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SCORPION-xisa [38]
3 years ago
15

A longitudinal wave is shown. Which label identifies a rarefaction?

Physics
1 answer:
aev [14]3 years ago
9 0

Answer:

B.

Explanation:

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Find the acceleration of a car with the mass of 1,200 kg and a force of
Mashutka [201]

Answer:9.17 m/s^2

Explanation:

mass=1200kg

Force=11 x 10^3 N

Acceleration=force ➗ mass

Acceleration=11 x 10^3 ➗ 1200

Acceleration=9.17

Acceleration=9.17 m/s^2

3 0
2 years ago
The potential difference between the plates of a capacitor is 145 V. Midway between the plates, a proton and an electron are rel
aniked [119]

Answer:

= 2.52 x 10^ 6 m/s        

Explanation:

The force that acts on charged particles between capacitor plates =

F = (q) (Δv)  ÷ d

Here,  d = distance between the two plates

          q = charge of the charged particle

         Δv = voltage

Normally, the force that makes both proton and electron released from rest, giving the charge acceleration is F=m X a. where m= mass and a = acceleration

Poting this equation with the first one, we have:

m X a =  (q) (Δv)  ÷ d

So, the acceleration of a proton when moving towards a negatively charged plate is

a = (q) (Δv)  ÷ (d) (m) {proton}

Likewise, the acceleration of an electron when moving towards a positively charged plate is

a = (q) (Δv)  ÷ (d) (m) {electron}

Dividing the proton acceleration formula by the electron acceleration formula we have:

a (proton) / a (electron) = m (proton) / m(electron)

inserting equation of motion to get distance, s

s = ut + 1/2 at^2

recall that electron travel distance, d/2

d/2 = 1/2 at^2

making t the subject of the formula

we have, t =√(d ÷ a(electron))

The distance of proton:

d/2 =  ut + 1/2 at^2 [proton}

put d/2 =  ut + 1/2 at^2 [proton} into t =√(d ÷ a(electron))

Initial speed, ui = √(d ÷ a(electron)) = (d/2) - (1/2) x (d) (a(proton) + a(electron))

since acceleration wasn't given in the question, lets use mass(elect

ron)  ÷ mass(proton) rather than use (a(proton) + a(electron))

Therefore, intial speed= 1/2√((e X Δv) ÷ m(electron)) (1- m(electron)/ m(proton))

   Note, e = 1.60 x 10^-19

           m(electron) = 9.11 X 10^-31

            m(proton) = 1.67  X 10^-27

Input these values into the formula above, initial speed, UI =  

           = 2.52 x 10^ 6 m/s          

7 0
3 years ago
The electron beam inside a television picture tube is 0.40 {\rm mm} in diameter and carries a current of 50 {\rm \mu A}. This el
Harrizon [31]

Answer:

A.3.13x10^14 electrons

B.330A/m²

C.9.11x10^5N/C

D. 0.23W

.pls see attached file for explanations

8 0
3 years ago
Your electric drill rotates initially at 5.21 rad/s. You slide the speed control and cause the drill to undergo constant angular
RUDIKE [14]

Answer:

The drill's angular displacement during that time interval is 24.17 rad.

Explanation:

Given;

initial angular velocity of the electric drill, \omega _i = 5.21 rad/s

angular acceleration of the electric drill, α = 0.311 rad/s²

time of motion of the electric drill, t = 4.13 s

The angular displacement of the electric drill at the given time interval is calculated as;

\theta = \omega _i t \ + \ \frac{1}{2}\alpha t^2\\\\\theta = (5.21 \ \times \ 4.13) \ + \ \frac{1}{2}(0.311)(4.13)^2\\\\\theta = (21.5173 ) \ + \ (2.6524)\\\\\theta =24.17 \ rad

Therefore, the drill's angular displacement during that time interval is 24.17 rad.

6 0
3 years ago
Evidence that the cosmic background radiation really is the remnant of a Big Bang comes from predicting characteristics of remna
mario62 [17]

Answer:

B) The cosmic background radiation is expected to contain spectral lines of hydrogen and helium, and it does.

Explanation:

3 0
2 years ago
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