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masya89
1 month ago
11

A company designed and sells an ultrasonic​ receiver, which detects sounds unable to be heard by the human ear. The receiver can

detect mechanical and electrical sounds such as leaking​ gases, air,​ corona, and motor friction noises. It can also be used to hear​ bats, insects, and even beading water. The receiver is 1818 in. deep. The focus is 6 in.6 in. from the vertex. Find the diameter of the outside edge of the receiver.

Physics
1 answer:
Softa [3K]1 month ago
6 0
The diameter at the outer edge of the receiver is derived from the given parameters within the schematic free body diagram. Here, point A indicates the vertex where the receiver is situated, S is designated as the focus, BP denotes the radius of the outer edge, and BC is twice the radius (the diameter). With AS as 4 inches and AP as 18 inches, and by utilizing the corresponding parabolic equation, the resulting diameter BC is calculated as 2r.
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1 month ago
What is the gauge pressure of the water right at the point p, where the needle meets the wider chamber of the syringe? neglect t
Yuliya22 [3333]

Details that are not provided: the problem figure is included.

We can address the exercise by applying Poiseuille's law. This law indicates that for a fluid flowing in a laminar manner within a confined pipe,

\Delta P = \frac{8 \mu L Q}{\pi r^4}

where:

\Delta P represents the pressure difference across the two ends

\mu denotes the viscosity of the fluid

L signifies the length of the pipe

Q=Av indicates the volumetric flow rate, where A=\pi r^2 is the cross-sectional area of the tube and v refers to the fluid's velocity

r stands for the pipe's radius.

This law can be utilized for the needle, allowing us to compute the pressure difference between point P and the needle's end. In this scenario, we have:

\mu=0.001 Pa/s is the dynamic viscosity of water at 20^{\circ}

L=4.0 cm=0.04 m

Q=Av=\pi r^2 v= \pi (1 \cdot 10^{-3}m)^2 \cdot 10 m/s =3.14 \cdot 10^{-5} m^3/s

and r=1 mm=0.001 m

Substituting these values into the formula yields:

\Delta P = 3200 Pa

This pressure difference specifies the value between point P and the needle's termination. As the end of the needle is under atmospheric pressure, the gauge pressure at point P, relative to atmospheric pressure, is exactly 3200 Pa.

8 0
3 months ago
If a person weighs 882 N on the surface of the earth, at what altitude above the earth’s surface must they be for their weight t
Sav [3153]

Para calcular el peso utilizamos la fórmula:
F=m*g=882N

Cuando tenemos dos cuerpos, empleamos la fórmula de la gravedad general:
F=G* \frac{ M_{p}*M_{E} }{ r^{2} }
Donde:
G = constante de gravedad = 6.67* 10^{-11} m^{3} kg^{-1} s^{-2}
Mp = masa de la persona = 882 / 9.81= 89.9kg
ME = masa de la Tierra = 5.97* 10^{24} kg
r = distancia entre la Tierra y la persona

A partir de estas dos fórmulas, deducimos que los lados izquierdos son equivalentes, por lo tanto, los lados derechos también deben serlo.
m*g=G* \frac{ M_{p}*M_{E} }{ r^{2} }

Resolviendo esta ecuación para r:
r= \sqrt{G* \frac{ M_{p}*M_{E}}{M_{p}*g}
r=6371116m = 6371.116km

Esta es la distancia desde el centro de la Tierra. El radio de la Tierra es 6370km y la altura sobre la superficie es 6371.116 - 6370 = 1.116km o 1116m.

3 0
2 months ago
Calculate the minimum average power output necessary for a person to run up a 12.0 m long hillside, which is inclined at 25.0° a
Ostrovityanka [3204]

Answer:

Power output, P = 924.15 watts

Explanation:

We have the following parameters:

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Elapsed time, t = 3 s

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h=l\ sin\theta

h=12\times \ sin(25)

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P=\dfrac{E}{t}

P=\dfrac{mgh}{t}

P=\dfrac{55.8\times 9.8\times 5.07}{3}

P = 924.15 watts

This indicates that a minimum average power output of 924.15 watts is essential for an individual to ascend this elevation. Thus, this is the answer sought.

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kicyunya [3294]
The correct answer is -10.
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1 month ago
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