Question

calculate the missing variables using the necessary and rules.

1. what does newtons 2nd law of motion state?

force (f)
f = m x a
newtons (n)
a water - filled balloon with an overall mass of 1kg undergoes an acceleration of 2m/s². how much force is being applied to the balloon?
mass of the rocket 0.5 kg
a theme park ride carriage with a mass of 1000kg needs to be accelerated at 0.5 m/s² along a track. what force is required to move the carriage?
a tractor carrying hay bales has a total mass of 9,100 kg. what force is required to reach an acceleration of 4 m/s²?
acceleration
mass (m)
m = f/a
kilograms (kg)
a skyrocket is launched with a force of 10n and accelerates at 20 m/s². what is the mass of this skyrocket?
acceleration of the coal - wagon is m/s
a 7.5 n force is applied to a football, generating an initial acceleration of 12 m/s². calculate the mass of the football.
mass of the football: 0.625
a cannonball accelerates at 1000 m/s² from an applied force of 5000n what is the mass of the cannonball?
mass of the cannon - ball 5kg
acceleration (a)
a = f/m
meters per second (m/s²)
a loaded coal wagon with a total mass of 20,000 kg is pushed by a force of 300,000 n. what is the wagons acceleration?
force required for the theme - park ride carriage 500n
a spring balance is used to launch a 0.05 kg foam ball with a force of 5n. what acceleration is produced?
a 1000 kg satellite in space needs a course correction. to achieve this, one of its rocket motors is fired to apply 100n of force as thrust. what will the acceleration be due to this thrust? 0.1 m/s²

Explanation:

Explicación:

Paso 1: Fórmula de la segunda ley de Newton

La segunda ley de Newton es $F = m\times a$, donde $F$ es la fuerza en newtons (N), $m$ es la masa en kilogramos (kg) y $a$ es la aceleración en metros por segundo al cuadrado ($m/s^{2}$).

Caso 1: Globos llenos de agua

Dado $m = 1$ kg y $a=2$ $m/s^{2}$, usamos $F = m\times a$.
$F=1\times2 = 2$ N

Caso 2: Cohete

Dado $F = 10$ N y $a = 20$ $m/s^{2}$, usamos $m=\frac{F}{a}$.
$m=\frac{10}{20}=0.5$ kg

Caso 3: Vagón de carbón

Dado $F = 300000$ N y $m = 20000$ kg, usamos $a=\frac{F}{m}$.
$a=\frac{300000}{20000}=15$ $m/s^{2}$

Caso 4: Carruaje de parque de diversiones

Dado $m = 1000$ kg y $a = 0.5$ $m/s^{2}$, usamos $F = m\times a$.
$F=1000\times0.5 = 500$ N

Caso 5: Fútbol

Dado $F = 7.5$ N y $a = 12$ $m/s^{2}$, usamos $m=\frac{F}{a}$.
$m=\frac{7.5}{12}=0.625$ kg

Caso 6: Pelota de espuma

Dado $F = 5$ N y $m = 0.05$ kg, usamos $a=\frac{F}{m}$.
$a=\frac{5}{0.05}=100$ $m/s^{2}$

Caso 7: Tractor

Dado $m = 9100$ kg y $a = 4$ $m/s^{2}$, usamos $F = m\times a$.
$F=9100\times4=36400$ N

Caso 8: Bala de cañón

Dado $F = 50000$ N y $a = 1000$ $m/s^{2}$, usamos $m=\frac{F}{a}$.
$m=\frac{50000}{1000}=50$ kg

Caso 9: Satélite

Dado $F = 100$ N y $m = 1000$ kg, usamos $a=\frac{F}{m}$.
$a=\frac{100}{1000}=0.1$ $m/s^{2}$

Respuesta:

• Caso 1: $F = 2$ N
• Caso 2: $m = 0.5$ kg
• Caso 3: $a = 15$ $m/s^{2}$
• Caso 4: $F = 500$ N
• Caso 5: $m = 0.625$ kg
• Caso 6: $a = 100$ $m/s^{2}$
• Caso 7: $F = 36400$ N
• Caso 8: $m = 50$ kg
• Caso 9: $a = 0.1$ $m/s^{2}$

Answer:

Explicación:

Paso 1: Fórmula de la segunda ley de Newton

La segunda ley de Newton es $F = m\times a$, donde $F$ es la fuerza en newtons (N), $m$ es la masa en kilogramos (kg) y $a$ es la aceleración en metros por segundo al cuadrado ($m/s^{2}$).

Caso 1: Globos llenos de agua

Dado $m = 1$ kg y $a=2$ $m/s^{2}$, usamos $F = m\times a$.
$F=1\times2 = 2$ N

Caso 2: Cohete

Dado $F = 10$ N y $a = 20$ $m/s^{2}$, usamos $m=\frac{F}{a}$.
$m=\frac{10}{20}=0.5$ kg

Caso 3: Vagón de carbón

Dado $F = 300000$ N y $m = 20000$ kg, usamos $a=\frac{F}{m}$.
$a=\frac{300000}{20000}=15$ $m/s^{2}$

Caso 4: Carruaje de parque de diversiones

Dado $m = 1000$ kg y $a = 0.5$ $m/s^{2}$, usamos $F = m\times a$.
$F=1000\times0.5 = 500$ N

Caso 5: Fútbol

Dado $F = 7.5$ N y $a = 12$ $m/s^{2}$, usamos $m=\frac{F}{a}$.
$m=\frac{7.5}{12}=0.625$ kg

Caso 6: Pelota de espuma

Dado $F = 5$ N y $m = 0.05$ kg, usamos $a=\frac{F}{m}$.
$a=\frac{5}{0.05}=100$ $m/s^{2}$

Caso 7: Tractor

Dado $m = 9100$ kg y $a = 4$ $m/s^{2}$, usamos $F = m\times a$.
$F=9100\times4=36400$ N

Caso 8: Bala de cañón

Dado $F = 50000$ N y $a = 1000$ $m/s^{2}$, usamos $m=\frac{F}{a}$.
$m=\frac{50000}{1000}=50$ kg

Caso 9: Satélite

Dado $F = 100$ N y $m = 1000$ kg, usamos $a=\frac{F}{m}$.
$a=\frac{100}{1000}=0.1$ $m/s^{2}$

Respuesta:

• Caso 1: $F = 2$ N
• Caso 2: $m = 0.5$ kg
• Caso 3: $a = 15$ $m/s^{2}$
• Caso 4: $F = 500$ N
• Caso 5: $m = 0.625$ kg
• Caso 6: $a = 100$ $m/s^{2}$
• Caso 7: $F = 36400$ N
• Caso 8: $m = 50$ kg
• Caso 9: $a = 0.1$ $m/s^{2}$