The downlink frequency is lower than the uplink frequency. True or False

A pressure wave is what type of wave

Alex777 [14]

propagated disturbance is a variation

5 0

3 years ago

Someone can help me?

boyakko [2]

<h3>When the object is placed at a further distance from the center of the mirror's curvature (twice the focal length), we will get a thumbnail</h3><h3 /><h3>position of the image from the mirror; Between focus and center of curvature of the mirror (double focal length)</h3><h3 /><h3> picture description; real, inverted, mini</h3>

<h3>* This picture is to draw the rays, just replace the candle with an apple .</h3>

<h3>Do you want me to write it in Spanish to help you?? ^_^</h3>

I hope I helped you^_^

<h3 />

8 0

3 years ago

A diffraction grating is illuminated simultaneously with red light of wavelength 670 nm and light of an unknown wavelength. The

marusya05 [52]

Answer:

dsin∅ = m× λ

so, dsin∅red = 3(670nm)

also, dsin∅? =5λ?

however ,if they overlap then dsin∅red = dsin∅?

3(670nm) /5 =402nm

∴λ = 402nm

Explanation:

4 0

3 years ago

6) Find the speed a spherical raindrop would attain by falling from 4.00 km. Do this:a) In the absence of air dragb) In the pres

sleet_krkn [62]

We are asked to determine the velocity of a rain drop if it falls from 4 km.

To do that we will use the following formula:

$2ah=v_f^2-v_0^2$

Where:

$\begin{gathered} a=\text{ acceleration} \\ h=\text{ height} \\ v_f,v_0=\text{ final and initial velocity} \end{gathered}$

If we assume the initial velocity to be 0 we get:

$2ah=v_f^2$

The acceleration is the acceleration due to gravity:

$2gh=v_f^2$

Now, we take the square root to both sides:

$\sqrt{2gh}=v_f$

Now, we substitute the values:

$\sqrt{2(9.8\frac{m}{s^2})(4000m)}=v_f$

solving the operations:

$280\frac{m}{s}=v$

Therefore, the velocity without air drag is 280 m/s.

Part B. we are asked to determine the velocity if there is air drag. To do that we will use the following formula:

$F_d=\frac{1}{2}C\rho_{air}Av^2$

Where:

$\begin{gathered} F_d=drag\text{ force} \\ C=\text{ constant} \\ \rho_{air}=\text{ density of air} \\ A=\text{ area} \\ v=\text{ velocity} \end{gathered}$

We need to determine the drag force. To do that we will use the following free-body diagram:

Since the velocity that the raindrop reaches is the terminal velocity and its a constant velocity this means that the acceleration is zero and therefore the forces are balanced: