Ask question

Ask question

All categories

Ray Of Light [21]

3 years ago

9

Some radar systems detect the size and shape of objects such as aircraft and geological terrain. Approximately what is the small

est observable detail (in m) utilizing 495 MHz radar?

1 answer:

tatiyna3 years ago

5 0

Answer:

<em>0.61 m</em>

<em></em>

Explanation:

The smallest observable length by the radar must be at least equal to or greater than the wavelength of the radar.

using the relationship

c = fλ

where

c is the speed of light in vacuum = 3 x 10^8 m/s

f is the frequency of the wave = 495 MHz = 4.95 x 10^8 Hz

λ is the wavelength = ?

λ = c/f = (3 x 10^8)/(4.95 x 10^8) = <em>0.61 m</em>

You might be interested in

__________ is the study of things continually moving or at rest.

bogdanovich [222]

Force

Explanation:

so force is your answer see why b/c

4 0

3 years ago

. A child drops a ball from a window. The ball strikes the ground in 3.0 seconds. What is the. A child drops a ball from a windo

Vsevolod [243]

Answer:

30m/s

Explanation:

Use formula for height calculation: Time = √2h/√g
From here calculate height which comes out to be: 45m(if g is taken 10)
Then use formula for velocity calculation: v = √2gh

The velocity comes out to be 30m/s

3 0

3 years ago

magine an astronaut on an extrasolar planet, standing on a sheer cliff 50.0 m high. She is so happy to be on a different planet,

Mama L [17]

Answer:

$\Delta t=(\frac{20}{g'}+\sqrt{\frac{400}{g'^2}+\frac{100}{g'} } )-(\frac{20}{g}+\sqrt{\frac{400}{g^2}+\frac{100}{g} } )$

Explanation:

Given:

height above which the rock is thrown up, $\Delta h=50\ m$

initial velocity of projection, $u=20\ m.s^{-1}$

let the gravity on the other planet be g'

The time taken by the rock to reach the top height on the exoplanet:

$v=u+g'.t'$

where:

$v=$ final velocity at the top height = 0 $m.s^{-1}$

$0=20-g'.t'$ (-ve sign to indicate that acceleration acts opposite to the velocity)

$t'=\frac{20}{g'}\ s$

The time taken by the rock to reach the top height on the earth:

$v=u+g.t$

$0=20-g.t$

$t=\frac{20}{g} \ s$

Height reached by the rock above the point of throwing on the exoplanet:

$v^2=u^2+2g'.h'$

where:

$v=$ final velocity at the top height = 0 $m.s^{-1}$

$0^2=20^2-2\times g'.h'$

$h'=\frac{200}{g'}\ m$

Height reached by the rock above the point of throwing on the earth:

$v^2=u^2+2g.h$

$0^2=20^2-2g.h$

$h=\frac{200}{g}\ m$

The time taken by the rock to fall from the highest point to the ground on the exoplanet:

$(50+h')=u.t_f'+\frac{1}{2} g'.t_f'^2$ (during falling it falls below the cliff)

here:

$u=$ initial velocity= 0 $m.s^{-1}$

$\frac{200}{g'}+50 =0+\frac{1}{2} g'.t_f'^2$

$t_f'^2=\frac{400}{g'^2}+\frac{100}{g'}$

$t_f'=\sqrt{\frac{400}{g'^2}+\frac{100}{g'} }$

Similarly on earth:

$t_f=\sqrt{\frac{400}{g^2}+\frac{100}{g} }$

Now the required time difference:

$\Delta t=(t'+t_f')-(t+t_f)$

$\Delta t=(\frac{20}{g'}+\sqrt{\frac{400}{g'^2}+\frac{100}{g'} } )-(\frac{20}{g}+\sqrt{\frac{400}{g^2}+\frac{100}{g} } )$

3 0

3 years ago

How could a food web based in water be connected to a food web based on land? Give an example.

Serggg [28]

Land animals could eat water animals For example, a bear could eat salmon from the river and that could be connected to organisms from water to land

4 0

4 years ago

Read 2 more answers

A 60.0-kg skier with an initial speed of 12.0 m/s coasts up a 2.50-mhigh rise and angle is 35 degrees.Find her final speed at th

Shalnov [3]

Wt. = Fg = m*g = 60kg * 9.8N/kg=588 N.=
<span>Wt. of skier. </span>

<span>Fp=588*sin35 = 337 N.=Force parallel to </span>
<span>incline. </span>
<span>Fv = 588*cos35 = 482 N. = Force perpendicular to incline. </span>

<span>Fk = u*Fv = 0.08 * 482 = 38.5 N. = Force </span>
<span>of kinetic friction. </span>
<span>d =h/sinA = 2.5/sin35 = 4.36 m. </span>

<span>Ek + Ep = Ekmax - Fk*d </span>
<span>Ek = Ekmax-Ep-Fk*d </span>
<span>Ek=0.5*60*12^2-588*2.5-38.5*4.36=2682 J. </span>
<span>Ek = 0.5m*V^2 = 2682 J. </span>
<span>30*V^2 = 2682 </span>
<span>V^2 = 89.4 </span>
<span>V = 9.5 m/s = Final velocity.</span>

4 0

4 years ago

Read 2 more answers