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Novay_Z [31]
3 years ago
12

Which environment is least likely to support protists

Physics
1 answer:
max2010maxim [7]3 years ago
3 0

Answer:

A: Soil

Explanation:

Protists need a moist environment to survive, and shallow ponds, oceans, and blood is all moist. So, the answer would be the soil, because that is the least moist environment out of these options.

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Two parallel plates 19 cm on a side are given equal and opposite charges of magnitude 2.0 ✕ 10^−9 C. The plates are 1.8 mm apart
Romashka [77]

Answer:

E   = 5291.00 N/C

Explanation:

Expression for capacitance is

C = \frac{\epsilon  A}{d}

where

A is area of square plate

D = DISTANCE BETWEEN THE PLATE

C = \frac{\epsilon\times(19\times 10^{-2})^2}{1.5\times 10^{-3}}

C = 24.06 \epsilon

C = 24.06\times 8.854\times 10^{-12} F

C =2.1\times 10^{-10} F

We know that capacitrnce and charge is related as

V = \frac{Q}{C}

 = \frac{2\tiimes 10^{-9}}{2.\times 10^{-10}}

v = 9.523 V

Electric field is given as

E = \frac{V}{d}

   = \frac{9.52}{1.8*10^{-3}}

E   = 5291.00 V/m

E   = 5291.00 N/C

5 0
3 years ago
A jogger runs at a constant rate of 10.0 m every 2.0 seconds. The jogger starts at the origin and runs in the positive direction
Elis [28]

Answer:

(a) 25 m

(b) 75 m

Explanation:

Given that the jogger runs at a constant rate of 10.0 m every 2.0 seconds.

So, the speed of the jogger,

v=\frac{10}{2}=5m/s\;\cdots(i)

Let d be the distance covered by him in time, t s.

As distance=(speed) x (time)

So, d=vt

From equation (i)

\Rightarrow d=5t\;\cdots(ii)

As the jogger starts from origin, so, the distance, d, also represents the position of the jogger at the time t s.

The position-time graph has been shown.

(a) From equation (ii), for t=5.0 s

d=5\times 5=25 m

So, the jogger is at a distance of 25 m from the origin.

(b) Similarly, for t=15.0 s

d=5\times 15=75 m

So, the jogger is at a distance of 75 m from the origin.

8 0
3 years ago
A wave of infrared light has a speed of 6 m/s and a wavelength of 12 m. What is the frequency of this wave?
givi [52]
I'll be happy to solve the problem using the information that
you gave in the question, but I have to tell you that this wave
is not infrared light.

If it was a wave of infrared, then its speed would be close
to 300,000,000 m/s, not 6 m/s, and its wavelength would be
less than 0.001 meter, not 12 meters.

For the wave you described . . .

             Frequency  =  (speed)  /  (wavelength)

                                 =  (6 m/s)  /  (12 m)

                                 =      0.5 / sec

                                 =      0.5 Hz .  

(If it were an infrared wave, then its frequency would be
greater than 300,000,000,000 Hz.)
5 0
3 years ago
Read 2 more answers
A tall, cylindrical chimney falls over when its base is ruptured. Treat the chimney as a thin rod of length 53.2 m. At the insta
rusak2 [61]
From an energy balance, we can use this formula to solve for the angular speed of the chimney

ω^2 = 3g / h sin θ

Substituting the given values:
ω^2 = 3 (9.81) / 53.2 sin 34.1
ω^2 = 0.987 /s

The formula for radial acceleration is:
a = rω^2

So,
a = 53.2 (0.987) = 52.494 /s^2

The linear velocity is:
v^2 = ar
v^2 = 52.949 (53.2) = 2816.887

The tangential acceleration is:
a = r v^2
a = 53.2 (2816.887)
a = 149858.378 m/s^2

If the tangential acceleration is equal to g:
g = r^2 3g / sin θ
Solving for θ
θ = 67° 
7 0
3 years ago
Analogue signals transmit information for such things as _____________.
ivann1987 [24]

Transmission of information in ANY form can be done digitally
or analoguely.

Beginning about 30 years ago, everything slowly started changing
to digital.  Today, all commercial satellite communication, all optical
fiber communication, all internet communication, all computer
communication, all commercial cable communication, all commercial
television, and much of the telephone system, are all digital. 

On your computer ... .pdf,  .jpg, .mp3  etc.  are all digital methods of
moving and storing information.

AM and FM radio are an interesting subject.  They're all still analog.
They could easily be changed to all digital, and it would be a big
improvement, both for the broadcasters and for the listeners. 
BUT ... every AM and FM radio that anybody has now would be
obsolete.   Every single radio would either need to be replaced,
OR you'd need to add a digital decoder to every radio, like we
had to do with our TV sets a few years ago when television
suddenly became all digital.  With AM and FM radios, the decoders
would be bigger, and would cost more, than most of the radios.

And that's why commercial radio broadcasting is still analog.
 
7 0
3 years ago
Read 2 more answers
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