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4 years ago

What is true about isolines on a weather map?

Physics

2 answers:

vlabodo [156]4 years ago

6 0

Answer: Hi, isolines refer to lines where some quantity maintains constant.

For example, an "istoremal line" is a line where the temperature is constant, an "isobaric line" is a line where the pressure is constant.

Where constant means that, in the two examples i give you, in all the line the temperature or the pressure has the same value.

A thing that you usually can see in a whether map, is that the isolines are "closed" curves. And other thing, maybe more obvious, is that a line cant pass over other line (because this will mean that a point in the space has 2 different values of something).

Law Incorporation [45]4 years ago

5 0

<span>All isolines, or iso-intensity lines, connect points having equal values.</span>

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A meter stick moves parallel to its axis with speed of 0.96 c relative to you. What would you measure for the length of the stic

Fed [463]

Answer:

The length of the stick is 0.28 m.

The time the stick take to move is 0.97 ns.

Explanation:

Given that,

Relative speed of stick v= 0.96 c

Speed of light $c= 2.99793\times10^{8}\ m/s$

Proper length of stick = 1 m

We need to calculate the length of the stick

Using formula of length

$\Delta l=\Delta l_{0}\sqrt{(1-\dfrac{v^2}{c^2})}$

Put the value into the formula

$\Delta l=1\sqrt{1-\dfrac{(0.96)^2c^2}{c^2}}$

$\Delta l=1\sqrt{1-(0.96)^2}$

$\Delta l=0.28\ m$

We need to calculate the time the stick take to move

Using formula of time

$t=\dfrac{\Delta l}{v}$

Put the value into the formula

$t=\dfrac{0.28}{0.96\times(2.99793\times10^{8})}$

$t=9.72\times10^{-10}\ sec$

$t=0.97\ ns$

Hence, The length of the stick is 0.28 m.

The time the stick take to move is 0.97 ns.

7 0

3 years ago

In the first stage of a two-stage Carnot engine, energy is absorbed as heat Q1 at temperature T1 = 500 K, work W1 is done, and e

Nataly [62]

Answer:

Efficiency = 52%

Explanation:

Given:

First stage

heat absorbed, Q₁ at temperature T₁ = 500 K

Heat released, Q₂ at temperature T₂ = 430 K

and the work done is W₁

Second stage

Heat released, Q₂ at temperature T₂ = 430 K

Heat released, Q₃ at temperature T₃ = 240 K

and the work done is W₂

Total work done, W = W₁ + W₂

Now,

The efficiency is given as:

$\eta=\frac{\textup{Total\ work\ done}}{\textup{Energy\ provided}}$

Work done = change in heat

thus,

W₁ = Q₁ - Q₂

W₂ = Q₂ - Q₃

Thus,

$\eta=\frac{(Q_1-Q_2)\ +\ (Q_2-Q_3)}{Q_1}}$

$\eta=1-\frac{(Q_1-Q_3)}{Q_1}}$

$\eta=1-\frac{(Q_3)}{Q_1}}$

also,

$\frac{Q_1}{T_1}=\frac{Q_2}{T_2}=\frac{Q_3}{T_3}$

$\frac{T_3}{T_1}=\frac{Q_3}{Q_1}$

thus,

$\eta=1-\frac{(T_3)}{T_1}}$

thus,

$\eta=1-\frac{(240\ K)}{500\ K}}$

$\eta=0.52$

Efficiency = 52%

8 0

4 years ago

A harmonic wave is traveling along a rope. the oscillator that generates the wave completes 46.0 vibrations in 26.0 s. a given c

Rudik [331]

The crest of the wave has traveled 415 cm in 11.7s. then the crest should have travelled 415/11.7 cm in one second i.e. 35.47cm. Then the crest should have travelled 26*35.47 cm in 26 seconds i.e. 922.2cm which has 46 vibrations in it.
so for length of each vibration (Wavelength) is 20 cm

6 0

3 years ago

Which equations could be used as is, or rearranged to calculate for frequency of a wave? Check all that apply.

amm1812

-- Equations #2 and #6 are both the same equation,
and are both correct.

-- If you divide each side by 'wavelength', you get Equation #4,
which is also correct.

-- If you divide each side by 'frequency', you get Equation #3,
which is also correct.
With some work, you can rearrange this one and use it to calculate
frequency.

Summary:

-- Equations #2, #3, #4, and #6 are all correct statements,
and can be used to find frequency.

-- Equations #1 and #5 are incorrect statements.

7 0

3 years ago

Read 2 more answers

What is commonly used to run a central heating system?

kvv77 [185]

Electrical energy is and 100% correct answer

4 0

3 years ago

Read 2 more answers