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sergiy2304 [10]
3 years ago
11

Which term describes the energy of position?

Physics
2 answers:
choli [55]3 years ago
8 0
B. Gravitational potential energy
scoray [572]3 years ago
3 0

Answer:

B

Explanation:

You might be interested in
Over a period of one year, how much of the overall (night) sky would a Bellingham observer be able to see?
nika2105 [10]

Answer:

The overall sky a Bellingham observer would see is about 75-100% over the period of 1 year.

Explanation:

Due to following factors, the overall sky visibility will fall in this range.

  1. The location of Bellingham on Earth is in the Northern hemisphere over the equator.
  2. The tilt of the earth cannot allow the visibility of the complete sky.
  3. The light pollution and aerosol concentrations has reduced the sky visibility at night.
6 0
4 years ago
An emergency vehicle blowing its siren is moving
Georgia [21]

The frequency produced by the siren is 631.12 Hz

<h3>Doppler effect</h3>

The variation in frequency when a source of sound moves relative to an observer is determined by the doppler effect.

<h3>Frequency of observer</h3>

So, the frequency of the observer  f' = (v ± v')f/(v ± v") where

f' = 590 Hz

f = frequency of source or siren ,

v = speed of sound = 330 m/s,

v' = speed of observer = 0 m/s (since you are stationary) and

v" = speed of source = 23 m/s

Since the source moves away from the detector, the sign in the denominator is positive and v' = 0 m/s

So, f' = (v + 0)f/(v + v")

f' = vf/(v + v")

Since, we require the frequency of the source, make f subject of the formula, we have

<h3>Frequency of siren</h3>

f = (v + v")f'/v

Substituting the values of the variables into the equation, we have

f = (v + v")f'/v

f' = (330 m/s + 23 m/s)  × 590 Hz/330 m/s

f' = 353 m/s × 590 Hz/330 m/s

f' = 208270 m/sHz/330 m/s

f' = 631.12 Hz

The frequency produced by the siren is 631.12 Hz

Learn more about doppler effect here:

brainly.com/question/2169203

5 0
2 years ago
PLZ HELP ON #22-26!!!! <br><br>Please explain why and how you got your answer.
AleksAgata [21]
22. a - (vf^2 - vi^2)/(2d) 
a = (0 - 23^2)/(170) 
a = -3.1 m/s^2

23. Find the time (t) to reach 33 m/s at 3 m/s^2
33-0/t = 3 
33 = 3t 
t = 11 sec to reach 33 m/s^2
Find the av velocuty: 33+0/2 = 16.5 m/s
Dist = 16.5 * 11 = 181.5 meters to each 33m/s speed. Runway has to be at least this long. 

24. The sprinter starts from rest. The average acceleration is found from: 
(Vf)^2 = (Vi)^2 = 2as ---> a = (Vf)^2 - (Vi)^2/2s = (11.5m/s)^2-0/2(15.0m) = 4.408m/s^2 estimated: 4.41m/s^2
The elapsed time is found by solving
Vf = Vi + at ----> t = vf-vi/a = 11.5m/s-0/4.408m/s^2 = 2.61s

25. Acceleration of car = v-u/t = 0ms^-1-21.0ms^-1/6.00s = -3.50ms^-2
S = v^2 - u^2/2a = (0ms^-1)^2-(21.0ms^-1)^2/2*-3.50ms^-2 = 63.0m 

26. Assuming a constant deceleration of 7.00 m/s^2
final velocity, v = 0m/s 
acceleration, a = -7.00m/s^2
displacement, s - 92m 
Using v^2 = u^2 - 2as 
0^2 - u^2 + 2 (-7.00) (92) 
initial velocity, u = sqrt (1288) = 35.9 m/s 
This is the speed pf the car just bore braking. 

I hope this helps!! 

5 0
3 years ago
A converging lens of focal length 20 cm is placed in contact with a diverging lens of focal length 30 cm. The focal length of th
kramer

Answer:

The focal lenth (F) =+10.0cm

Explanation:

The formular for combined focal length (F) is given as;

1/f = 1/f_{1} + 1/f_{2}

In this  question,

F1 = 20cm

F2  = -30cm

Plugging the values into the formuar above,

1/f = 1/20 + 1/(-30)

1/f = 0.05 - 0.033[tex]1/f = -0.017f = [tex]1/ -0.017

f = 58.82cm

i.e. the combination behaves as a converging lens (because of the postive sign) of focal length 58.82cm .

7 0
3 years ago
A vacuum tube diode consists of concentric cylindrical electrodes, the negative cathode and the positive anode. Because of the a
DENIUS [597]

Answer:

   C = 4,174 10³ V / m^{3/4} ,  E = 7.19 10² / ∛x,    E = 1.5  10³ N/C

Explanation:

For this exercise we can calculate the value of the constant and the electric field produced,

Let's start by calculating the value of the constant C

           V = C x^{4/3}

           C = V / x^{4/3}

            C = 220 / (11 10⁻²)^{4/3}

            C = 4,174 10³ V / m^{3/4}

To calculate the electric field we use the expression

            V = E dx

             E = dx / V

             E = ∫ dx / C x^{4/3}

            E = 1 / C  x^{-1/3} / (- 1/3)

            E = 1 / C (-3 / x^{1/3})

We evaluate from the lower limit x = 0 E = E₀ = 0 to the upper limit x = x, E = E

            E = 3 / C     (0- (-1 / x^{1/3}))

            E = 3 / 4,174 10³   (1 / x^{1/3})

           E = 7.19 10² / ∛x

for x = 0.110 cm

          E = 7.19 10² /∛0.11

          E = 1.5  10³ N/C

6 0
4 years ago
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