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2 years ago

Frequency is measured in hertz. Which option best describes hertz? (1 point)

Physics

2 answers:

Rasek [7]2 years ago

6 0

Answer:

A) the number of vibrations produced per minute

Explanation:

Hope this helps.

Nadya [2.5K]2 years ago

3 0

Answer:

☑ the number of vibrations produced per minute

Explanation:

» The formula for frequency is as below;

${ \tt{frequency = \frac{n}{t} }} \\$

n is number of vibrations
t is the period of each vibration

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Reto: Aníbal desea construir una carpa de base hexagonal de 1,40 m de altura. El vértice de la base es de 0,80m y la longitud de

Rudiy27

Answer:

1.- para cubrir la superficie lateral 4.32 metros²

2.- Area de la base 2.15 metros²

3.- Volumen 1 m³

4.- Area total 6.47 metros²

Explanation:

El área lateral sera calcular el area de seis triangulos iguales cuya base es

0.80 de base x 6 lados = 4.80 metros perimetro de la base

4.80 perimetro de base x 1.80 arista lateral / 2 = 4.32m2

Area de la base:

Perímetro x Apotema / 2

Siendo la Apotema la altura de los triangulos que componen un hexagono calculada utilizando el teorema de pitágoras:

$\sqrt{0.8^2 + (0.8/2)^2} = Apotema$

Apotema = 0.894427191

Area: 4.80 x 0.894427191 / 2 = 2.146625258

sumando el area de la base mas el area lateral se obtiene el area total

2.15 + 4.32 = 6.47 metros

Volumen de la pirámide:

Area de la Base x Altura / 3

2.15 x 1.40 / 3 = 1.00333 m3

3 0

3 years ago

Assuming that Bernoulli's equation applies, compute the volume of water ΔV that flows across the exit of the pipe in 1.00 s . In

OLEGan [10]

Answer:

discharge rate (Q) = 0.2005 m^{3} / s

Explanation:

if you read the question you would see that some requirements are missing, by using search engines, you can get the complete question as stated below:

Water flows steadily from an open tank as shown in the figure. (Figure 1) The elevation of point 1 is 10.0m , and the elevation of points 2 and 3 is 2.00 m . The cross-sectional area at point 2 is 4.80x10-2m ; at point 3, where the water is discharged, it is 1.60x10-2m. The cross-sectional area of the tank is very large compared with the cross-sectional area of the pipe. Part A Assuming that Bernoulli's equation applies, compute the volume of water DeltaV that flows across the exit of the pipe in 1.00 s . In other words, find the discharge rate \Delta V/Delta t. Express your answer numerically in cubic meters per second.

solution:

time = 1 s

elevation of point 1 (z1) = 10 m

elevation of point 2 (z2) = 2 m

elevation of point 3 (z3) = 2 m

cross section area of point 2 = 4.8 x 10^{2} m

cross section area of point 3 = 1.6 x 10^{2} m

acceleration due to gravity (g) = 9.8 m/s^{2}

find the discharge rate at point 3 which is the exit pipe.

discharge rate (Q) = A3 x V3

where A3 is the cross sectional area at point 3 and V3 is the velocity of the fluid and can be gotten by applying Bernoulli's equation below

$\frac{P1}{ρg} + \frac{V1^{2} }{2g} + Z1 = \frac{P3}{ρg} + \frac{V3^{2} }{2g} + Z3$

pressure at point 1 (P1) is the same as pressure at point 3 (P3), and at point 1, the velocity (V1) = 0. therefore the equation now becomes

$\frac{P1}{ρg} + Z1 = \frac{P1}{ρg} + \frac{V3^{2} }{2g} + Z3$

$Z1 = \frac{V3^{2} }{2g} + Z3$

V3 = $\sqrt{2g(Z1-Z3)}$

V3 = $\sqrt{2 x 9.8 x (10 - 3)}$

V3 = 12.53 m/s

discharge rate (Q) = A3 x V3 = 1.6 x 10^{-2} x 12.53

discharge rate (Q) = 0.2005 m^{3} / s

8 0

3 years ago

Physics 10

Dominik [7]

Answer:

Option D (N/A-m)(m)(m/s)

Explanation:

(N/A-m)(m)(m/s) is the required dimensional analysis for calculation of emi

3 0

2 years ago

Engineers and science fiction writers have proposed designing space stations in the shape of a rotating wheel or ring, which wou

Gennadij [26K]

Answer:

a. The station is rotating at $1.496 \frac{rev}{min}$

b. the rotation needed is $2.8502 \frac{rev}{min}$

Explanation:

We know that the centripetal acceleration is

$a_{c}= \omega ^2 r$

where $\omega$ is the rotational speed and r is the radius. As the centripetal acceleration is feel like an centrifugal acceleration in the rotating frame of reference (be careful, as the rotating frame of reference is <u>NOT INERTIAL,</u> the centrifugal force is a fictitious force, the real force is the centripetal).

The rotational speed is :

$2.7 \frac{m}{s^2} = \omega ^2 * 110 \ m$

$\omega ^2 = \frac{2.7 \frac{m}{s^2}} {110 \ m}$

$\omega = \sqrt{ 0.02454 \frac{rad^2}{s^2} }$

$\omega = 0.1567 \frac{rad}{s}$

Knowing that there are $2\pi \ rad$ in a revolution and 60 seconds in a minute.

$\omega = 0.1567 \frac{rad}{s} \frac{1 \ rev}{2\pi \ rad} \frac{60 \ s}{1 \ min}$

$\omega = 1.496 \frac{rev}{min}$

The rotational speed needed is :

$9.8 \frac{m}{s^2} = \omega ^2 * 110 \ m$

$\omega ^2 = \frac{9.8 \frac{m}{s^2}} {110 \ m}$

$\omega = \sqrt{ 0.08909 \frac{rad^2}{s^2} }$

$\omega = 0.2985 \frac{rad}{s}$

Knowing that there are $2\pi \ rad$ in a revolution and 60 seconds in a minute.

$\omega = 0.2985 \frac{rev}{min} \frac{1 \ rev}{2\pi \ rad} \frac{60 \ s}{1 \ min}$

$\omega = 2.8502 \frac{rev}{min}$

3 0

3 years ago

Read 2 more answers

A runner runs 300 m at an average speed of 3.0 m/s. She then runs another 300m at an average

Kaylis [27]

Answer:

B. 4 m/s

Explanation:

v=d/t

Running for 300 m at 3 m/s takes 100 seconds and running at 300 m at 6 m/s takes 50 seconds. 100 s + 50 s = 150 s (total time). Total distance is 600 m, so 600 m/ 150 s = 4 m/s.

3 0

3 years ago