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3 years ago

5

A sinusoidal transverse wave travels along a long, stretched string. The amplitude of this wave is 0.08190.0819 m, its frequency

is 2.292.29 Hz, and its wavelength is 1.871.87 m. (a) What is the shortest transverse distance between a maximum and a minimum of the wave

1 answer:

Anvisha [2.4K]3 years ago

5 0

Answer:

The shortest transverse distance between a maximum and a minimum of the wave is 0.1638 m.

Explanation:

Given that,

Amplitude = 0.08190 m

Frequency = 2.29 Hz

Wavelength = 1.87 m

(a). We need to calculate the shortest transverse distance between a maximum and a minimum of the wave

Using formula of distance

$d=2A$

Where, d = distance

A = amplitude

Put the value into the formula

$d=2\times0.08190$

$d=0.1638\ m$

Hence, The shortest transverse distance between a maximum and a minimum of the wave is 0.1638 m.

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Convert 0.700 to scientific notation

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Answer:

7 × 10^-1

Explanation:

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3 years ago

How is the capacitance of a capacitor related to the charge stored on the capacitor and the potential difference across the capa

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Answer:

Capacitance is the ratio of the charge to the potential difference. How is the charge stored on a capacitor related to the capacitance of the capacitor and the potential difference across the capacitor? The charge equals the product of the capacitance and the potential difference.

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2 years ago

Newton's Law of Cooling states that the rate of cooling of an object is proportional to the temperature difference between the o

julsineya [31]

Answer:

1) $k=-\frac{1}{46}\approx -0.02$

2) $\textit{Limiting value} = 24$

3) $T(10)\approx \frac{494611944}{6436343}\approx 76.8$

Explanation:

First of all note that

$T_s=24$ is the surroundings temperature, the temperature of the room where the cup of coffee is. Then, the differential equation is:

$\frac{dT}{dt}=k(T-24)$

Also, note that all units are in degrees celsius and minutes. Then, we don't have to convert units. Let's not write units explicitly from now on.

Explanation

1) We have that

$\textit{rate of cooling}=\frac{dT}{dt}=1,\quad T=70$

at some point - the exact time at which this is true doesn't really play any role because the equation doesn't have t on the right hand side. Then, from the equation we get

$1=-(70-24)=46k\Rightarrow k=-\frac{1}{46}\approx -0.02$

The minus comes from considering the temperature must decrease. With this value we can write the equation more explicitely:

$\frac{dT}{dt}=-\frac{1}{46}(T-24)$

2) The coffee is cooling off as time goes by, and it won't get any cooler than 24 degrees celsius because that's the temperature of the room. Then, in the long run, the temperature of the coffee is 24 degrees celsius.

3) Remember that Euler's method consists of using an initial exact measurement to predict what will happen in the future, approximately. There is a formula to make those predictions an it depends on the time step they gave us. Let's compute things first and then I tell you the equations we used.

In this case we know that we start with a 90 degrees celsius cup of coffee, or, in terms of math,

$T(0)=90$

Then, we can predict:

$T(2)\approx 90+2\left[-\frac{1}{46}(90-24)\right]=\frac{2004}{23}\approx 87.1$

Let's use fractions so we don't lose accuracy from now. With this number we can make an approximation of the temperature after 2 more seconds:

$T(4)\approx \frac{2004}{23}+2\left[-\frac{1}{46}\left(\frac{2004}{23}-24\right)\right]=\frac{44640}{529}\approx 84.4$

and then

$T(6)\approx \frac{44640}{529}+2\left[-\frac{1}{46}\left(\frac{44640}{529}-24\right)\right]=\frac{994776}{12167}\approx 81.8$

and then

$T(8)\approx \frac{994776}{12167}+2\left[-\frac{1}{46}\left(\frac{994776}{12167}-24\right)\right]=\frac{22177080}{279841}\approx 79.2$

and finally, the number we wanted to find:

$T(10)\approx \frac{22177080}{279841}+2\left[-\frac{1}{46}\left(\frac{22177080}{279841}-24\right)\right]=\frac{494611944}{6436343}\approx 76.8$

I hope you noticed the pattern to compute the next prediction:

$\textit{next prediction} = \textit{previous one (or exact value if it's the first step)}\\+ h\ast(\textit{right hand side of the differential equation at the previous one})$

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Answer:

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Explanation:

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3 years ago

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Where,

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Answer:

an object with a constant acceleration always have:

A. changing velocity

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3 years ago

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