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rosijanka [135]
3 years ago
7

AYUDAAA PORFAVOR

Physics
1 answer:
Sholpan [36]3 years ago
5 0

Queremos crear un diagrama general para calcular el área de un triangulo.

Este será algo como:

  1. Definir variables
  2. Pedirle al usuario que introduzca los valores deseados (de las variables).
  3. Leer los valores deseados y asignarlo a la variable correspondiente.
  4. Realizar la operación para calcular el área.
  5. Mostrar en pantalla el resultado.

Como naturalmente habra algunas variaciones segun el programa que utilicemos, lo voy a escribir de forma bastante general.

Primero definamos nuestras variables:

Por ejemple, en fortran usariamos algo como:

real:: B, H, A

Donde B será la variable que usaremos para la base, H para la altura, y A para el área.

Luego tenemos que escribir en pantalla algo que le diga al usario que debe introducir la base y el area.

Luego el programa debe ser capaz de leer ese input.

con algo de la forma:

B = read*input 1

H = read*input 2

Una vez tenemos definidas las variables, simplemente calculamos el área del triangulo:

A = H*B/2

Finalmente la podemos mostrar en pantalla con algo como:

print(A).

Lo que nos mostraría el valor del área.

Concluyendo, el diagrama en general sería:

  1. Definir variables
  2. Pedirle al usuario que introduzca los valores deseados (de las variables).
  3. Leer los valores deseados y asignarlo a la variable correspondiente.
  4. Realizar la operación para calcular el área.
  5. Mostrar en pantalla el resultado.

Si quieres aprender más, puedes leer:

brainly.com/question/21949109

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Answer:

393.6m/s

Explanation:

Given parameters:

Acceleration  = 8.5m/s²

Distance  = 300m

Final velocity  = 400m/s

Unknown:

Initial velocity  = ?

Solution:

To solve this problem, we use the expression below;

             v² = u²  + 2as

v is the final velocity

u is the initial velocity

a is the acceleration

s is the distance

      So;

               v²  - 2as = u²

        u²   = v²   - 2as

        u²  = 400²   - (2 x 8.5 x 300)  

         u   = 393.6m/s

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A
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Maurinko [17]
Here is the first question

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4.5 cm3 of water is boiled at atmospheric pressure to become 4048.3 cm3 of steam, also at atmospheric pressure. Calculate the wo
Mrrafil [7]

1. 408.4 J

The work done by a gas is given by:

W=p\Delta V

where

p is the gas pressure

\Delta V is the change in volume of the gas

In this problem,

p=1.01\cdot 10^5 Pa (atmospheric pressure)

\Delta V=4048.3 cm^3 - 4.5 cm^3 =4043.8 cm^3 = 4043.8 \cdot 10^{-6}m^3 is the change in volume

So, the work done is

W=(1.01\cdot 10^5 Pa)(4043.8 \cdot 10^{-6}m^3)=408.4 J

2. 10170 J

The amount of heat added to the water to completely boil it is equal to the latent heat of vaporization:

Q = m \lambda_v

where

m is the mass of the water

\lambda_v = 2.26\cdot 10^6 J/kg is the specific latent heat of vaporization

The initial volume of water is

V_i = 4.5 cm^3 = 4.5\cdot 10^{-6}m^3

and the water density is

\rho = 1000 kg/m^3

So the water mass is

m=\rho V_i = (1000 kg/m^3)(4.5\cdot 10^{-6}m^3)=4.5\cdot 10^{-3}kg

So, the amount of heat added to the water is

Q=(4.5\cdot 10^{-3}kg)(2.26\cdot 10^6 J/kg)=10,170 J

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4 years ago
When using a calorimeter, the initial temperature of a metal is 70.4C. The initial temperature of the water is 23.6C. At the end
Sunny_sXe [5.5K]

1) 29.8 C

At the beginning, the metal is at higher temperature (70.4 C) while the water is at lower temperature (23.6 C). When they are put in contact, the metal transfers heat to the water, until they reach thermal equilibrium: at thermal equilibrium the two objects (the metal and the water have same temperature). Therefore, since the temperature of the water at thermal equilibrium is 29.8 C, the final temperature of the metal must be the same (29.8 C).

2) 6.2 C

The temperature change of the water is given by the difference between its final temperature and its initial temperature:

\Delta T = T_f - T_i

where

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Substituting into the formula,

\Delta T=29.8 C-23.6 C=6.2 C

And the positive sign means that the temperature of the water has increased.

3) -40.6 C

The temperature change of the metal is given by the difference between its final temperature and its initial temperature:

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Substituting into the formula,

\Delta T=29.8 C-70.4 C=-40.6 C

And the negative sign means the temperature of the metal has decreased.

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