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2 years ago

7

Subsurface ocean currents continually circulate from the warm waters near the equator to the colder waters in other parts of the

world. What is the main cause of these currents?

1 answer:

soldi70 [24.7K]2 years ago

8 0

<span>the main cause of these currents is the </span>differences in density of ocean water.

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A battleship launches a shell horizontally at 100 m/s from the ship’s deck that’s 50 m above the water. The shell is intended to

Annette [7]

Answer:

The shell will land 10.18m away from the buoy.

Explanation:

In order to solve this problem, we must first do a sketch of what the problem looks like (see attached picture).

Now, there are two cases, one with the tailwind and another with the tailwind. In both cases the shell would have the same vertical initial velocity and acceleration, therefore the shell would hit the water in the same amount of time. So we need to first find the time it takes the shell to hit the water.

In order to do so we can use the following equation:

$y_{f}=y_{0}+V_{0}t+\frac{1}{2}at^{2}$

now, we know that the final height and the initial velocity are to be zero, so we can simplify the equation like this:

$0=y_{0}+\frac{1}{2}at^{2}$

and solve for t:

$t=\sqrt{\frac{-2y_0}{a}}$

now we can substitute the values:

$t=\sqrt{\frac{-2(50m)}{-9.81m/t^2}}$

t=3.19s

Since it takes 3.19s for the shell to hit the water, that's the amount of time it spends flying horizontally.

So we can consider the shell to move at a constant speed if there was no tailwind, so we can find the distance from the ship to point A to be:

$x_{A}=V_{x}t$

$x_{A}=(100m/s)(3.19)$

$x_{A}=319m$

We can now find the distance between the ship to point B, which is the point the ball falls due to the tailwind. Since the movement will be accelerated in this scenario, we can find the distance by using the following formula:

$x_{f}=V_{x0}t+\frac{1}{2}a_{x}t^{2}$

So we can substitute the given values:

$x_{f}=(100m/s)(3.19s)+\frac{1}{2}(2m/s^{2})(3.19s)^{2}$

Which yields:

$x_{f}=329.18m$

so now we can use the A and B points to find by how far the shell missed the buoy:

Distance=329.18m-319m=10.18m

So the shell missed the buoy by 10.18m.

8 0

2 years ago

Differences between Pressure and upthrust

Angelina_Jolie [31]

Answer:

Pressure is equal to the ratio of thrust to the area in contact. Upthrust is a force exerted by the fluids on an object placed in the fluid . Upthrust acts in upward direction.

4 0

2 years ago

An electrical conductor is an element with __________ electrons in its outer orbit.

Setler [38]

An electric conductor is an element with free electrons in its outer orbit

5 0

2 years ago

Which of the following statements is FALSE?

Levart [38]

A thrust fault is a reverse fault with an extremely high dip (close to 90°). This is the false statement.

Answer: Option D

<u>Explanation:</u>

Faults are the fracture or fracture zone occurring on the rocks. These fractures can travel through the rocks leading to massive destruction. So, depending upon the direction of their travel, the faults can be classified as normal, reverse and strike slip fault. Also, the angle of dip along the fault is one of the important criteria for determining the type of faults.

There is dip-slip fault which has its movement along the vertical fault plane while the strike slip fault will be in horizontal direction. Similarly, an oblique fault will be acting in both vertical and the horizontal direction. So, the fourth statement related to thrust fault is false as in reverse fault or thrust fault the dip will be shallow and not high.

5 0

2 years ago

A torsional pendulum consists of a disk of mass 450 g and radius 3.5 cm, hanging from a wire. If the disk is given an initial an

Montano1993 [528]

To solve this problem we will use the kinematic equations of angular motion, starting from the definition of angular velocity in terms of frequency, to verify the angular displacement and its respective derivative, let's start:

$\omega = 2\pi f$

$\omega = 2\pi (2.5)$

$\omega = 5\pi rad/s$

The angular displacement is given as the form:

$\theta (t) = \theta_0 cos(\omega t)$

In the equlibrium we have to $t=0, \theta(t) = \theta_0$ and in the given position we have to

$\theta(t) = \theta_0 cos(5\pi t)$

Derived the expression we will have the equivalent to angular velocity

$\frac{d\theta}{dt} = 2.7rad/s$

Replacing,

$\theta_0(sin(5\pi t))5\pi = 2.7$

Finally

$\theta_0 = \frac{2.7}{5\pi}rad = 9.848\°$

Therefore the maximum angular displacement is 9.848°

6 0

2 years ago