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3 years ago

What are the contour lines on a contour map?

2 answers:

7 0

In cartography, a contour line (often just called a "contour") joins points of equal elevation (height) above a given level, such as mean sea level. A contour map is a map illustrated with contour lines, for example a topographic map, which thus shows valleys and hills, and the steepness of slopes.-hope that helped:)

dalvyx [7]3 years ago

5 0

Contour lines are lines that signifies the elevation on a mountain or hill

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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:

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

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

3 years ago

(33%) Problem 3: Two cars collide at an icy intersection and stick together afterward. The first car has a mass of 1100 kg and i

expeople1 [14]

Answer:

The final velocity of the cars is 8.38 m/s at an angle 39.1° south of west.

Explanation:

Given that,

Mass of first car = 1100 kg

Velocity of first car = 8.5 m/s

Mass of second car = 650 kg

Velocity of second car = 17.5 m/s

Suppose we need to find the final velocity of the cars and direction of the cars.

We need to calculate the velocity of the car in west direction

Using conservation of momentum in west direction

$m_{f}v_{f}+m_{s}v_{s}= (m_{f}+m_{s})v_{x}$

$v_{x}=\dfrac{m_{f}v_{f}+m_{s}v_{s}}{(m_{f}+m_{s})}$

Put the value into the formula

$v_{x}=\dfrac{1100\times0+650\times17.5}{1100+650}$

$v_{x}=6.5\ m/s$

We need to calculate the velocity of the car in south direction

Using conservation of momentum in south direction

$m_{f}v_{f}+m_{s}v_{s}= (m_{f}+m_{s})v_{y}$

$v_{y}=\dfrac{m_{f}v_{f}+m_{s}v_{s}}{(m_{f}+m_{s})}$

Put the value into the formula

$v_{y}=\dfrac{1100\times8.5+650\times0}{1100+650}$

$v_{y}=5.3\ m/s$

We need to calculate the final velocity of the cars

Using formula of velocity

$v_{eq}=\sqrt{(6.5)^2+(5.3)^2}$

$v_{eq}=8.38\ m/s$

We need to calculate the direction

Using formula of direction

$\tan\theta=\dfrac{v_{y}}{v_{x}}$

Put the value into the formula

$\tan\theta=\dfrac{5.3}{6.5}$

$\theta=\tan^{-1}(\dfrac{5.3}{6.5})$

$\theta=39.1^{\circ}$

Hence, The final velocity of the cars is 8.38 m/s at an angle 39.1° south of west.

8 0

3 years ago

Andy is entering early adolescence. although he used to be a big help to his mother by doing household chores, he now refuses, s

o-na [289]

Answer:

a. gender intensification

Explanation:

This theory states that in the adolescence both girls and boys are pressured to follow gender roles established culturally.

8 0

3 years ago

Researchers condition a flatworm to contract when exposed to light by repeatedly pairing the light with electric shock. the elec

Serga [27]

Negative Reinforcement

5 0

3 years ago

Find the length (in m) of an organ pipe closed at one end that produces a fundamental frequency of 494 Hz when air temperature i

elena-14-01-66 [18.8K]

Answer:

0.173 m.

Explanation:

The fundamental frequency of a closed pipe is given as

fc = v/4l .................. Equation 1

Where fc = fundamental frequency of a closed pipe, v = speed of sound l = length of the pipe.

Making l the subject of the equation,

l = v/4fc ................ Equation 2

also

v = 331.5×0.6T ................. Equation 3

Where T = temperature in °C, T = 18.0 °c

Substitute into equation 3

v = 331.5+0.6(18)

v = 331.5+10.8

v = 342.3 m/s.

Also given: fc = 494 Hz,

Substitute into equation 2

l = 342.3/(4×494)

l = 342.3/1976

l =0.173 m.

Hence the length of the organ pipe = 0.173 m.

7 0

3 years ago