A train traveled 280km in 3.5 hours. what was the average speed in m/s?

Consider two antennas separated by 9.00 m that radiate in phase at 120 MHz, as described in Exercise 35.3. A receiver placed 150

alexgriva [62]

Answer:

$\phi=4.52 rad$

Explanation:

From the question we are told that

Distance b/e antenna's $d=9.00m$

Frequency of antenna Radiation $F_r=120 MHz \approx 120*10^6Hz$

Distance from receiver $d_r=150m$

Intensity of Receiver $i= 10$

Distance difference of the receiver b/w antenna's $(r^2-r^1)=1.8m$

Generally the equation for Phase difference $\phi$ is mathematically given by

$\phi=\frac{2\pi}{\frac{c}{f_r}} *(r^2-r^1)$

$\phi=\frac{2*\pi}{\frac{3*10^{8}}{120*10^6}} *1.8$

$\phi=\frac{4\pi}{5} *1.8$

Therefore phase difference f between the two radio waves produced by this path difference is given as

$\phi=4.52 rad$

7 0

2 years ago

Una persona de 80 kg de masa corre a una velocidad cuya magnitud es de 9m/s. Cual es la magnitud de la cantidad de movimiento? Q

natka813 [3]

Responder:

13,01 m / s

Explicación:

Paso uno:

datos dados

masa de la persona 1 m = 80 kg

velocidad de la persona 1 v = 9 m / s

masa de la persona 2 M = 55kg

velocidad de la persona 2 v =?

Segundo paso:

la expresión del impulso se da como

P = mv

para la primera persona, el impulso es

P = 80 * 9

P = 720N

Paso tres:

queremos que la segunda persona tenga el mismo impulso que la primera, por lo que la velocidad debe ser

720 = 55v

v = 720/55

v = 13,09

v = 13,01 m / s

Por lo tanto, la magnitud de la velocidad debe ser 13.01 m / s.

4 0

3 years ago

If a statement is true, select true. if it's false, select false. <br>

Elena L [17]

3 is false 2 is true and the rest true

7 0

3 years ago

Read 2 more answers

A charge is divided q1 and (q-q1)what will be the ratio of q/q1 so that force between the two parts placed at a given distance i

Arturiano [62]

Answer:

$q / q_{1} = 2$ , assuming that $q_{1}$ and $(q - q_{1})$ are point charges.

Explanation:

Let $k$ denote the coulomb constant. Let $r$ denote the distance between the two point charges. In this question, neither $k$ and $r$ depend on the value of $q_{1}$ .

By Coulomb's Law, the magnitude of electrostatic force between $q_{1}$ and $(q - q_{1})$ would be:

$\begin{aligned}F &= \frac{k\, q_{1}\, (q - q_{1})}{r^{2}} \\ &= \frac{k}{r^{2}}\, (q\, q_{1} - {q_{1}}^{2})\end{aligned}$ .

Find the first and second derivative of $F$ with respect to $q_{1}$ . (Note that $0 < q_{1} < q$ .)

First derivative:

$\begin{aligned}\frac{d}{d q_{1}}[F] &= \frac{d}{d q_{1}} \left[\frac{k}{r^{2}}\, (q\, q_{1} - {q_{1}}^{2})\right] \\ &= \frac{k}{r^{2}}\, \left[\frac{d}{d q_{1}} [q\, q_{1}] - \frac{d}{d q_{1}}[{q_{1}}^{2}]\right]\\ &= \frac{k}{r^{2}}\, (q - 2\, q_{1})\end{aligned}$ .

Second derivative:

$\begin{aligned}\frac{d^{2}}{{d q_{1}}^{2}}[F] &= \frac{d}{d q_{1}} \left[\frac{k}{r^{2}}\, (q - 2\, q_{1})\right] \\ &= \frac{(-2)\, k}{r^{2}}\end{aligned}$ .

The value of the coulomb constant $k$ is greater than $0$ . Thus, the value of the second derivative of $F$ with respect to $q_{1}$ would be negative for all real $r$ . $F\!$ would be convex over all $q_{1}$ .

By the convexity of $\! F$ with respect to $\! q_{1} \!$ , there would be a unique $q_{1}$ that globally maximizes $F$ . The first derivative of $F\!$ with respect to $q_{1}\!$ should be $0$ for that particular $\! q_{1}$ . In other words:

$\displaystyle \frac{k}{r^{2}}\, (q - 2\, q_{1}) = 0$ <em>.</em>

$2\, q_{1} = q$ .

$q_{1} = q / 2$ .

In other words, the force between the two point charges would be maximized when the charge is evenly split:

$\begin{aligned} \frac{q}{q_{1}} &= \frac{q}{q / 2} = 2\end{aligned}$ .

3 0

2 years ago

Kyra and Pari are timing how long it takes for 1 g of sugar to dissolve in hot water. Kyra records a time of 24.3 seconds. Pari

Hoochie [10]

<h3>Answer:</h3><h3>we can say that:-</h3>

A reading with more no of significant figures is considered to be more precise.
Kyra recorded a reading of 24.3 sec. Since all non 0 digits are considered to be significant this reading has 3 significant figures.
Pari recorded a reading of 24 sec. Since all non 0 digits are considered to be significant this reading has 3 significant figures.

<h3>hence we can say that kyra's reading has more significant figures nd so it is more precise.</h3>

7 0

2 years ago