calculate the missing variables using the necessary and rules.\n1. what does newtons 2nd law of motion…

calculate the missing variables using the necessary and rules.\n1. what does newtons 2nd law of motion state?\nforce (f)\nf = m x a\nnewtons (n)\na water - filled balloon with an overall mass of 1kg undergoes an acceleration of 2m/s². how much force is being applied to the balloon?\nmass of the rocket 0.5 kg\na 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?\na tractor carrying hay bales has a total mass of 9,100 kg. what force is required to reach an acceleration of 4 m/s²?\nacceleration\nmass (m)\nm = f/a\nkilograms (kg)\na skyrocket is launched with a force of 10n and accelerates at 20 m/s². what is the mass of this skyrocket?\nacceleration of the coal - wagon is m/s\na 7.5 n force is applied to a football, generating an initial acceleration of 12 m/s². calculate the mass of the football.\nmass of the football: 0.625\na cannonball accelerates at 1000 m/s² from an applied force of 5000n what is the mass of the cannonball?\nmass of the cannon - ball 5kg\nacceleration (a)\na = f/m\nmeters per second (m/s²)\na 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?\nforce required for the theme - park ride carriage 500n\na spring balance is used to launch a 0.05 kg foam ball with a force of 5n. what acceleration is produced?\na 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

calculate the missing variables using the necessary and rules.\n1. what does newtons 2nd law of motion state?\nforce (f)\nf = m x a\nnewtons (n)\na water - filled balloon with an overall mass of 1kg undergoes an acceleration of 2m/s². how much force is being applied to the balloon?\nmass of the rocket 0.5 kg\na 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?\na tractor carrying hay bales has a total mass of 9,100 kg. what force is required to reach an acceleration of 4 m/s²?\nacceleration\nmass (m)\nm = f/a\nkilograms (kg)\na skyrocket is launched with a force of 10n and accelerates at 20 m/s². what is the mass of this skyrocket?\nacceleration of the coal - wagon is m/s\na 7.5 n force is applied to a football, generating an initial acceleration of 12 m/s². calculate the mass of the football.\nmass of the football: 0.625\na cannonball accelerates at 1000 m/s² from an applied force of 5000n what is the mass of the cannonball?\nmass of the cannon - ball 5kg\nacceleration (a)\na = f/m\nmeters per second (m/s²)\na 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?\nforce required for the theme - park ride carriage 500n\na spring balance is used to launch a 0.05 kg foam ball with a force of 5n. what acceleration is produced?\na 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

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: $a = 0.1$ $m/s^{2}$