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marishachu [46]

3 years ago

5

The advantage of a synchronous orbit is that the satellite can: fly over the entire earth transmit over the entire earth receive

continuous signals from a ground transmitter all of the above

1 answer:

dsp733 years ago

3 0

Answer:

<u>receive continuous signals from a ground transmitter</u>

Explanation:

Remember a <u>synchronous orbit is synchronous in the sense that it has equal rotational period to the body orbiting having the same direction of rotation as that body</u>. An example of this is the Moon and the planet earth.

Thus it allows continuous signals to be sent from a ground transmitter and received by a space object.

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The wall of a large room is covered with acoustic tile in which small holes are drilled 4.6 mm from center to center. How far ca

Zielflug [23.3K]

Answer:

L= 27.42m

Explanation:

We have the variables

$D=4.6*10^{-3}m$

$d= 4*10^{-3}m$

$\lambda = 550*10^{-9}m$

Use the relation,

$L=\frac{D}{\theta_R} = \frac{D_d}{1.22\lambda}$

$L= \frac{(4.6*10^{-3}m)(4*10^{-3}m)}{1.22(550*10^{-9}m)}$

$L= 27.42m$

8 0

4 years ago

Where does every piece of matter begin?

Margarita [4]

Every piece of matter begins “Out of this world”

6 0

3 years ago

A mass of gas has a volume of 4m3, a temperature of 290k, and an absolute pressure of 475 kpa. When the gas is allowed to expand

motikmotik

Given:

V1 = 4m3

T1 = 290k

P1 = 475 kpa = 475000 Pa

V2 = 6.5m3

T2 = 277K

Required:

P

Solution:

n = PV/RT

n = (475000 Pa)(4m3) / (8.314 Pa-m3/mol-K)(290k)

n = 788 moles

P = nRT/V

P = (788 moles)(8.314 Pa-m3/mol-K)(277K)/(6.5m3)

P = 279,204 Pa or 279 kPa

4 0

3 years ago

Read 2 more answers

A parachutist bails out and freely falls 63 m. Then the parachute opens, and thereafter she decelerates at 1.5 m/s2. She reaches

yaroslaw [1]

Answer:

(a) The parachutist spent 24.84 secs in air

(b) The height the fall begins is 472 m

Explanation:

Here is the complete question:

A parachutist bails out and freely falls 63 m. Then the parachute opens, and thereafter she decelerates at 1.5 m/s2. She reaches the ground with a speed of 3.3 m/s. (a) How long is the parachutist in the air (b) At what height does the fall begin?

Explanation:

From one of the equations of kinematics for free fall

$H = ut - \frac{1}{2}gt^{2}$

Where H is the height

u is the initial velocity

t is the time

and g is the acceleration due to gravity (Take g = 9.8 m/s2)

Now, we can find the time spent before the parachute opens.

u = 0 m/s (we assume the parachutist starts from rest)

H = - 63 m

∴ $-63 = 0(t) - \frac{1}{2}(0.98) t^{2} \\-63 = -4.9 t\\t^{2} = \frac{63}{4.9} \\t^{2} = 12.86\\t = \sqrt{12.86} \\t = 3.59 s$

This the time spent before the parachute opens

Also from one of the equations of kinematics for free fall

$v = u - gt$

where v is the final velocity

We can determine the final velocity before the parachute opens and she starts to decelerate

∴ $v = - 9.8(3.59)\\v = - 35.18m/s$

Now, we will calculate the time spent after the parachute opens

From one of the equation of kinematics for linear motion,

$v = u + at$

Here, the initial velocity will be the final velocity just before the parachute opens, that is

$u = - 35.18 m/s$

From the question,

$v = - 3.3 m/s$

$a = 1.5 m/s^{2}$

We then get

$-3.3 = - 35.18 + (1.5)t$

$-3.3 + 35.18 = 1.5t\\31.88 = 1.5t$

$t = \frac{31.88}{1.5}$

$t = 21.25 secs$

(a) To determine how long the parachutist is in the air,

That is sum of the time used when falling freely and the time used after the parachute opens

= 3.59 secs + 21.25 secs

24.84 secs

Hence, the parachutist spent 24.84 secs in air

(b) To determine what height the fall begins

First, we will calculate the height from which the parachute opens

From one of the equation of kinematics for linear motion,

$x = ut + \frac{1}{2}at^{2} \\$

$x = -35.18(21.25) + \frac{1}{2}(1.5)(21.25)^{2} \\x = -747.58 + 338.67\\x = -408.91m\\$

$x$ ≅ $- 409m$

Hence, the height the fall begins is 63m + 409m

= 472 m

3 0

3 years ago

A cyclist racing in the keiran is riding at the top of the track at 5m/s. Then he sprints downhill in the sprinting to the finis

kumpel [21]

The cyclist is moving by uniformly accelerated motion, with an initial velocity of $v_i=5~m/s$ and an acceleration of $a=9~m/s^2$ .
The acceleration is given by
$a= \frac{v_f-v_i}{\Delta t}$
where $v_f$ is the final velocity and $\Delta t$ is the time between the end and the beginning of the motion, and in our case it is 1.75 s. Therefore, from this relationship we can find the final velocity:
$v_f=v_i + a \Delta t=5~m/s+9~m/s^2 \cdot 1.75~s=20.75~m/s$

6 0

4 years ago