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2 years ago

_ drives the energy and motion of waves?

Physics

2 answers:

Papessa [141]2 years ago

7 0

Answer:

Wind-driven waves, or surface waves,

Explanation:

____ [38]2 years ago

5 0

Answer:

Kinetic Energy(?)

<u><em>I hope this helped at all. sorry if it didn't</em></u>

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Which of the following is a correct formula of speed

satela [25.4K]

Answer:s= d/t

Explanation:

7 0

3 years ago

The 1.18-kg uniform slender bar rotates freely about a horizontal axis through O. The system is released from rest when it is in

sattari [20]

Answer:

k = 11,564 N / m, w = 6.06 rad / s

Explanation:

In this exercise we have a horizontal bar and a vertical spring not stretched, the bar is released, which due to the force of gravity begins to descend, in the position of Tea = 46º it is in equilibrium;

let's apply the equilibrium condition at this point

Axis y

W_{y} - Fr = 0

Fr = k y

let's use trigonometry for the weight, we assume that the angle is measured with respect to the horizontal

sin 46 = $W_{y}$ / W

W_{y} = W sin 46

we substitute

mg sin 46 = k y

k = mg / y sin 46

If the length of the bar is L

sin 46 = y / L

y = L sin46

we substitute

k = mg / L sin 46 sin 46

k = mg / L

for an explicit calculation the length of the bar must be known, for example L = 1 m

k = 1.18 9.8 / 1

k = 11,564 N / m

With this value we look for the angular velocity for the point tea = 30º

let's use the conservation of mechanical energy

starting point, higher

Em₀ = U = mgy

end point. Point at 30º

$Em_{f}$ = K -Ke = ½ I w² - ½ k y²

em₀ = Em_{f}

mgy = ½ I w² - ½ k y²

w = √ (mgy + ½ ky²) 2 / I

the height by 30º

sin 30 = y / L

y = L sin 30

y = 0.5 m

the moment of inertia of a bar that rotates at one end is

I = ⅓ mL 2

I = ½ 1.18 12

I = 0.3933 kg m²

let's calculate

w = Ra (1.18 9.8 0.5 + ½ 11,564 0.5 2) 2 / 0.3933)

w = 6.06 rad / s

7 0

2 years ago

A cube has a drag coefficient of 0.8. What would be the terminal velocity of a sugar cube 1 cm on a side in air ( = 1.2 kg/mº)?

anzhelika [568]

0.495 m/s

Explanation

the formula for the terminal velocity is given by:

$\begin{gathered} v=\sqrt[]{\frac{2mg}{\sigma AC}} \\ \text{where} \\ \end{gathered}$

m is the mass

g is 9.81 m/s²

ρ is density

A is area

C is the drag coefficient

then

Step 1

Let's find the mass

$\begin{gathered} \sigma=\frac{m}{v} \\ m=\sigma\cdot v \\ \text{mass}=(2\cdot10^3\frac{\operatorname{kg}}{m^3})\cdot(0.01m)^3 \\ \text{mass}=(2\cdot10^3\frac{\operatorname{kg}}{m^3})\cdot(1\cdot10^{-6}) \\ \text{mass}=2\cdot10^{-3}\operatorname{kg} \\ \text{mass}=0.002\text{ kg } \\ \text{Area}=(0.01\text{ m}\cdot0.01m)=0.0001m^2 \end{gathered}$

now, replace

$\begin{gathered} v=\sqrt[]{\frac{2mg}{\sigma AC}} \\ v=\sqrt[]{\frac{2(0.002kg)(9.81\text{ }\frac{m}{s^2})}{(2\cdot10^3\frac{\operatorname{kg}}{m^3})(0.0001m^2)0.8}} \\ v=\sqrt[]{\frac{0.03924\frac{\operatorname{kg}m}{s^2}}{0.16\frac{\operatorname{kg}}{m^{}}}} \\ v=\sqrt[]{0.2452\frac{m^2}{s^2}} \\ v=0.495\text{ m/s} \end{gathered}$

hence, the answer is 0.495 m/s