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Alekssandra [29.7K]
3 years ago
7

You are holding one end of a horizontal stretched string. Flicking your wrist will send a pulse down the string. Which actions w

ill make the pulse travel faster
Physics
1 answer:
kumpel [21]3 years ago
5 0

Answer:

Use a lighter string of the same length, under the same tension.

Stretch the string tighter to increase the tension

Explanation:

The wave speed depends on propertices of the medium, not on how you generate the wave. For a string

Increasing the tension or decreasing the linear density (lighter string) will increase the wave speed.

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A wire is wrapped around a piece of iron, and then electricity is run through the wire. What happens to the iron?
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7 0
3 years ago
Dave is moving 3 m/s when he crashes his bike into a wall, which stops him in 0.6 seconds. If Dave and his bike have a mass of 9
Arisa [49]
Force = change of momentum / time taken
Force = (90x3)/0.6
7 0
3 years ago
In your own words, discuss how tides are monitored. Describe the old and new methods of monitoring tides.
harkovskaia [24]
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6 0
4 years ago
An Atwood's machine consists of two different masses, both hanging vertically and connected by an ideal string which passes over
Tasya [4]

Answer:

V₁ = √ (gy / 3)

Explanation:

For this exercise we will use the concepts of mechanical energy, for which we define energy n the initial point and the point of average height and / 2

Starting point

    Em₀ = U₁ + U₂

    Em₀ = m₁ g y₁ + m₂ g y₂

Let's place the reference system at the point where the mass m1 is

     y₁ = 0

    y₂ = y

    Em₀ = m₂ g y = 2 m₁ g y

End point, at height yf = y / 2

    E_{mf} = K₁ + U₁ + K₂ + U₂

    E_{mf} = ½ m₁ v₁² + ½ m₂ v₂² + m₁ g y_{f} + m₂ g y_{f}

Since the masses are joined by a rope, they must have the same speed

     E_{mf} = ½ (m₁ + m₂) v₁² + (m₁ + m₂) g y_{f}

   E_{mf}= ½ (m₁ + 2m₁) v₁² + (m₁ + 2m₁) g y_{f}

How energy is conserved

   Em₀ =  E_{mf}

   2 m₁ g y = ½ (m₁ + 2m₁) v₁² + (m₁ + 2m₁) g y_{f}

   2 m₁ g y = ½ (3m₁) v₁² + (3m₁) g y / 2

   3/2 v₁² = 2 g y -3/2 g y

   3/2 v₁² = ½ g y

   V₁ = √ (gy / 3)

5 0
3 years ago
Alice's friends Bob and Charlie are having a race to a distant star 10 light years away. Alice is the race official who stays on
hodyreva [135]

Solution :

The distance between the starting point and the end point, L_0 = 10 light years

But due to the relativistic motion of Bob and Charlie, the distance will be reduced following the Lorentz contraction. The contracted length will be different since they are moving with different speeds.

For Bob,

Speed of Bob's rocket with respect to Alice, L_b = 0.7  \ c

So the distance appeared to Bob due to the length contraction,

$L_b=L_0\sqrt{1-\frac{V_b^2}{c^2}$

$L_b=10\times \sqrt{1-0.49} \ Ly$

    $=7.1 \ Ly$

Therefore, the time required to finish the race by Bob is

$t_b = \frac{L_b}{V_b}$

  $=\frac{7.1 \ c}{0.7 \ c}$

  = 10.143 year

For Charlie,

Speed of Charlie's rocket with respect to Alice, L_c = 0.866 \ c

So the distance appeared to Charlie due to the length contraction,

$L_b=L_0\sqrt{1-\frac{V_c^2}{c^2}$

$L_b=10\times \sqrt{1-0.75} \ Ly$

    $=5 \ Ly$

The time required to finish the race by Charlie is

$t_b = \frac{L_c}{V_c}$

  $=\frac{5 \ c}{0.866 \ c}$

  = 5.77 year

5 0
3 years ago
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