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Artemon [7]
3 years ago
7

When two atoms approach each other the electron of each are attracted to the protons of the other. This is known as

Physics
1 answer:
vlada-n [284]3 years ago
6 0

Answer:

electrostatic force

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Radda [10]

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7 0
3 years ago
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Two cars are initially separated by 2500 m and traveling towards each other. One car travels at 4.5 m/s and the second car trave
mylen [45]

Answer:

714.285s

Explanation:

use relative velocity

8-4.5 = 3.5m/s

x = 2500m

2500/3.5 = 714.285s = 700s (with sig figs)

7 0
3 years ago
How much energy goes toward kinetic energy
kirill [66]

kinetic is moving

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8 0
3 years ago
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You are an engineer in charge of designing a new generation of elevators for a prospective upgrade to the Empire State Building.
DanielleElmas [232]

Answer:

   t_total = 23.757 s

Explanation:

This is a kinematics exercise.

Let's start by calculating the distance and has to reach the limit speed of

v = 18.8 m / s

         v = v₀ + a t₁

the elevator starts with zero speed

         v = a t₁

         t₁ = v / a

         t₁ = 18.8 / 2.40

         t₁ = 7.833 s

in this time he runs

         y₁ = v₀ t₁ + ½ a t₁²

         y₁ = ½ a t₁²

         y₁ = ½ 2.40 7.833²

         y₁ = 73.627 m

This is the time and distance traveled until reaching the maximum speed, which will be constant throughout the rest of the trip.

           x_total = x₁ + x₂

           x₂ = x_total - x₁

           x₂ = 373 - 73,627

           x₂ = 299.373 m

this distance travels at constant speed,

           v = x₂ / t₂

           t₂ = x₂ / v

           t₂ = 299.373 / 18.8

           t₂ = 15.92 s

therefore the total travel time is

           t_total = t₁ + t₂

           t_total = 7.833 + 15.92

           t_total = 23.757 s

6 0
3 years ago
In unit-vector notation, what is the torque about the origin on a particle located at coordinates (0 m, −3.0 m, 2.0 m) due to fo
irinina [24]

Answer:

The torque about the origin is 2.0Nm\hat{i}-8.0Nm\hat{j}-12.0Nm\hat{k}

Explanation:

Torque \overrightarrow{\tau} is the cross  product between force \overrightarrow{F} and vector position \overrightarrow{r} respect a fixed point (in our case the origin):

\overrightarrow{\tau}=\overrightarrow{r}\times\overrightarrow{F}

There are multiple ways to calculate a cross product but we're going to use most common method, finding the determinant of the matrix:

\overrightarrow{r}\times\overrightarrow{F} =-\left[\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k}\\ F1_{x} & F1_{y} & F1_{z}\\ r_{x} & r_{y} & r_{z}\end{array}\right]

\overrightarrow{r}\times\overrightarrow{F} =-((F1_{y}r_{z}-F1_{z}r_{y})\hat{i}-(F1_{x}r_{z}-F1_{z}r_{x})\hat{j}+(F1_{x}r_{y}-F1_{y}r_{x})\hat{k})

\overrightarrow{r}\times\overrightarrow{F} =-((0(2.0m)-0(-3.0m))\hat{i}-((4.0N)(2.0m)-(0)(0))\hat{j}+((4.0N)(-3.0m)-0(0))\hat{k})

\overrightarrow{r}\times\overrightarrow{F}=-2.0Nm\hat{i}+8.0Nm\hat{j}+12.0Nm\hat{k}=\overrightarrow{\tau}

4 0
3 years ago
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