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exis [7]
2 years ago
5

When two trains, going in opposite directions, are passing on tracks that are laid out close together, the train cars can often

be seen to be leaning in toward one another where they are in proximity. How might the air passing through the narrow gap separating the two trains contribute to the observed attraction between their cars?
Physics
2 answers:
Afina-wow [57]2 years ago
6 0
<h3>Answer:</h3>

"<u>Therefore, the speed of the air decreases, the pressure acting on the two trains increases (by Bernoulli's principle), and the two trains lean in toward one another</u>."

daser333 [38]2 years ago
6 0
When there is air in between two objects, it pulls them together because of the pressure of the air
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Answer:

a) V = 195.70 m/s

b) f=3.02 × 10⁻⁴ Hz

c) T = 3311.25 seconds

Explanation:

Given:

Wavelength, λ = 646 Km = 646000 m

Distance traveled = 3410 Km = 3410000 m

Time = 4.84 h = 4.84 × 3600 s = 17424 seconds

a) The speed (V) of the wave is given as

V = distance / time

V = 3410000 m/ 17424 seconds

or

V = 195.70 m/s

b) The frequency (f) of the wave is given as:

f = V / λ

f= 195.70 / 646000

f=3.02 × 10⁻⁴ Hz

c) The time period (T)  is given as:

T = 1/ f

T = 1/ (3.02 × 10⁻⁴) Hz

T = 3311.25 seconds

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When an object radiates heat, the strength of this radiation far from the object decreases when distance from the source increas
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The source of cosmic background radiation filled the entire universe.

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D:The source of cosmic background radiation filled the entire universe.

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2 years ago
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What is the value of R2 in this parallel circuit? (5 stars)
o-na [289]

Answer:

20 Ω

Explanation:

Voltage, current, and resistance are related by Ohm's law:

V = IR

40 V = (4 A) R

R = 10 Ω

The total resistance of the circuit is 10 Ω.

Resistors in parallel have a total resistance of:

1/R = 1/R₁ + 1/R₂

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2 years ago
The barricade at the end of a subway line has a large spring designed to compress 2.00 m when stopping a 1.10 ✕ 105 kg train mov
Mrac [35]

Answer:

(a) k = 1684.38 N/m = 1.684 KN/m

(b) Vi = 0.105 m/s

(c) F = 1010.62 N = 1.01 KN

Explanation:

(a)

First, we find the deceleration of the car. For that purpose we use 3rd equation of motion:

2as = Vf² - Vi²

a = (Vf² - Vi²)/2s

where,

a = deceleration = ?

Vf = final velocity = 0 m/s (since, train finally stops)

Vi = Initial Velocity = 0.35 m/s

s = distance covered by train before stopping = 2 m

Therefore,

a = [(0 m/s)² - (0.35 m/s)²]/(2)(2 m)

a = 0.0306 m/s²

Now, we calculate the force applied on spring by train:

F = ma

F = (1.1 x 10⁵ kg)(0.0306 m/s²)

F = 3368.75 N

Now, for force constant, we use Hooke's Law:

F = kΔx

where,

k = Force Constant = ?

Δx = Compression = 2 m

Therefore.

3368.75 N = k(2 m)

k = (3368.75 N)/(2 m)

<u>k = 1684.38 N/m = 1.684 KN/m</u>

<u></u>

<u>(</u>c<u>)</u>

Applying Hooke's Law with:

Δx  = 0.6 m

F = (1684.38 N/m)(0.6 m)

<u>F = 1010.62 N = 1.01 KN</u>

<u></u>

(b)

Now, the acceleration required for this force is:

F = ma

1010.62 N = (1.1 kg)a

a = 1010.62 N/1.1 x 10⁵ kg

a = 0.0092 m/s²

Now, we find initial velocity of train by using 3rd equation of motion:

2as = Vf² - Vi²

a = (Vf² - Vi²)/2s

where,

a = deceleration = -0.0092 m/s² (negative sign due to deceleration)

Vf = final velocity = 0 m/s (since, train finally stops)

Vi = Initial Velocity = ?

s = distance covered by train before stopping = 0.6 m

Therefore,

-0.0092 m/s² = [(0 m/s)² - Vi²]/(2)(0.6 m)

Vi = √(0.0092 m/s²)(1.2 m)

<u>Vi = 0.105 m/s</u>

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hope this helps

6 0
3 years ago
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