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3 years ago

Why are ocean currents important?

1 answer:

4 0

For the considerably longer periods– decades to millennia – which are relevant for climate change, the slightly larger heat capacity of the deep ocean is important. Ocean currents and mixing by winds and waves can transport and redistribute heat to deeper ocean layers.

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What net force would be needed to accelerate the box at 2.3 m/s^2

Akimi4 [234]

The pertinent equation here is F=ma. You haven't shared the mass of the box, so I will use M to represent that mass.

Then F = M(2.3 m/s^2) (answer)

6 0

3 years ago

Read 2 more answers

If the speed of a car is 20m/s .How long does it take to cover a distance of 1km?

3241004551 [841]

Answer: 50 seconds

Explanation:

1km=1000m

Distance/speed=time

1000/20=50

50 seconds

7 0

3 years ago

Read 2 more answers

A steel ball rolls with a constant velocity on a tabletop 0.950 m high it rolls off and hit the ground 0.352 m from the edge of

sp2606 [1]

Answer:

0.799 m/s if air resistance is negligible.

Explanation:

For how long is the ball in the air?

Acceleration is constant. The change in the ball's height $\Delta h$ depends on the square of the time:

$\displaystyle \Delta h = \frac{1}{2} \;g\cdot t^{2} + v_0\cdot t$ ,

where

$\Delta h$ is the change in the ball's height.
$g$ is the acceleration due to gravity.
$t$ is the time for which the ball is in the air.
$v_0$ is the initial vertical velocity of the ball.

The height of the ball decreases, so this value should be the opposite of the height of the table relative to the ground. $\Delta h = -0.950\;\text{m}$ .
Gravity pulls objects toward the earth, so $g$ is also negative. $g \approx -9.81\;\text{m}\cdot\text{s}^{-2}$ near the surface of the earth.
Assume that the table is flat. The vertical velocity of the ball will be zero until it falls off the edge. As a result, $v_0 = 0$ .

Solve for $t$ .

$\displaystyle \Delta h = \frac{1}{2} \;g\cdot t^{2} + v_0\cdot t$ ;

$\displaystyle -0.950 = \frac{1}{2} \times (-9.81) \cdot t^{2}$ ;

$\displaystyle t^{2} =\frac{-0.950}{1/2 \times (-9.81)}$ ;

$t \approx 0.440315\;\text{s}$ .

What's the initial horizontal velocity of the ball?

Horizontal displacement of the ball: $\Delta x = 0.352\;\text{m}$ ;
Time taken: $\Delta t = 0.440315\;\text{s}$

Assume that air resistance is negligible. Only gravity is acting on the ball when it falls from the tabletop. The horizontal velocity of the ball will not change while the ball is in the air. In other words, the ball will move away from the table at the same speed at which it rolls towards the edge.

$\begin{aligned}\text{Rolling Velocity}&=\text{Horizontal Velocity} \\&= \text{Average Horizontal Velocity}\\ &=\frac{\Delta x}{\Delta t}=\frac{0.352\;\text{m}}{0.440315\;\text{s}}=0.0799\;\text{m}\cdot\text{s}^{-1}\end{aligned}$ .

Both values from the question come with 3 significant figures. Keep more significant figures than that during the calculation and round the final result to the same number of significant figures.

3 0

3 years ago

A sprinter with a mass of 80 kg accelerates from 0 m/s to 9 m/s in 3 s. What is the runner's acceleration?

sertanlavr [38]

Acceleration = (change in speed) / (time for the change) = 9/3 = 3 m/s² .

His mass makes no difference.

5 0

3 years ago

The Hubble Space Telescope in orbit above the Earth has a 2.4 m circular aperture. The telescope has equipment for detecting ult

strojnjashka [21]

Answer:

Option d

The minimum angular separation between two objects that the Hubble Space Telescope can resolve is $4.8x10^{-8}rad$ .

Explanation:

The resulting image in a telescope that will be gotten from an object is a diffraction pattern instead of a perfect point (point spread function (PSF)).

That diffraction pattern is gotten because the light encounters different obstacles on its path inside the telescope (interacts with the walls and edges of the instrument).

The diffraction pattern is composed by a central disk, called Airy disk, and diffraction rings.

The angular resolution is defined as the minimal separation at which two sources can be resolved one for another, or in other words, when the distance between the two diffraction pattern maxima is greater than the radius of the Airy disk.

The angular resolution can be determined in analytical way by means of the Rayleigh criterion.

$\theta = 1.22\frac{\lambda}{D}$ (1)