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ankoles [38]
4 years ago
10

What is a characteristic of a default static route? ​?

Physics
1 answer:
Vaselesa [24]4 years ago
7 0

A default route is an exceptional kind of static route because the subnet and network you are identifying as the target for the traffic that it would equal to all zeros. The characteristic of a default static route would be it classifies the gateway IP address to which the router directs all IP packets for which it does not have a knowledge or static route.

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Are small molecular units joined together in large molecules
inysia [295]
<span>Polymers are small molecular units joined together in large molecules.

hope this helps!</span>
7 0
3 years ago
On a distant planet, a rock falls in 48.4 s from the top of a 1.10e+02 m cliff to the planet surface below. What is the accelera
pav-90 [236]

this can be solve using the formala of free fall

t = sqrt( 2y/ g)

where t is the time of fall

y is the height

g is the acceleration due to gravity

48.4 s = sqrt (2 (1.10e+02 m)/ g)

G = 0.0930 m/s2

The velocity at impact

V = sqrt(2gy)

= sqrt( 2 ( 0.0930 m/s2)( 1.10e+02 m)

V = 4.523 m/s

<span> </span>

8 0
4 years ago
<img src="https://tex.z-dn.net/?f=2.25%20%5Ctimes%2030" id="TexFormula1" title="2.25 \times 30" alt="2.25 \times 30" align="absm
Paraphin [41]
67.5 is the answer i believe!!
7 0
3 years ago
Read 2 more answers
A student is studying simple harmonic motion of a spring. She conducts an experiment where she measures the amplitude and period
myrzilka [38]

Answer:

5.9*10^{-4}m

Explanation:

to find the uncertainty of the displacement it is necessary to compute the uncertainty for the angular frequency:

\omega=\frac{2\pi}{T}=\frac{2\pi}{0.40s}=15.707rad/s\\\\\frac{d \omega}{\omega}=\frac{dT}{T}\\\\d\omega=\omega \frac{dT}{T}=(15.707rad/s)\frac{0.020s}{0.40s}=0.785rad/s

then, you can calculate the uncertainty in angular displacement:

\theta=\omega t\\\\\frac{d\theta}{\theta}=\sqrt{(\frac{d\omega}{\omega})^2+(\frac{dt}{t})^2}\\\\d\theta=\theta\sqrt{(\frac{d\omega}{\omega})^2+(\frac{dt}{t})^2}=0.0422

finally, by using:

y=Acos(\omega t)\\\\dy=dAcos(\omega t)d(\omega t)=(dA)cos(\theta)d\theta=(0.002m)cos(0.785)(0.0422)\\\\dy=5.9*10^{-4}m

7 0
3 years ago
A homing device left its home to deliver a message. On​ takeoff, the device encountered a tailwind of 14 mph and made the delive
andre [41]

Answer:

speed in still air is 42 mph

Explanation:

given data

tail wind speed = 14 mph

time t1 = 5 min = 5/60 hr = 1/12 hr

return time t2 = 10 min = 10 /60 hr = 1/6 hr

to find out

how fast they fly

solution

we consider here speed in still air is x

then

speed in wind will be = x + 14

and against wind speed = x - 14

so

distance = speed × time   .................1

so in wind distance = ( x + 14 ) × 1/12    ...................2

and against wind distance = ( x - 14 ) × 1/6    ..................3

so from equation 2 and 3

( x + 14 ) × 1/12 = ( x - 14 ) × 1/6

x = 42

so speed in still air is 42 mph

3 0
4 years ago
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