PLEASE HELP I HAVE BEEN STUCK ON THIS FOR A WHILE.

AlexFokin [52]

Answer:

B ( I think so?)

Explanation:

Ocean currents are driven by wind, water density differences, and tides. Oceanic currents describe the movement of water from one location to another.

Coastal currents are intricately tied to winds, waves, and land formations. Winds that blow along the shoreline—longshore winds—affect waves and, therefore, currents.

3 0

4 years ago

Water drops from the nozzle of a shower onto the floor 81 inches below. The drops fall at regular intervals of time, the first d

Alexxx [7]

Answer:

0.91437 m

0.22859 m

Explanation:

g = Acceleration due to gravity = 9.81 m/s² = a

$s=81\ inches=81\times 0.0254=2.0574\ m$

$s=ut+\frac{1}{2}at^2\\\Rightarrow 2.0574=0t+\frac{1}{2}\times 9.81\times t^2\\\Rightarrow t=\sqrt{\frac{2.0574\times 2}{9.81}}\\\Rightarrow t=0.64764\ s$

When the time intervals are equal, if four drops are falling then we have 3 time intervals.

So, the time interval is

$t'=\dfrac{t}{3}\\\Rightarrow t'=\dfrac{0.64764}{3}\\\Rightarrow t'=0.21588\ s$

For second drop time is given by

$t''=2t'\\\Rightarrow t''=2\times 0.21588\\\Rightarrow t''=0.43176\ s$

Distance from second drop

$s=ut+\dfrac{1}{2}at^2\\\Rightarrow y''=ut''+\dfrac{1}{2}at''^2\\\Rightarrow s=0\times t+\dfrac{1}{2}\times 9.81\times 0.43176^2\\\Rightarrow s=0.91437\ m$

Distance from second drop is 0.91437 m

Distance from third drop

$s=ut+\dfrac{1}{2}at^2\\\Rightarrow y''=ut'+\dfrac{1}{2}at'^2\\\Rightarrow s=0\times t+\dfrac{1}{2}\times 9.81\times 0.21588^2\\\Rightarrow s=0.22859\ m$

Distance from third drop is 0.22859 m

6 0

3 years ago

I can’t figure this out!!! Answer what you can , please.

tigry1 [53]

a/b. The ball has velocity vector at time $t$

$\vec v=(v_x,v_y)=(v_0\cos63^\circ,v_0\sin63^\circ-gt)$

where $v_0=16\dfrac{\rm m}{\rm s}$ is the ball's initial speed and $g=9.8\dfrac{\rm m}{\mathrm s^2}$ .

c. At its highest point, the ball has 0 vertical speed. This occurs when

$v_0\sin63^\circ-gt=0\implies t=1.5\,\mathrm s$

d. Recall that

${v_y}^2-{v_{0y}}^2=-2g\Delta y$

so that at its highest point,

$0^2-(v_0\sin63^\circ)^2=-2g\Delta y\implies\Delta y=10\,\mathrm m$

e. This is just twice the time it takes for the ball to reach its maximum height, $t=2.9\,\mathrm s$ .

f. The ball's horizontal position after time $t$ is

$v_0\cos63^\circ\,t$

so that after the time found in part (f), the ball has traveled

$v_0\cos63^\circ(2.9\,\mathrm s)=11\,\mathrm m$

4 0

4 years ago

When you approach a light source that in turn is moving towards you, your speed relative to the emitted light waves _____.?

Natalka [10]

This is a trick question:

The Doppler effect states that as you move closer to the source, the frequency of light(or sound/waves in general) increases, but technically the speed of light is always the same speed, even if you are moving at the speed of light.

Thus, the answer would be something along the lines of <u>don't change</u>.

7 0

4 years ago

Read 2 more answers

Are predators or their prey more likely to be successful?

mihalych1998 [28]

Predator because they are the one’s preying on the animals, the chances of the prey getting away is slim.

3 0

3 years ago

Read 2 more answers