1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Molodets [167]
4 years ago
6

Some life forms move so slowly that their movement cannot be detected by normal observation. The best way to determine any movem

ent is by placing a mark at some point, and observing any change in position from it. Why is this true?
Physics
2 answers:
romanna [79]4 years ago
5 0
Not sure the precise concept of "normal observation", but I assume that is observed by "eyes".

Eye observation is basically macroscopic, but when you use a mark, which can be regarded as a point of mass, then it goes to microscopic.

Mark is a reference point which you can compare the relative position change, but with your eyes, first you cannot notice microscopic changes, second the eyes cannot precisely set a stable reference point.
kirza4 [7]4 years ago
3 0

The mark is a reference point.

You might be interested in
A particular coaxial cable is comprised of inner and outer conductors having radii 1 mm and 3 mm respectively, separated by air.
noname [10]

Answer:

The value is  \rho_s  =  4.026 *10^{-6} \  C/m^2

Explanation:

From the question we are told that

   The radius of the inner conductor  is  r_1 = 1 \ mm =  0.001 \ m

    The radius of the outer conductor is  r_2 = 3 \ mm = 0.003 \  m

    The potential at the outer conductor is  V = 1.5 kV  =  1.5 *10^{3} \  V

Generally the capacitance per length of the capacitor like set up of the two conductors is

      C= \frac{2 * \pi * \epsilon_o }{ ln [\frac{r_2}{r_1} ]}

Here \epsilon_o is the permitivity of free space with value  \epsilon_o =  8.85*10^{-12} C/(V \cdot m)

=>   C= \frac{2 *  3.142  * 8.85*10^{-12}  }{ ln [\frac{0.003}{0.001} ]}

=>   C= 50.6 *10^{-12} \  F/m

Generally given that the potential  of the outer conductor with respect to the inner conductor is positive it then mean that the outer conductor is positively charge

Generally the line  charge density of the outer  conductor is mathematically represented as

      \rho_l  =  C *  V

=>   \rho_d  =  50.6*10^{-12} *  1.5*10^{3}

=>   \rho_d  =  7.59*10^{-8} \  C/m

Generally the surface charge density is mathematically represented as

        \rho_s  =  \frac{\rho_l }{2 \pi * r_2 }    here 2 \pi r = (circumference \ of \ outer \  conductor  )

=>    \rho_s  =  \frac{7.59 *10^{-8} }{2* 3.142 * 0.003 }

=>    \rho_s  =  4.026 *10^{-6} \  C/m^2

3 0
3 years ago
A quarterback throws a football with an angle of elevation 55° and speed 60 ft/s. Find the horizontal and vertical components o
EastWind [94]

Answer:

The horizontal component of the velocity vector is;

vh = 34.4 ft/s

The vertical component of the velocity vector is;

vy = 49.1 ft/s

Explanation:

Given;

Velocity of football v = 60 ft/s

Angle of elevation ∅ = 55°

The horizontal component of the velocity vector is;

vh = vcos∅

Substituting the values;

vh = 60cos55°

vh = 34.41458618106 ft/s

vh = 34.4 ft/s

The vertical component of the velocity vector is;

vy = vsin∅

Substituting the values;

vy = 60sin55°

vy = 49.14912265733 ft/s

vy = 49.1 ft/s

7 0
3 years ago
If R = 20 Ω, what is the equivalent resistance between points A and B in the figure?​
Paraphin [41]

Answer:

Option C. 70 Ω

Explanation:

Data obtained from the question include:

Resistor (R) = 20 Ω

From diagram given ABOVE, we observed the following

1. R and R are in parallel connections.

2. 2R and 2R are in parallel connections.

3. 4R and 4R are in parallel connections.

Next, we shall determine the equivalent resistance in each case.

This is illustrated below:

1. Determination of the equivalent resistance for R and R parallel connections.

R = 20 Ω

Equivalent R = (R×R) /(R+R)

Equivalent R = (20 × 20) /(20 + 20)

Equivalent R = 400/40

Equivalent R = 10 Ω

2. Determination of the equivalent resistance for 2R and 2R parallel connections.

R = 20 Ω

2R = 2 × 20 = 40 Ω

Equivalent 2R = (2R×2R) /(2R+2R)

Equivalent 2R = (40 × 40) /(40 + 40)

Equivalent 2R = 1600/80

Equivalent 2R = 20 Ω

3. Determination of the equivalent resistance for 4R and 4R parallel connections.

R = 20 Ω

4R = 4 × 20 = 80 Ω

Equivalent 4R = (4R×4R) /(4R+4R)

Equivalent 4R = (80 × 80) /(80 + 80)

Equivalent 4R = 6400/160

Equivalent 4R = 40 Ω

Thus, the equivalence of R, 2R and 4R are now in series connections. We can obtain the equivalent resistance in the circuit as follow:

Equivalent of R = 10 Ω

Equivalent of 2R = 20 Ω

Equivalent of 4R = 40 Ω

Equivalent =?

Equivalent = Equivalent of (R + 2R + 4R)

Equivalent = 10 + 20 + 40

Equivalent = 70 Ω

Therefore, the equivalent resistance between point A and B is 70 Ω.

6 0
3 years ago
The velocity selector in in a mass spectrometer consists of a uniform magnetic field oriented at 90 degrees to a uniform electri
NeTakaya

Answer:

50k/h is the answer to iy

3 0
3 years ago
A ball falls from a height of 25 cm what is the same distance described in meters? Someone help ASAP THIS IS DO TONIGHT
posledela

.25 meters

divide the value by 100 to get your answer.

5 0
3 years ago
Read 2 more answers
Other questions:
  • Is mass on a ramp related to acceleration?
    15·1 answer
  • What happens to the amount of energy present when it is moved from one form to another?
    7·1 answer
  • Tracy scuffs her socked feet across a carpet. When she touches a doorknob, she gets a small shock. How did Tracy become charged?
    9·2 answers
  • A 500N car is pulled up to 20m plank to a flat form 5m above the ground by an effort of 150N parallel to the plank. Calculate Ve
    9·1 answer
  • Select the correct answer
    8·2 answers
  • Can you answer my question please go to my profile ​
    6·2 answers
  • 17. How long does it take a giraffe running at a speed of 33 m/s to run 200 meters !
    12·1 answer
  • In can expermient why the can is chrushed​
    12·1 answer
  • A 1.1-cm-diameter pipe widens to 2.34 cm, then narrows to 4.9 mm. Liquid flows through the first segment at a speed of 3.57 m/s.
    8·1 answer
  • Slove these problem physics ​
    15·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!