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Luda [366]
3 years ago
5

PLEASE HELP!! Newtons first law states that if an objects velocity is changing a __________ must be acting on it. Question 1 opt

ions: Force Mass Acceleration Interia
Physics
1 answer:
Anika [276]3 years ago
5 0

Newtons first law states that if an objects velocity is changing a <u>force</u> must be acting on it.

Explanation:

Newton's first law of motion states that:

"An object at rest (or in uniform motion) will remain at rest (or will continue moving with the same velocity) unless acted upon an unbalanced force"

We can apply this law to a daily life example:

  • Take a book at rest on a table: the forces acting on the book are balanced. If we do not apply any other force, we know that the book will remain at rest: this is exactly what is summarized in Newton's first law.
  • Take a space probe moving in the interstellar space, very far from any planet or source of gravitational force. Since there are no forces acting on the proble, the probe will continue moving at the same velocity (same speed and same direction) forever, unless stopped by a new force acting on it.

This means that in order to put an object at rest in motion, or to stop an object already in motion, or to change its velocity, an unbalanced force needs to be applied: otherwise, the object will continue having the same velocity (which can be either zero or non-zero), so it will continue having  same speed and same direction.

Learn more about Newton laws of motion:

brainly.com/question/11411375

brainly.com/question/1971321

brainly.com/question/2286502

brainly.com/question/2562700

#LearnwithBrainly

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The electric field between two parallel plates is uniform, with magnitude 646 N/C. A proton is held stationary at the positive p
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Answer:

The distance from the positive plate at which the two pass each other is 0.0023 cm.

Explanation:

We need to find the acceleration of each particle first. Let's use the electric force equation.

F=Eq

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m_{p}a_{p}=Eq_{p}

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<u>For the electron</u>

m_{e}a_{e}=Eq_{e}

a_{e}=\frac{Eq_{e}}{m_{e}}

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Now we know that the plate separation is 4.26 cm or 0.0426 m. The travel distance of the proton plus the travel distance of the electron is 0.0426 m.

x_{p}+x_{e}=0.0426

Both of them have an initial speed equal to zero. So we have:

\frac{1}{2}a_{p}t^{2}+\frac{1}{2}a_{e}t^{2}=0.0426

t^{2}(a_{p}+a_{e})=2*0.0426

t^{2}=\frac{2*0.0426}{a_{p}+a_{e}}

t=\sqrt{\frac{2*0.0426}{6.19*10^{10}+1.14*10^{14}}}

t=2.73*10^{-8}\: s    

With this time we can find the distance from the positive plate (x(p)).

x_{p}=\frac{1}{2}a_{p}t^{2}

x_{p}=\frac{1}{2}6.19*10^{10}*(2.73*10^{-8})^{2}

x_{p}=0.0023\: cm

Therefore, the distance from the positive plate at which the two pass each other is 0.0023 cm.

I hope it helps you!

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