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3 years ago

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Imagine that while you and a passenger are in a deep-diving submersible in the North Pacific near Alaska’s Aleutian Islands, you

encounter a long, narrow depression on the ocean floor. Your passenger asks whether you think it is a submarine canyon, a rift valley, or a deep-ocean trench. How would you respond? Explain your response.

1 answer:

mamaluj [8]3 years ago

6 0

Answer:

I would say its a deep ocean trench

Explanation:

This is because deep ocean trenches are found at the deepest part of the ocean and also at Pacific ocean margins or Rim where subduction usually occurs and Aleutian islands are part of the Pacific Rim

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Kobotan [32]

a) $0.4 m/s^2$

We can find the acceleration of the box by using the suvat equation:

$s=ut+\frac{1}{2}at^2$

where

s is the distance travelled

u is the initial velocity

t is the time

a is the acceleration

For the box in the problem,

s = 16 m

t = 9 s

u = 0 (it starts from rest)

Solving for a, we find the acceleration:

$a=\frac{2s}{t^2}=\frac{2(16)}{9^2}=0.4 m/s^2$

C) Tension force in the chain: 446 N

In order to find the normal force, we have to write the equation of the forces along the vertical direction.

We have 3 forces acting along this direction on the box:

- The normal force, N, upward

- The force of gravity, mg, downward (where m= mass of the box and g = acceleration of gravity)

- The vertical component of the tension of in the chain, $T sin \theta$ , upward

So the equation of the forces along the vertical direction is

$N+Tsin \theta - mg = 0$ (1)

Along the horizontal direction, instead, we have the following equation:

$T cos \theta - \mu N = ma$ (2)

where

$T cos \theta$ is the horizontal component of the tension in the chain

$\mu N$ is the frictional force

a is the acceleration

From (1) we write

$N=mg-T sin \theta$

And substituting into (2),

$T cos \theta - \mu (mg-T sin \theta) = ma\\T cos \theta - \mu mg + \mu T sin \theta = ma\\T = \frac{ma+\mu mg}{cos \theta + \mu sin \theta}$

And substituting:

m = 100 kg

$\theta=25^{\circ}$

$\mu=0.46$

$a=0.4 m/s^2$

$g=9.8 m/s^2$

We find the tension in the chain:

$T = \frac{(100)(0.4)+(0.46)(100)(9.8)}{cos 25 + 0.46 sin 25}=446 N$

B) Normal force: 792 N

We can now find the normal force by using again equation (1):

$N+Tsin \theta - mg = 0$

And substituting:

T = 446 N

m = 100 kg

$\theta=25^{\circ}$

$g=9.8 m/s^2$

We find:

$N=mg-T sin \theta=(100)(9.8)-(446)(sin 25)=792 N$

3 0

3 years ago

A vertical tube one meter long is open at the top. It is filled with 50 cm of water. If the velocity of sound is 344 m/s, what w

emmainna [20.7K]

Answer:

The fundamental resonance frequency is 172 Hz.

Explanation:

Given;

velocity of sound, v = 344 m/s

total length of tube, Lt = 1 m = 100 cm

height of water, hw = 50 cm

length of air column, L = Lt - hw = 100 cm - 50 cm = 50 cm

For a tube open at the top (closed pipe), the fundamental wavelength is given as;

Node to anti-node (N ---- A) : L = λ / 4

λ = 4L

λ = 4 (50 cm)

λ = 200 cm = 2 m

The fundamental resonance frequency is given by;

$f_0 = \frac{v}{\lambda}\\\\f_0 = \frac{344}{2}\\\\f_0 = 172 \ Hz\\\\$

Therefore, the fundamental resonance frequency is 172 Hz.

6 0

2 years ago

A car is traveling around a horizontal circular track with radius r = 210 m at a constant speed v = 23 m/s as shown. The angle θ

Fantom [35]

Centripetal acceleration of car is given by formula

$a_c = \frac{v^2}{R}$

now plug in the values in this

$a_c = \frac{23^2}{210}$

$a_c = 2.52 m/s^2$

Part b)

At position A we have

x component of acceleration is given as

$a_x = -a_c cos23$

$a_x = -2.32 m/s^2$

Part c)

At position A we have

y component of acceleration is given as

$a_y = -a_c sin23$

$a_y = -0.98 m/s^2$

Part d)

At position B we have

x component of acceleration is given as

$a_x = -a_c cos53$

$a_x = -1.52 m/s^2$

Part e)

At position B we have

Y component of acceleration is given as

$a_y = a_c sin53$

$a_y = 2.01 m/s^2$

6 0

3 years ago

Please help me out. Convert 28 pints to cups. Using one step conversion.

KengaRu [80]

Answer:

56

Explanation:

6 0

3 years ago

A spring has a natural length of 0.5 m and was stretched by 0.02 m. if the spring had a resultant energy of 0.5 j what is the sp

Anna71 [15]

$\textbf{2500 }\dfrac{\textbf{kg}}{\textbf{s}^{\textbf{2}}}$

Explanation:

Natural length of a spring is $0.5\text{ }m$ . The spring is streched by $0.02\text{ }m$ . The resultant energy of the spring is $0.5\text{ }J$ .

The potential energy of an ideal spring with spring constant $k$ and elongation $x$ is given by $\dfrac{1}{2}kx^{2}$ .

So, in the current problem, the natural length of the spring is not required to find the spring constant $k$ .

$\text{Potential Energy in the spring = }\dfrac{1}{2}kx^{2}\\0.5\text{ }J\text{ }=\text{ }\dfrac{1}{2}k(0.02\text{ }m)^{2}\\k\times0.0004\text{ }m^{2}\text{ }=\text{ }1\text{ }J\text{ }=\text{ }1\text{ }kg\frac{m^{2}}{s^{2}}\\k\text{ }=\text{ }\dfrac{1\text{ }kg\dfrac{m^{2}}{s^{2}}}{0.0004\text{ }m^{2}}\text{ }=\text{ }2500\text{ }\frac{kg}{s^{2}}$

∴ The spring constant of the spring = $2500\text{ }\frac{kg}{s^{2}}$

4 0

3 years ago