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iVinArrow [24]
3 years ago
14

You are asked to design a horizontal curve with a 40-degree central angle (' = 40) for a two-lane road with 11-ft lanes. the des

ign speed is 70 mi/h and superelevation is limited to 0.06 f
a. True
b. Falset. give the radius, degree of curvature, and length of curve that you would recommend. 3.43 for the h
Physics
1 answer:
Alex17521 [72]3 years ago
3 0
The answer is <span>b. False. give the radius, degree of curvature, and length of curve that you would recommend. 3.43 for the h. Give it a try man
 </span>
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Write true if the statement is false change the underlined word to make the statement true. Write our answers on a separate shee
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Answer:

false

Explanation:

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A parallel-plate vacuum capacitor is connected to a battery and charged until the stored electric energy is . The battery is rem
Viktor [21]

Answer:

A

The energy dissipated in the resistor {U_k} = \frac{U}{k}

B

The energy dissipated in the resistor{U_k} = kU

Explanation:

In order to gain a good understanding of the solution above it is necessary to understand that the concept required to solve the question is energy stored in the parallel plate capacitor.

Initially, take the first case. In that, according to the formula for energy stored in parallel plate capacitor with the dielectric inserted between the two plates, find the energy stored. Then, find the energy stored in the parallel plate capacitor when no dielectric is present. Then, write the equation of energy stored in the capacitor with the dielectric present in the form of the energy stored in the capacitor without the dielectric present. The equation must not be in the form of voltage as battery is removed in this case.

For part B, use the equation of the energy dissipated in the resistor. Write it in the form of the equation for energy stored in the parallel plate capacitor without dielectric in it. The equation must be in the form of voltage as battery is kept connected. Looking at the fundamentals

The energy stored in the parallel plate capacitor with the dielectric is given by,

                 U _k = \frac{1}{2} \frac{q ^2}{kC}

Here, the energy stored in the capacitor will be equal to the energy dissipated in the resistor. In this equation, Uk is the energy dissipated in the resistor, q is charge, k is the dielectric constant, and C is the capacitance.

Now, the equation of the energy stored in the parallel plate capacitor without dielectric is,

​ U= \frac{1}{2} \frac{q ^2}{C}

In this equation, U is the energy stored in the parallel plate capacitor without dielectric, q is charge, and C is the capacitance.

For part B, the battery is still connected. Thus, the equation q = CV is used to modify the above equation.

Thus, the energy stored in the parallel plate capacitor with the dielectric is given by,

U_ k = \frac{1}{2} \frac{k ^{2} C^ 2 V ^2}{kC} \\\\= \frac{1}{2}  kCV ^2

In this equation, Uk is the energy dissipated in the resistor, V is voltage, k is the dielectric constant, and C is the capacitance.

The equation of the energy stored in the parallel plate capacitor without dielectric is,

U= \frac{1}{2} \frac{C^ 2 V ^2}{C} \\\\= \frac{1}{2} CV ^2

In this equation, U is the energy dissipated in the resistor, V is voltage, k is the dielectric constant, and C is the capacitance.

(A)

The equation for energy dissipated in the resistor is,

 U _k = \frac{1}{2} \frac{q ^2}{kC}

Substitute U = \frac{1}{2}\frac{{{q^2}}}{C}  in the equation of {U_k}

U _k = \frac{1}{2} (\frac{1}{k} )\frac{q ^2}{C} \\\\= (\frac{1}{k} ) \frac{q^2}{C}\\\\ U_{k} = \frac{U}{k}

Note :

If the resistance relates to the capacitor, the energy stored in the capacitor is dissipated through the resistance. Thus, by substituting the equation of U, the expression is found out.

(B)

The equation for energy dissipated in the resistor is

U_{k} = \frac{1}{2}kCV^2

Here, V is voltage in the circuit.

Substitute U =\frac{1}{2} CV^2 in the equation of {U_k}

So,

        U_{k} = \frac{1}{2} kCV^2\\

       = k(\frac{1}{2} CV^2)

       U_{k} = kU

4 0
2 years ago
An astronaut has a mass of 50.0 kg on earth. what is her mass on the moon, where gravity is 1 6 that on earth?
never [62]

I believe the correct gravity on the moon is 1/6 of Earth. Take note there is a difference between 1 6 and 1/6.

HOWEVER, we should realize that the trick here is that the question asks about the MASS of the astronaut and not his weight. Mass is an inherent property of an object, it is unaffected by external factors such as gravity. What will change as the astronaut moves from Earth to the moon is his weight, which has the formula: weight = mass times gravity.

<span>Therefore if he has a mass of 50 kg on Earth, then he will also have a mass of 50 kg on moon.</span>

6 0
3 years ago
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velikii [3]
This problem must be solved using a sketch. I attached an illustration of the problem.
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Law of reflection tells us that light ray reflects off the surface at the same angle in which it falls on it( i attached another illustration of this).
Now we can write tangens equations:
tan(\theta)=\frac{h-0.8}{1}\\&#10;tan(\theta)=\frac{1.8-h}{2.8}\\&#10;\frac{h-0.8}{1}=\frac{1.8-h}{2.8}
We solve for h:
\frac{h-0.8}{1}=\frac{1.8-h}{2.8}\\&#10;2.8h-0.8\cdot 2.8=1.8-h\\&#10;3.8h=1.8+2.24\\&#10;h=\frac{4.04}{3.8}\\&#10;h=1.06m


6 0
3 years ago
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