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3 years ago

Why does Newton's first law refer only to an unbalanced force and not to a balanced force

Physics

2 answers:

SVETLANKA909090 [29]3 years ago

4 0

Answer:

we will first consider a physics book at rest on a tabletop. There are two forces acting upon the book. One force - the Earth's gravitational pull - exerts a downward force. The other force - the push of the table on the book (sometimes referred to as a normal force) - pushes upward on the book.

Jet001 [13]3 years ago

3 0

Answer:

Explanation:

Newton's first law states that an object will not change its motion unless acted on by an unbalanced force.

If an object was acted on by a balanced force, these forces cancel out resulting in a net force of 0 newtons, which doesn't change anything for the object.

For example: A balanced force acts on a box. One man pushes the box with a force of 5 N to the right. Another man is on the other side pushing with 5 N to the left. Calculating the net force, (5-5=0) you get that the net force is 0 N. Because the net force is 0, this situation is a balanced force. Basically, the force cancels out leaving no force to act on the object, so the motion is not changed.

However, an unbalanced force results in a change to an object's motion. For example, if there was only one man pushing the box 5N to the right, the total net force would be 5 N to the right (5-0=5). Which mean the man pushing does not get cancelled out and actually affects the box by pushing it to the right by 5N.

Hope that clarifies things a little.

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3 years ago

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3 years ago

The driver of a car traveling at 31.3 m/s applies the brakes and undergoes a constant deceleration of 1.6 m/s2.How many revoluti

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Answer:

$R=156.99\operatorname{Re}vs$

Explanation: The equations used are as follows:

$\begin{gathered} x(t)=x_o+v_ot+\frac{1}{2}at^2\Rightarrow(1) \\ v(t)=v_o+at\Rightarrow(2) \end{gathered}$

By using equation (2), the time needed for the car to come to rest is calculated as follows:

$\begin{gathered} v(t)=(31.3ms^{-1})_{}+(-1.6ms^{-2})t=0 \\ t=\frac{31.3ms^{-1}}{1.6ms^{-2}}=19.56s \\ t=19.563s \end{gathered}$

By using equation (1), The total distance traveled in that time would be as:

$\begin{gathered} x(19.563s)=_{}(31.3ms^{-1})\cdot(19.563s)+\frac{1}{2}(-1.6ms^{-2})\cdot(19.563s)^2\Rightarrow(1) \\ x(19.563s)=612.31-306.17=306.14m \\ \therefore\Rightarrow \\ x(19.563s)=306.14m \end{gathered}$

The revolutions taken by the tire before the car comes to rest would be:

$\begin{gathered} C=2\pi\cdot(0.31m)=1.95m \\ R=\frac{x(19.563s)}{C}=\frac{306.14m}{1.95m}=156.99\operatorname{Re}v \\ R=156.99\operatorname{Re}vs \end{gathered}$

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