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snow_lady [41]
3 years ago
6

How does Newton’s first second and third laws apply to eating your breakfast

Physics
2 answers:
____ [38]3 years ago
6 0

Answer:

Newton's third law states that for every action (force) in nature there is an equal and opposite reaction. In other words, if object A exerts a force on object B, then object B also exerts an equal and opposite force on object A. Notice that the forces are exerted on different objects. However much you push your fork in the food, that much a dent it will cause.

Explanation:

ankoles [38]3 years ago
3 0

Answer:

¡I hope it helps you! :) .................

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Which of the following is an example of applying a force?
maw [93]
The correct response would be C. Carpenter hammering a nail. The carpenter is applying a force as he or she is hitting the surface of the nail with the hammer.
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3 years ago
A proton moving along the positive x-axis enters an electric field that is directed along the positive y-axis. What is the direc
katrin2010 [14]

Answer:

Direction of the electric force acting on the proton after it enters the electric field is along the positive y-axis.

Explanation:

In an electrical field electrical field lines goes from positive to negative. Therefore, electrical force vector is on exactly the same way with electrical field lines for a proton. Which means if the electrical field directed along the positive y-axis, electrical force for proton will be on the same way.

8 0
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
lisov135 [29]

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}

3 0
2 years ago
Is there a definite end to our atmosphere?
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7 0
3 years ago
How does acceleration values differ mathematically from deceleration values?
Serggg [28]
If a problem says the acceleration is some positive value than solve using that value, a negative acceleration is said to be deceleration. E.g. a car decelerating at 10 m/sec can be said to be accelerating at -10 m/sec.

If a problem states decelerates at A, then use -A for acceleration in the classic equations which are for acceleration. If a problem says accelerates at a negative value like -A the use -A as the value for acceleration, it can also be said to be decelerating at A.
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