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3 years ago

Which of the following is a zone found in the open ocean

Physics

2 answers:

Svetlanka [38]3 years ago

5 0

What are your answer choices?

horrorfan [7]3 years ago

5 0

The answer to this question is D.

The neritic zone (option a) is a shallow area of ocean (about 200m in depth) which is essentially the area where the ocean interacts with the coast, although in very rare cases this can be further out to sea dependent on the oceanic plate.

The surface zone (option b) is, as the name suggests, where the ocean and the air meet and therefore not exclusive to the "open ocean" as specified in the question.

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The equation for gear ratio

Ymorist [56]

In a gear train with two gears, the gear ratio is defined as follows
$R= \frac{\omega _A}{\omega _B}$

where $\omega _A$ is the angular velocity of the input gear while $\omega _B$ is the angular velocity of the output gear.

This can be rewritten as a function of the number of teeth of the gears. In fact, the angular velocity of a gear is inversely proportional to the radius r of the gear:
$\omega = \frac{v}{r}$
But the radius is proportional to the number of teeth N of the gear. Therefore we can rewrite the gear ratio also as
$R= \frac{\omega _A}{\omega _B} = \frac{r_B}{r_A} = \frac{N_B}{N_A}$

4 0

3 years ago

A Tennis ball falls from a height 40m above the ground the ball rebounds

worty [1.4K]

If the ball is dropped with no initial velocity, then its velocity v at time t before it hits the ground is

v = -g t

where g = 9.80 m/s² is the magnitude of acceleration due to gravity.

Its height y is

y = 40 m - 1/2 g t²

The ball is dropped from a 40 m height, so that it takes

0 = 40 m - 1/2 g t²

==> t = √(80/g) s ≈ 2.86 s

for it to reach the ground, after which time it attains a velocity of

v = -g (√(80/g) s)

==> v = -√(80g) m/s ≈ -28.0 m/s

During the next bounce, the ball's speed is halved, so its height is given by

y = (14 m/s) t - 1/2 g t²

Solve y = 0 for t to see how long it's airborne during this bounce:

0 = (14 m/s) t - 1/2 g t²

0 = t (14 m/s - 1/2 g t)

==> t = 28/g s ≈ 2.86 s

So the ball completes 2 bounces within approximately 5.72 s, which means that after 5 s the ball has a height of

y = (14 m/s) (5 s - 2.86 s) - 1/2 g (5 s - 2.86 s)²

==> (i) y ≈ 7.5 m

(ii) The ball will technically keep bouncing forever, since the speed of the ball is only getting halved each time it bounces. But y will converge to 0 as t gets arbitrarily larger. We can't realistically answer this question without being given some threshold for deciding when the ball is perfectly still.

During the first bounce, the ball starts with velocity 14 m/s, so the second bounce begins with 7 m/s, and the third with 3.5 m/s. The ball's height during this bounce is

y = (3.5 m/s) t - 1/2 g t²

Solve y = 0 for t :

0 = (3.5 m/s) t - 1/2 g t²

0 = t (3.5 m/s - 1/2 g t)

==> (iii) t = 7/g m/s ≈ 0.714 s

As we showed earlier, the ball is in the air for 2.86 s before hitting the ground for the first time, then in the air for another 2.86 s (total 5.72 s) before bouncing a second time. At the point, the ball starts with an initial velocity of 7 m/s, so its velocity at time t after 5.72 s (but before reaching the ground again) would be

v = 7 m/s - g t

At 6 s, the ball has velocity

(iv) v = 7 m/s - g (6 s - 5.72 s) ≈ 4.26 m/s

4 0

3 years ago

A 100-m-long wire carrying a current of 4.0 A will be accompanied by a magnetic field of what strength at a distance of 0.050 m

solong [7]

Answer:

1.6 x 10^-5 T

Explanation:

i = 4 A

r = 0.05 m

The magnetic field due to long wire at a distance r is given by

$B = \frac{\mu _{0}}{4\pi }\times \frac{2i}{r}$

B = 10^-7 x 2 x 4 / 0.05

B = 1.6 x 10^-5 T

3 0

3 years ago

If a baseball pitch leaves the pitcher's hand horizontally at a velocity of 150 km/h by what percent will the pull of gravity ch

Slav-nsk [51]

0.52% First, let's convert that speed into m/s. 150 km/h * 1000 m/km / 3600 s/h = 41.667 m/s Now let's see how much time gravity has to work on the ball. Divide the distance by the speed. 18 m / 41.667 m/s = 0.431996544 s Now multiply that time by the gravitational acceleration to see what the vertical component to the ball's speed that gravity adds. 0.431996544 s * 9.8 m/s^2 = 4.233566131 m/s Use the pythagorean theorem to get the new velocity of the ball. sqrt(41.667^2 + 4.234^2) = 41.882 m/s Finally, let's see what the difference is (41.882 - 41.667)/41.667 = 0.005159959 = 0.5159959% Rounding to 2 figures, gives 0.52%

8 0

3 years ago

PY85

aliya0001 [1]

Answer:

460 g

Explanation:

Heat lost by the warm water = heat gained by the cold water

-mCΔT = mCΔT

-m (4.184 J/g/K) (37°C − 85°C) = (1000 g) (4.184 J/g/K) (37°C − 15°C)

-m (37°C − 85°C) = (1000 g) (37°C − 15°C)

-m (-48°C) = (1000 g) (22°C)

m = 458 g

Rounded to two significant figures, you need a mass of 460 g of water.

3 0

3 years ago