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swat32
3 years ago
14

Imagine that you were given a ship equipped with a device that could measure how deep the water is. You are asked to use this sh

ip to find a subduction zone. What would you look for as you sailed around the world's oceans?
Physics
1 answer:
alexdok [17]3 years ago
3 0

Answer:

Deep ocean trenches

Explanation:

Subduction zones are Earth's lithosphere (the crust and as well as the upper non-convecting part of the upper mantle) gravitational collapsing sites. Along convergent plate boundaries, subduction zones exist where one sheet of oceanic lithosphere merges with any other layer. So, in order to find  depth of water in the sea, we can find these subduction zones. The best places are a deep ocean- trench like the Mariana Trench in the pacific ocean.

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My answer :

Explanation:

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3 years ago
A hanging weight, with a mass of m1 = 0.365 kg, is attached by a string to a block with mass m2 = 0.825 kg as shown in the figur
morpeh [17]

The speed of the block after it has moved the given distance away from the initial position is 1.1 m/s.

<h3>Angular Speed of the pulley </h3>

The angular speed of the pulley after the block m1 fall through a distance, d, is obatined from conservation of energy and it is given as;

K.E = P.E

\frac{1}{2} mv^2 + \frac{1}{2} I\omega^2 = mgh\\\\\frac{1}{2} m_2v_0^2 + \frac{1}{2} \omega^2(m_1R^2_2 + m_2R_2^2) + \frac{1}{2} \omega^2( \frac{1}{2} MR_1^2 + \frac{1}{2} MR_2^2) = m_1gd- \mu_km_2gd\\\\\frac{1}{2} m_2v_0^2 + \frac{1}{2} \omega^2[R_2^2(m_1 + m_2)+ \frac{1}{2} M(R_1^2 + R_2^2)] = gd(m_1 - \mu_k m_2)\\\\

\frac{1}{2} m_2v_0 + \frac{1}{4} \omega^2[2R_2^2(m_1 + m_2) + M(R^2_1 + R^2_2)] = gd(m_1 - \mu_k m_2)\\\\2m_2v_0 + \omega^2 [2R_2^2(m_1 + m_2) + M(R^2_1 + R^2_2)] = 4gd(m_1 - \mu_k m_2)\\\\\omega^2 [2R_2^2(m_1 + m_2) + M(R^2_1 + R^2_2)] =  4gd(m_1 - \mu_k m_2) - 2m_2v_0^2\\\\\omega^2 = \frac{ 4gd(m_1 - \mu_k m_2) - 2m_2v_0^2}{2R_2^2(m_1 + m_2) + M(R^2_1 + R^2_2)} \\\\\omega = \sqrt{\frac{ 4gd(m_1 - \mu_k m_2) - 2m_2v_0^2}{2R_2^2(m_1 + m_2) + M(R^2_1 + R^2_2)}} \\\\

Substitute the given parameters and solve for the angular speed;

\omega = \sqrt{\frac{ 4(9.8)(0.7)(0.365 \ - \ 0.25\times 0.825) - 2(0.825)(0.82)^2}{2(0.03)^2(0.365 \ + \ 0.825)\  \ +\  \ 0.35(0.02^2\  + \ 0.03^2)}} \\\\\omega = \sqrt{\frac{3.25}{0.00214\ + \ 0.000455 } } \\\\\omega = 35.39 \ rad/s

<h3>Linear speed of the block</h3>

The linear speed of the block after travelling 0.7 m;

v = ωR₂

v = 35.39 x 0.03

v = 1.1 m/s

Thus, the speed of the block after it has moved the given distance away from the initial position is 1.1 m/s.

Learn more about conservation of energy here: brainly.com/question/24772394

5 0
2 years ago
What is the displacement of the car between t=1s and t=4s
tensa zangetsu [6.8K]

Answer:

Option C. 30 m

Explanation:

From the graph given in the question above,

At t = 1 s,

The displacement of the car is 10 m

At t = 4 s

The displacement of the car is 40 m

Thus, we can simply calculate the displacement of the car between t = 1 and t = 4 by calculating the difference in the displacement at the various time. This is illustrated below:

Displacement at t = 1 s (d1) = 10 m

Displacement at t= 4 s (d2) = 40

Displacement between t = 1 and t = 4 (ΔD) =?

ΔD = d2 – d1

ΔD = 40 – 10

ΔD = 30 m.

Therefore, the displacement of the car between t = 1 and t = 4 is 30 m.

4 0
3 years ago
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