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DanielleElmas [232]
3 years ago
13

In beachville, the last high tide occurred at 1:00 p.m. the next high tide will happen around _____. 1:25

Physics
2 answers:
Vladimir [108]3 years ago
8 0
At Around 1:25 a.m. it will happen
xxTIMURxx [149]3 years ago
4 0
The answer is 1:25am
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you nose out another runner to win 100.000 m dash. if your total time for the race was 13.800 s and you aced out the other runne
sammy [17]

Your average speed was

(100 m) / (13.8 s) = 7.25 m/s .

If you finished 0.001s ahead of him, then at your average speed, that corresponds to

(7.25 m/s) x (0.001 s) = 0.00725 m

That's 7.25 millimeters ... about 0.28 of an inch !

NOTE:. I think this is only valid if your speed was a constant ~7.25 m/s all the way.

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Is the stomach just below the waist?
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3 years ago
Fariza wears a red hat in her school play. On stage, she is lit by a spotlight shining only green light. When our eyes receive n
FromTheMoon [43]

Answer:It's how the color mixes together . Red and green both makes a dark color

Explanation:

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3 years ago
A plate of uniform areal density is bounded by the four curves: where and are in meters. Point has coordinates and . What is the
Natali5045456 [20]

The question is incomplete. The complete question is :

A plate of uniform areal density $\rho = 2 \ kg/m^2$ is bounded by the four curves:

$y = -x^2+4x-5m$

$y = x^2+4x+6m$

$x=1 \ m$

$x=2 \ m$

where x and y are in meters. Point $P$ has coordinates $P_x=1 \ m$ and $P_y=-2 \ m$. What is the moment of inertia $I_P$ of the plate about the point $P$ ?

Solution :

Given :

$y = -x^2+4x-5$

$y = x^2+4x+6$

$x=1 $

$x=2 $

and $\rho = 2 \ kg/m^2$ , $P_x=1 \ $ , $P_y=-2 \ $.

So,

$dI = dmr^2$

$dI = \rho \ dA  \ r^2$  ,           $r=\sqrt{(x-1)^2+(y+2)^2}$

$dI = (\rho)((x-1)^2+(y+2)^2)dx \ dy$

$I= 2 \int_1^2 \int_{-x^2+4x-5}^{x^2+4x+6}((x-1)^2+(y+2)^2) dy \ dx$

$I= 2 \int_1^2 \int_{-x^2+4x-5}^{x^2+4x+6}(x-1)^2+(y+2)^2 \  dy \ dx$

$I=2 \int_1^2 \left( \left[ (x-1)^2y+\frac{(y+2)^3}{3}\right]_{-x^2+4x-5}^{x^2+4x+6}\right) \ dx$

$I=2 \int_1^2 (x-1)^2 (2x^2+11)+\frac{1}{3}\left((x^2+4x+6+2)^3-(-x^2+4x-5+2)^3 \ dx$

$I=\frac{32027}{21} \times 2$

  $= 3050.19 \ kg \ m^2$

So the moment of inertia is  $3050.19 \ kg \ m^2$.

4 0
3 years ago
Which of the following creates an adhesive force that prevents separation of the parietal and visceral pleurae during ventilatio
Diano4ka-milaya [45]

Answer:

Negative intrapleural pressure is the correct answer

Explanation:

Intrapleural pressure is more subatmospheric in the uppermost part of the thorax than in the lowermost parts in the standing horse.

Air moves from a region of higher pressure to one of lower pressure.  Therefore, for air to be moved into or out of the lungs, a pressure difference between the atmosphere and the alveoli must be established. If there is no pressure difference, no airflow will occur.

Under normal circumstances, inspiration is accomplished by causing alveolar pressure to fall below atmospheric pressure. When the mechanics of breathing are being discussed, atmospheric pressure is conventionally referred to as 0 cm H2O, so lowering alveolar pressure below atmospheric pressure is known as negative-pressure breathing.

3 0
3 years ago
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