the answer is 1a as rearrange gives I = v divided by r

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3 years ago

Two charges are located in the x – y plane. If ????1=−4.10 nC and is located at (x=0.00 m,y=0.600 m) , and the second charge has

faust18 [17]

Answer:

The x-component of the electric field at the origin = -11.74 N/C.

The y-component of the electric field at the origin = 97.41 N/C.

Explanation:

<u>Given:</u>

Charge on first charged particle, $q_1=-4.10\ nC=-4.10\times 10^{-9}\ C.$
Charge on the second charged particle, $q_2=3.80\ nC=3.80\times 10^{-9}\ C.$

Position of the first charge = $(x_1=0.00\ m,\ y_1=0.600\ m).$
Position of the second charge = $(x_2=1.50\ m,\ y_2=0.650\ m).$

The electric field at a point due to a charge $q$ at a point $r$ distance away is given by

$\vec E = \dfrac{kq}{|\vec r|^2}\ \hat r.$

where,

$k$ = Coulomb's constant, having value $\rm 8.99\times 10^9\ Nm^2/C^2.$

$\vec r$ = position vector of the point where the electric field is to be found with respect to the position of the charge $q$ .

$\hat r$ = unit vector along $\vec r$ .

The electric field at the origin due to first charge is given by

$\vec E_1 = \dfrac{kq_1}{|\vec r_1|^2}\ \hat r_1.$

$\vec r_1$ is the position vector of the origin with respect to the position of the first charge.

Assuming, $\hat i,\ \hat j$ are the units vectors along x and y axes respectively.

$\vec r_1=(0-x_1)\hat i+(0-y_1)\hat j\\=(0-0)\hat i+(0-0.6)\hat j\\=-0.6\hat j.\\\\|\vec r_1| = 0.6\ m.\\\hat r_1=\dfrac{\vec r_1}{|\vec r_1|}=\dfrac{0.6\ \hat j}{0.6}=-\hat j.$

Using these values,

$\vec E_1 = \dfrac{(8.99\times 10^9)\times (-4.10\times 10^{-9})}{(0.6)^2}\ (-\hat j)=1.025\times 10^2\ N/C\ \hat j.$

The electric field at the origin due to the second charge is given by

$\vec E_2 = \dfrac{kq_2}{|\vec r_2|^2}\ \hat r_2.$

$\vec r_2$ is the position vector of the origin with respect to the position of the second charge.

$\vec r_2=(0-x_2)\hat i+(0-y_2)\hat j\\=(0-1.50)\hat i+(0-0.650)\hat j\\=-1.5\hat i-0.65\hat j.\\\\|\vec r_2| = \sqrt{(-1.5)^2+(-0.65)^2}=1.635\ m.\\\hat r_2=\dfrac{\vec r_2}{|\vec r_2|}=\dfrac{-1.5\hat i-0.65\hat j}{1.634}=-0.918\ \hat i-0.398\hat j.$

Using these values,

$\vec E_2= \dfrac{(8.99\times 10^9)\times (3.80\times 10^{-9})}{(1.635)^2}(-0.918\ \hat i-0.398\hat j) =-11.74\ \hat i-5.09\ \hat j\ N/C.$

The net electric field at the origin due to both the charges is given by

$\vec E = \vec E_1+\vec E_2\\=(102.5\ \hat j)+(-11.74\ \hat i-5.09\ \hat j)\\=-11.74\ \hat i+(102.5-5.09)\hat j\\=(-11.74\ \hat i+97.41\ \hat j)\ N/C.$

Thus,

x-component of the electric field at the origin = -11.74 N/C.

y-component of the electric field at the origin = 97.41 N/C.

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ehidna [41]

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