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3 years ago

6. Why does sound travel faster through a steel bar than through a swimming pool filled with water?

1 answer:

3 0

I understand that sound travels faster in water then in air. Water is a liquid, and air is gas.

Water still has the ability to roll the molecules over each other (so water can flow), it has some flexibility.

But I do not understand how a solid that is inflexible can make sound waves travel faster then in a flexible liquid.

In fact, sound waves travel over 17 times faster through steel than through air.

Sound waves travel over four times faster in water than it would in air.

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Why is friction useful between your feet and the ground?

mixer [17]

Answer:

It allows you to walk faster.

Explanation:

It is the same force that allows you to accelerate forward when you run. Your planted foot can grip the ground and push backward, which causes the ground to push forward on your foot. We call this grip type of friction, where the surfaces are prevented from slipping across each other, a static frictional force.

3 0

2 years ago

Read 2 more answers

A rocket is fired straight up. It contains two stages (Stage 1 and Stage 2) of solid rocket fuel that are designed to burn for 1

Archy [21]

Answer:

a) y= 3.5 10³ m, b) t = 64 s

Explanation:

a) For this exercise we use the vertical launch kinematics equation

Stage 1

y₁ = y₀ + v₀ t + ½ a t²

y₁ = 0 + 0 + ½ a₁ t²

Let's calculate

y₁ = ½ 16 10²

y₁ = 800 m

At the end of this stage it has a speed

v₁ = vo + a₁ t₁

v₁ = 0 + 16 10

v₁ = 160 m / s

Stage 2

y₂ = y₁ + v₁ (t-t₀) + ½ a₂ (t-t₀)²

y₂ = 800 + 150 5 + ½ 11 5²

y₂ = 1092.5 m

Speed is

v₂ = v₁ + a₂ t

v₂ = 160 + 11 5

v₂ = 215 m / s

The rocket continues to follow until the speed reaches zero (v₃ = 0)

v₃² = v₂² - 2 g y₃

0 = v₂² - 2g y₃

y₃ = v₂² / 2g

y₃ = 215²/2 9.8

y₃ = 2358.4 m

The total height is

y = y₃ + y₂

y = 2358.4 + 1092.5

y = 3450.9 m

y= 3.5 10³ m

b) Flight time is the time to go up plus the time to go down

Let's look for the time of stage 3

v₃ = v₂ - g t₃

v₃ = 0

t₃ = v₂ / g

t₃ = 215 / 9.8

t₃ = 21.94 s

The time to climb is

$t_{s}$ = t₁ + t₂ + t₃

t_{s} = 10+ 5+ 21.94

t_{s} = 36.94 s

The time to descend from the maximum height is

y = v₀ t - ½ g t²

When it starts to slow down it's zero

y = - ½ g t_{b}²

t_{b} = √-2y / g

t_{b} = √(- 2 (-3450.9) /9.8)

t_{b} = 26.54 s

Flight time is the rise time plus the descent date

t = t_{s} + t_{b}

t = 36.94 + 26.54

t =63.84 s

t = 64 s

3 0

3 years ago

Help wit these questions someone.

romanna [79]

In series circuit, Req = R₁ + R₂ + R₃ + ···

In parallel circuit, $\frac{1}{Req} = \frac{1}{R1} + \frac{1}{R2} + \frac{1}{R3} +...$

total resistance in the upper branch = R₂ + R₃ = R₂ + 2

$\frac{1}{Req} = \frac{1}{R2+R3} + \frac{1}{R1}$

$\frac{1}{4} = \frac{1}{R2+2} +\frac{1}{6}$

R₂ + 2 = 12

R₂ = 10Ω

$\frac{1}{Req} = \frac{1}{R2+R3} + \frac{1}{R1}$

$\frac{1}{Req} = \frac{1}{2+1} + \frac{1}{4}$

Req = 1.7Ω

7 0

3 years ago

Help meh in this question plzzz <br>

iragen [17]

The Moment of Inertia of the Disc is represented by $I = \frac{15}{32}\cdot M\cdot R^{2}$ . (Correct answer: A)

Let suppose that the Disk is a Rigid Body whose mass is uniformly distributed. The Moment of Inertia of the element is equal to the Moment of Inertia of the entire Disk minus the Moment of Inertia of the Hole, that is to say:

$I = I_{D} - I_{H}$ (1)

Where:

$I_{D}$ - Moment of inertia of the Disk.

$I_{H}$ - Moment of inertia of the Hole.

Then, this formula is expanded as follows:

$I = \frac{1}{2}\cdot M\cdot R^{2} - \frac{1}{2}\cdot m\cdot \left(\frac{1}{2}\cdot R^{2} \right)$ (1b)

Dimensionally speaking, Mass is directly proportional to the square of the Radius, then we derive the following expression for the Mass removed by the Hole ( $m$ ):

$\frac{m}{M} = \frac{R^{2}}{4\cdot R^{2}}$

$m = \frac{1}{2}\cdot M$

And the resulting equation is:

$I = \frac{1}{2}\cdot M\cdot R^{2} -\frac{1}{2}\cdot \left(\frac{1}{4}\cdot M \right) \cdot \left(\frac{1}{4}\cdot R^{2} \right)$

$I = \frac{1}{2} \cdot M\cdot R^{2} - \frac{1}{32}\cdot M\cdot R^{2}$

$I = \frac{15}{32}\cdot M\cdot R^{2}$

The moment of inertia of the Disc is represented by $I = \frac{15}{32}\cdot M\cdot R^{2}$ . (Correct answer: A)

Please see this question related to Moments of Inertia: brainly.com/question/15246709

5 0

3 years ago

Where are bar magnets the strongest?

Alecsey [184]

Answer:

The magnetic field is strongest at the center and weakest between the two poles just outside the bar magnet. The magnetic field lines are densest at the center and least dense between the two poles just outside the bar magnet.

Explanation:

3 0

2 years ago

Read 2 more answers