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Ostrovityanka [42]
3 years ago
8

Which of the following choices best represents a potential stressor?

Physics
1 answer:
Julli [10]3 years ago
6 0

an unexpected visitor as it begins making you feel uncomfortable but it's get calm


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Which of the following best describes longitudinal waves
Alex17521 [72]
Compression and rearfraction 
6 0
3 years ago
If an object looks blue, it reflects ____ waves. Question 4 options: blue red and yellow all but blue all
Masteriza [31]

Answer:

Blue

Explanation:

Because, That object absorbs all colours light waves but des not absorbs Blue. So The blue light is reflected because it is not absorbed

4 0
2 years ago
A que profundidad esta nadando una persona dentro de una alberca si la presión absoluta sobre ésta es de 156kPa?
lara31 [8.8K]

Answer:

La persona está nadando en la alberca a una profundida de 5.575 metros.

Explanation:

La presión absoluta (P_{tot}) experimentada por la persona es la suma de la presión atmosférica (P_{atm}) y la presión hidrostática de la columna de agua de la alberca (P_{h}), medidas en kilopascales. Es decir,

P_{tot} = P_{atm}+P_{h} (1)

P_{tot} = P_{atm} + \frac{\rho\cdot g \cdot z}{1000} (2)

Donde:

\rho - Densidad del fluido de la alberca, medida en kilogramos por metro cúbico.

g - Aceleración gravitacional, medida en metros por segundo al cuadrado.

z - Profundidad de la persona en la alberca, medida en metros.

Si sabemos que P_{atm} = 101.325\,kPa, \rho = 1000\,\frac{kg}{m^{3}}, g = 9.807\,\frac{m}{s^{2}} y P_{tot} = 156\,kPa, entonces la profundidad de la persona en la alberca es:

156 = 101.325 +\frac{(1000)\cdot (9.807)\cdot z}{1000}

54.675 = 9.807\cdot z

z = 5.575\,m

La persona está nadando en la alberca a una profundida de 5.575 metros.

5 0
3 years ago
PLEASE HELP WITH EXPLANATION! I WILL GIVE YOU BRAINLIEST!
DiKsa [7]

Answer:

divide the distance by the time and get average velocity in units of s. the detection to the left

6 0
3 years ago
A plane is flying east at 115 m/s. The wind accelerates it at 2.88 m/s^2 directly northwest. After 25.0s, what is the magnitude
kondaur [170]

Answer:

81.8 m/s

Explanation:

The initial velocity of the plane is:

v_0=115 m/s (toward east)

So, decomposing along the x- and y- directions:

v_{x0} = 115 m/s\\v_{y0} = 0

(we took east as positive x-direction and north as positive y-direction)

The acceleration is

a=2.88 m/s^2 (northwest, so the angle with the positive x-direction is 135 degrees)

Decomposing it along the two directions:

a_x = a cos 135^{\circ} = (2.88 m/s^2)(cos 135^{\circ})=-2.04 m/s^2\\a_y = a sin 135^{\circ} = (2.88 m/s^2)(sin 135^{\circ})=2.04 m/s^2

So the two components of the velocity after a time t = 25.0 s will be

v_x = v_{x0} + a_x t = 115 m/s + (-2.04 m/s^2)(25.0 s)=64 m/s\\v_y = v_{y0} + a_y t = 0 m/s + (2.04 m/s^2)(25.0 s)=51 m/s

So, the magnitude of the velocity of the plane will be

v=\sqrt[v_x^2+v_y^2}=\sqrt{(64 m/s)^2+(51 m/s)^2}=81.8 m/s

6 0
3 years ago
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