91影视

The acceleration of the object at constant force F is given by a = 10 m/s²

Since force is directly proportional to the acceleration of the object by Newton's Second law, it can be written as

$$\begin{gathered} F=ma \\ F鈭�a \end{gathered}$$

Consider the initial force to be F and initial acceleration be a.

If the force F' is halved, then the acceleration a' is also halved.

Therefore, the new acceleration is calculated as :

$$\begin{gathered} \frac{F}{F^{\text{'}}}=\frac{a}{a^{\text{'}}} \\ \frac{F}{F/2}=\frac{10m/s^2}{a^{\text{'}}} \\ {a}^{\text{'}}=5m/s^2 \end{gathered}$$

The formula to calculate the force acting on the object is given by F=ma

In terms of relation of acceleration and mass of the object, it can be written as

$$\begin{gathered} a=\frac{F}{m} \\ a鈭�\frac{1}{m} \end{gathered}$$

Consider the original acceleration be a= 10 m/s² and mass be m.

When the mass is halved, let the final mass be m'=m/2 and the acceleration be a' respectively.

The new acceleration of the object when the mass is doubled is given by

$$\begin{gathered} \frac{a^{\text{'}}}{a}=\frac{m}{m^{\text{'}}} \\ {a}^{\text{'}}=\frac{m}{m/2}脳10m/s^2 \\ =20m/s^2 \end{gathered}$$

Consider the original force, F, mass be m and acceleration of the object be a=10 m/s².

The new force be F' =F/2 and mass be m'=m/2 respectively.

The new acceleration when the force and the object鈥檚 mass are both doubled can be calculated as

$$\begin{gathered} F=ma \\ a=\frac{F}{m} \\ \frac{a^{\text{'}}}{a}=\frac{{\displaystyle \raisebox{1ex}{$F^{\text{'}}$}\!\left/ \!\raisebox{-1ex}{$m\text{'}$}\right.}}{{\displaystyle \raisebox{1ex}{$F$}\!\left/ \!\raisebox{-1ex}{$m$}\right.}} \\ =\frac{{\displaystyle \raisebox{1ex}{$F/2$}\!\left/ \!\raisebox{-1ex}{$m/2$}\right.}}{{\displaystyle \raisebox{1ex}{$F$}\!\left/ \!\raisebox{-1ex}{$m$}\right.}} \\ {a}^{\text{'}}=a \\ =10m/s^2 \end{gathered}$$

Consider the original force, F, mass be m and acceleration of the object be a=10 m/s².

The new force be F' =2F and mass be m'=m/2 respectively.

The new acceleration when the force is doubled and the object鈥檚 mass is halved can be calculated as:

$$\begin{gathered} F=ma \\ a=\frac{F}{m} \\ \frac{a^{\text{'}}}{a}=\frac{{\displaystyle \raisebox{1ex}{$F^{\text{'}}$}\!\left/ \!\raisebox{-1ex}{$m\text{'}$}\right.}}{{\displaystyle \raisebox{1ex}{$F$}\!\left/ \!\raisebox{-1ex}{$m$}\right.}} \\ =\frac{{\displaystyle \raisebox{1ex}{$(F/2)$}\!\left/ \!\raisebox{-1ex}{$\left({\displaystyle 2m}\right)$}\right.}}{{\displaystyle \raisebox{1ex}{$F$}\!\left/ \!\raisebox{-1ex}{$m$}\right.}} \\ {a}^{\text{'}}=a/4 \\ =10/4m/s^2 \\ =2.5m/s^2 \end{gathered}$$