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kykrilka [37]
2 years ago
13

A ski gondola is connected to the top of a hill by a steel cable of length 600 m and diameter 1.2 cm . As the gondola comes to t

he end of its run, it bumps into the terminal and sends a wave pulse along the cable. It is observed that it took 18 s for the pulse to return. Part A What is the speed of the pulse?
Physics
1 answer:
Gemiola [76]2 years ago
6 0

Answer:

v = 66.7 m/s

Explanation:

Given that,

The length of steel cable, L = 600 m

Diameter = 1.2 cm

It is observed that it took 18 s for the pulse to return.

The time taken to cover 600 m will be :

t = T/2

t = 9 s

Let v be the of the pulse. We know that,

v=\dfrac{L}{t}\\\\v=\dfrac{600}{9}\\\\v=66.7\ m/s

So, the speed of the pulse is equal to 66.7 m/s.

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iVinArrow [24]
energy associate with position or shape
5 0
3 years ago
R'=2 Ohm R"=1,5 Ohm R"'=2 Ohm​
Andrew [12]

Answer:

2.5 ohm

Explanation:

R' and R''' are parallel

So,

1/R1= 1/R' + 1/R'''

1/R1 = 1/2 + 1/2

1/R1 = 1

so,

R1= 1 ohm

Now R1 and R'' are in series

so,

R= R1 + R''

R= 1 + 1.5

R= 2.5 ohm

5 0
3 years ago
Write a complete scientific explanation to account for why the ball that was moving faster caused more flour to spread out.​
Harlamova29_29 [7]

Answer:

It is due to the large impulse is imparted on the flour.

Explanation:

A ball is moving faster.

When a ball is moving faster strikes to the flour, the change in momentum is large and thus the impulse imparted on the flour is large.

Impulse = change in momentum

So, as the flour experiences large impulse and large momentum so that the flour spreads out.

If the change in momentum is large so the flour spreads out is more.  

8 0
3 years ago
Tenses of<br>write=<br>read=<br><br>​
PilotLPTM [1.2K]

Answer:

present

Explanation:

read doesn't change but write is in present tense

7 0
3 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>

4 0
3 years ago
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