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2 years ago

There is a bell at the top of a tower that is 45m high. The bell mass is 190. Calculate its potential energy.

Physics

1 answer:

Umnica [9.8K]2 years ago

6 0

Answer:

83790 J.

Explanation:

Potential Energy: This can be defined as the energy possessed by a body, by vitue of it position in the gravitational field. The S.I unit of Potential Energy is Joules (J).

From the question,

P.E = mgh........................................... Equation 1

Where P.E = potential Energy of the bell, m = mass of the bell, h = height of the tower, g = Acceleration due to gravity.

Given: m = 190 kg, h = 45 m.

Constant: g = 9.8 m/s²

Substitute these values into equation 1

P.E = 190×45×9.8

P.E = 83790 J.

Hence the potential energy of the bell is 83790 J.

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At what point of thermometric scale does Kelvin scale reading coincide with Fahrenheit scale

amid [387]

Answer:

<em>At 574.59 Kelvin, the Fahrenheit temperature will be 574.59 °F.</em>

Explanation:

We first need to find a relation between the Kelvin scale and the Fahranheit scale. We'll use the Celsius scale to relate them.

The Kelvin and Celsius scales are related by the formula:

K = °C + 273.15

Solving for °C:

°C = K - 273.15

Besides, the Kelvin and Celsius scales are related by:

°C = 5 ⁄ 9(°F - 32)

Now we find a temperature, say X, where both scales coincide. Equating both formulas:

X - 273.15=5 ⁄ 9(X - 32)

Multiply by 9:

9X - 2,458.35 = 5X - 160

Simplifying:

4X = 2,458.35 - 160=2,298.35

Solving:

X =2,298.35 / 4 = 574.59

At 574.59 Kelvin, the Fahrenheit temperature will be 574.59 °F.

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3 years ago

Write a rule for the sequence. 3, -3, -9, -15. A. Start with 3 and add -6 repeatedly B. Start with -6 and add 3 repeatedly C. St

faust18 [17]

Substract two consecutive terms of the sequence to see if there is a common difference:

$\begin{gathered} (-3)-(3)=-3-3=-6 \\ (-9)-(-3)=-9+3=-6 \\ (-15)-(-9)=-15+9=-6 \end{gathered}$

As we can see, there is a common difference of -6.

Then, if a number of the sequence is given, the next one can be found by adding -6 (which is the same as subtracting 6).

Notice that the first term of the sequence is 3.

Then, the rule for the sequence is to start with 3 and add -6 repeatedly.

Therefore, the correct choice is option A) Start with 3 and add -6 repeatedly.

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1 year ago

Calcula el valor de la velocidad de las ondas sonoras en el agua sabiendo que su

dybincka [34]

La velocidad de las ondas sonoras es aproximadamente 1469,694 metros por segundo.
La longitud de onda de las ondas sonoras es 1,470 metros.

1) Inicialmente, debemos determinar la velocidad de las ondas sonoras a través del agua ( $v$ ), en metros por segundo:

$v = \sqrt{\frac{K}{\rho} }$ (1)

Donde:

$K$ - Módulo de compresibilidad, en newtons por metro cuadrado.
$\rho$ - Densidad del agua, en kilogramos por metro cúbico.

Si sabemos que $\rho = 1\times 10^{3}\,\frac{kg}{m^{3}}$ y $K = 2,16\times 10^{9}\,\frac{N}{m^{2}}$ , entonces la velocidad de las ondas sonoras es:

$v = \sqrt{\frac{2,16\times 10^{9}\,\frac{N}{m^{2}}}{1\times 10^{3}\,\frac{kg}{m^{3}} } }$

$v\approx 1469,694\,\frac{m}{s}$

La velocidad de las ondas sonoras es aproximadamente 1469,694 metros por segundo.

2) Luego, determinamos la longitud de onda ( $\lambda$ ), en metros, mediante la siguiente fórmula:

$\lambda = \frac{v}{f}$ (2)

Donde $f$ es la frecuencia de las ondas sonoras, en hertz.

Si sabemos que $v\approx 1469,694\,\frac{m}{s}$ y $f = 1000\,hz$ , entonces la longitud de onda de las ondas sonoras es:

$\lambda = \frac{1469,694\,\frac{m}{s} }{1000\,hz}$

$\lambda = 1,470\,m$

La longitud de onda de las ondas sonoras es 1,470 metros.

Para aprender más sobre las ondas sonoras, invitamos a ver esta pregunta verificada: brainly.com/question/1070238

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2 years ago

What is water that flows across eatlrtus surface

ira [324]

Water that flows across the surface is called a;

Runoff

That's when rain has saturated the ground to the point it cant hold anymore and it runs over the surface.

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3 years ago

Suppose you could fit 100 dimes, end to end, between your card with the pinhole and your dime-sized sunball. how many suns could

Naddika [18.5K]

Answer: 100 suns

Explanation:

We can solve this with the following relation:

$\frac{d}{x_{sunball-pinhole}}=\frac{D}{x_{sun-pinhole}}$

Where:

$d=17.91 mm =17.91(10)^{-3} m$ is the diameter of a dime

$D$ is the diameter of the Sun

$x_{sun-pinhole}=150,000,000 km=1.5(10)^{11} m$ is the distance between the Sun and the pinhole

$x_{sunball-pinhole}=100 d=1.791 m$ is the amount of dimes that fit in a distance between the sunball and the pinhole

Finding $D$ :

$D=\frac{d}{x_{sunball-pinhole}}x_{sun-pinhole}$

$D=\frac{17.91(10)^{-3} m}{1.791 m} 1.5(10)^{11} m$

$D=1.5(10)^{9} m$ This is roughly the diameter of the Sun

Now, the distance between the Earth and the Sun is one astronomical unit (1 AU), which is equal to:

$1 AU=149,597,870,700 m$

So, we have to divide this distance between $D$ in order to find how many suns could it fit in this distance:

$\frac{149,597,870,700 m}{1.5(10)^{9} m}=99.73 suns \approx 100 suns$

8 0

2 years ago