it depends on the atom . but in some cases it may explode
Queremos crear un diagrama general para calcular el área de un triangulo.
Este será algo como:
- Definir variables
- Pedirle al usuario que introduzca los valores deseados (de las variables).
- Leer los valores deseados y asignarlo a la variable correspondiente.
- Realizar la operación para calcular el área.
- 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:
- Definir variables
- Pedirle al usuario que introduzca los valores deseados (de las variables).
- Leer los valores deseados y asignarlo a la variable correspondiente.
- Realizar la operación para calcular el área.
- Mostrar en pantalla el resultado.
Si quieres aprender más, puedes leer:
brainly.com/question/21949109
Answer:
Option B. 5 nC
Explanation:
From the question given above, the following data were obtained:
Capicitance (C) = 100 pF
Potential difference (V) = 50 V
Quantity of charge (Q) =?
Next, we shall convert 100 pF to Farad (F). This can be obtained as follow:
1 pF = 1×10¯¹² F
Therefore,
100 pF = 100 pF × 1×10¯¹² F / 1 pF
100 pF = 1×10¯¹⁰ F
Next, we shall determine the quantity of charge. This can be obtained as follow:
Capicitance (C) = 1×10¯¹⁰ F
Potential difference (V) = 50 V
Quantity of charge (Q) =?
Q = CV
Q = 1×10¯¹⁰ × 50
Q = 5×10¯⁹ C
Finally, we shall convert 5×10¯⁹ C to nano coulomb (nC). This can be obtained as follow:
1 C = 1×10⁹ nC
Therefore,
5×10¯⁹ C = 5×10¯⁹ C × 1×10⁹ nC / 1 C
5×10¯⁹ C = 5 nC
Thus, the quantity of charge is 5 nC
Answer:
<em><u>mark</u></em><em><u> </u></em><em><u>me</u></em><em><u> </u></em><em><u>brianliest</u></em><em><u> </u></em><em><u>plz</u></em>
Explanation:
- Law of inertia, also called Newton's first law, postulate in physics that, if a body is at rest or moving at a constant speed in a straight line, it will remain at rest or keep moving in a straight line at constant speed unless it is acted upon by a force.
- Law of Inertia states that a body in a state of rest or uniform motion remains in the same state until and unless an external force acts on it.
- A body continues to be in its state of rest or in uniform motion along a straight line unless an external force is applied on it. This law is also called law of inertia.
Answer:
Fractional error = 0.17
Percent error = 17%
F = 112 ± 19 N
Explanation:
Plug in the values to find the force:
F = (3.5 kg) (20 m/s)² / (12.5 m) = 112 N
Find the fractional error:
ΔF/F = Δm/m + 2Δv/v + Δr/r
ΔF/F = 0.1/3.5 + 2(1/20) + 0.5/12.5
ΔF/F = 0.17
Multiply by 100% to find the percent error:
ΔF/F × 100% = 17%
Solve for the absolute error:
ΔF = 0.17 × 112 N = 19 N
Therefore, the force is:
F = 112 ± 19 N
