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Alla [95]
2 years ago
6

Of the following approximate conversions of celsius into fahrenheit, which is most accurate? a. 24°c almost-equals 75.9°f b. 2°c

almost-equals 35.8°f c. 34°c almost-equals 92.7°f d. 13°c almost-equals 55.0°f
Physics
1 answer:
Softa [21]2 years ago
4 0

The most accurate among the options is 13°c almost-equals 55.0°f, when you round down 55.4, you get 55.0

Hence, option D) is the most accurate.

<h3>What is Scale of Temperature?</h3>

Scale of temperature is simply a method used in calibrating the physical quantity of temperature in metrology.

Conversion of Celsius into Fahrenheit, Formula is expressed as;

Degree = (0 × 9/5) + 32 = Fahrenheit

First we check each option;

A) 24°c almost-equals 75.9°f

24°C = (24 × 9/5) + 32 = 75.2°F

B) 2°C almost-equals 35.8°f

2°C = (2 × 9/5) + 32 = 35.6°F

C) 34°C almost-equals 92.7°f

34°C = (34 × 9/5) + 32 = 93.2°F

D) 13°c almost-equals 55.0°f

13°C = (13 × 9/5) + 32 = 55.4°F

The most accurate among the options is 13°c almost-equals 55.0°f, when you round down 55.4, you get 55.0

Hence, option D) is the most accurate.

Learn more about temperature scales here: brainly.com/question/88395

#SPJ4

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3 years ago
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a brick is suspended above the ground at a height of 6.6 m. it has a mass of 5.3 kg. what is the potential energy of the brick
Svetradugi [14.3K]
The formula for potential energy is
E(p) = mgh

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Therefore energy = (5.3)(9.8)(6.6)
= 342.8 J

How did I get 9.8?
9.8 is the constant for gravity
8 0
3 years ago
consider the photoelectric effect experiment. in one experiment yellow light shines on a piece of potassium metal. a current is
kompoz [17]

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I = Q/t, as the charge increase , the current will also increase.

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A simple harmonic oscillator has amplitude 0.43 m and period 3.9 sec. What is the maximum acceleration?
rjkz [21]

Answer:

Maximum acceleration in the simple harmonic motion will be 0.854rad/sec^2              

Explanation:

We have given amplitude of simple harmonic motion is A = 0.43 m

Time period of the oscillation is T = 3.9 sec

We have to find the maximum acceleration

For this we have to find the angular frequency

Angular frequency will be equal to \omega =\frac{2\pi }{T}=\frac{2\times 3.14}{3.9}=1.61rad/sec

Maximum acceleration is given by a_{max}=\omega ^2A=1.61^2\times 0.43=0.854rad/sec^2

So maximum acceleration in the simple harmonic motion will be 0.854rad/sec^2

8 0
3 years ago
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Una prenda de 320gramos de ropa gira en el interior de una lavadora si dicha lavadora tiene 40 cm y gira con una frecuencia de 4
Nitella [24]

Answer:

Período del tambor: T = 0.25\,s, fuerza sobre la prenda: F \approx 80.852\,N, velocidad lineal del tambor: v \approx 10.053\,\frac{m}{s}, velocidad angular del tambor: \omega \approx 25.133\,\frac{rad}{s}.

Explanation:

La expresión tiene un error por omisión, su forma correcta queda descrita a continuación:

<em>"Una prenda de 320 gramos de ropa gira en el interior de una lavadora si dicha lavadora tiene un radio de 40 centímetros y gira con una frecuencia de 4 hertz. Halle </em><em>a)</em><em> el período, </em><em>b) </em><em>la velocidad angular, </em><em>c) </em><em>la fuerza con la que gira la prenda y </em><em>d) </em><em>la velocidad lineal de la lavadora."</em>

El tambor gira a velocidad angular constante (\omega), en radianes por segundo, lo cual significa que la prenda experimenta una aceleración centrífuga (a), en metros por segundo al cuadrado. En primer lugar, calculamos el período de rotación del tambor (T), en segundos:

T = \frac{1}{f} (1)

Donde f es la frecuencia, en hertz.

(f = 4\,hz)

T = \frac{1}{4\,hz}

T = 0.25\,s

Ahora determinamos la fuerza aplicada sobre la prenda (F), en newtons:

F = m\cdot a (2)

F = \frac{4\pi^{2}\cdot m \cdot r}{T^{2}} (2b)

Donde:

m - Masa de la prenda, en kilogramos.

r - Radio interior del tambor, en metros.

(m = 0.32\,kg, r = 0.4\,m, T = 0.25\,s)

F = \frac{4\pi^{2}\cdot (0.32\,kg)\cdot (0.4\,m)}{(0.25\,s)^{2}}

F \approx 80.852\,N

La velocidad lineal de la lavadora es:

v = \frac{2\pi\cdot r}{T} (3)

(r = 0.4\,m, T = 0.25\,s)

v = \frac{2\pi\cdot (0.4\,m)}{0.25\,s}

v \approx 10.053\,\frac{m}{s}

Y la velocidad angular del tambor de la lavadora:

\omega = \frac{2\pi}{T}

(T = 0.25\,s)

\omega = \frac{2\pi}{0.25\,s}

\omega \approx 25.133\,\frac{rad}{s}

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