A scientific theory is:

Define international employees

lora16 [44]

Answer:

An “international employee” is defined as an employee of Stanford University whose work site is located.

Explanation:

6 0

3 years ago

An airplane is flying in the direction 10° east of south at 701 km/hr. Find the component form ofthe velocity of the airplane, a

solniwko [45]

Answer:

The component form will be;

In the x-axis = 121.73 due west

In the y-axis = 690.35 due south

Explanation:

An image of the calculation has been attached

7 0

2 years ago

Four traveling waves are described by the following equations, where all quantities are measured in SI units and y represents th

agasfer [191]

Answer:

$T_1=T_3=\dfrac{2\pi}{21}$

$T_2=T_4=\dfrac{2\pi}{42}$

Explanation:

Wave 1, $y_1=0.12\ cos(3x-21t)$

Wave 2, $y_2=0.15\ sin(6x+42t)$

Wave 3, $y_3=0.13\ cos(6x+21t)$

Wave 4, $y_4=-0.27\ sin(3x-42t)$

The general equation of travelling wave is given by :

$y=A\ cos(kx\pm \omega t)$

The value of $\omega$ will remain the same if we take phase difference into account.

For first wave,

$\omega_1=21$

$\dfrac{2\pi }{T_1}=21$

$T_1=\dfrac{2\pi}{21}$

For second wave,

$\omega_2=42$

$\dfrac{2\pi }{T_2}=42$

$T_2=\dfrac{2\pi}{42}$

For the third wave,

$\omega_3=21$

$\dfrac{2\pi }{T_3}=21$

$T_3=\dfrac{2\pi}{21}$

For the fourth wave,

$\omega_4=42$

$\dfrac{2\pi }{T_4}=42$

$T_4=\dfrac{2\pi}{42}$

It is clear from above calculations that waves 1 and 3 have same time period. Also, wave 2 and 4 have same time period. Hence, this is the required solution.

3 0

2 years ago

Two charges are located in the x x – y y plane. If q 1 = − 2.75 nC q1=−2.75 nC and is located at ( x = 0.00 m , y = 0.600 m ) (x

ludmilkaskok [199]

Answer:

$\vec{E}=(-15.78\hat{i}+63.81\hat{j})\frac{N}{C}$

Explanation:

In this case we have to work with vectors. Firs of all we have to compute the angles between x axis and the r vector (which points the charges):

$\theta_1=tan^{-1}(\frac{0.6m}{0m})=90\° \\\\\theta_2=tan^{-1}(\frac{0.4m}{1.3m})=17.10\°$

the electric field has two components Ex and Ey. By considering the sign of the charges we obtain that:

$\vec{E} = E_x \hat{i}+E_y\hat{j}\\\\\vec{E}=(E_1cos\theta_1-E_2cos\theta_2)\hat{i}+(E_1sin\theta_1-E_2sin\theta_2)\hat{j}\\\\E_1=k\frac{q}{r_1^2}=(8.99*10^{9}\frac{N}{m^2C^2})\frac{2.75*10^{-9}C}{(0.6m)^2}=68.67\frac{N}{C}\\\\E_2=k\frac{q}{r_2^2}=(8.99*10^{9}\frac{N}{m^2C^2})\frac{3.40*10^{-9}C}{((1.3m)^2+(0.4m)^2)}=16.52\frac{N}{C}$

Hence, by replacing E1 and E2 we obtain:

$\vec{E}=[(68.67N/C)cos(90\°)-(16.52N/C)cos(17.10\°)]\hat{i}+[(68.67N/C)sin(90\°)-(16.52N/C)sin(17.10\°)]\hat{j}\\\\\vec{E}=(-15.78\hat{i}+63.81\hat{j})\frac{N}{C}$

hope this helps!!

3 0

2 years ago

Mercury is used as thermometeric liquid.Why?

garik1379 [7]

Mercury is used as thermometeric liquid because it is a good conductor of heat colorful, expansion rate is uniform and it doesn't wet the wall of capillary

4 0

2 years ago