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aleksandr82 [10.1K]
2 years ago
12

What is the displacement if a person traveled from point B to E?

Physics
1 answer:
nikitadnepr [17]2 years ago
3 0
<h3>Answers:</h3>
  1. displacement = 3 km
  2. displacement = 0 km

=====================================================

Explanation:

Problem 1

Focus on the y axis which is the distance axis.

Point B has a y coordinate of 2

Point E has a y coordinate of 5

The displacement is 5-2 = 3, which means the person has moved 3 km.

--------------------------

Problem 2

Both D and E have the same y coordinate. So the displacement is 0 (since 5-5 = 0). No change in distance has occurred. During segment DE, the person is resting. By the way it is impossible to go back to D since it's impossible to go back in time.

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Two charged particles separated by a distance of = 3 and experienced electrostatic forces of = 60 . What would be this force if
klemol [59]

Answer: 539.4 N

Explanation:

Let's begin by explaining that Coulomb's Law establishes the following:  

"The electrostatic force F_{E} between two point charges q_{1} and q_{2} is proportional to the product of the charges and inversely proportional to the square of the distance d that separates them, and has the direction of the line that joins them"

What is written above is expressed mathematically as follows:

F_{E}= K\frac{q_{1}.q_{2}}{d^{2}} (1)

Where:

F_{E}=60 N  is the electrostatic force

K=8.99(10)^{9} Nm^{2}/C^{2} is the Coulomb's constant  

q_{1} and q_{2} are the electric charges

d=3 m is the separation distance between the charges  

Then:

60 N= 8.99(10)^{9} Nm^{2}/C^{2}\frac{q_{1}.q_{2}}{(3 m)^{2}} (2)

Isolating q_{1} and q_{2}:

q_{1}q_{2}=6(10)^{-8} C^{2} (3)

Now, if we keep the same charges but we decrease the distance to d_{1}=1 m, (1) is rewritten as:

F_{E}=8.99(10)^{9} Nm^{2}/C^{2}\frac{6(10)^{-8} C^{2}}{(1 m)^{2}} (4)

Then, the new electrostatic force will be:

F_{E}= 539.4 N (5) As we can see, the electrostatic force is increased when we decrease the distance between the charges.

4 0
3 years ago
A rock from space in Earth's Atmosphere would be considered?
Gnesinka [82]

Answer:

A meteor mate

Explanation:

hope this helps

8 0
3 years ago
A train whose proper length is 1200 m passes at a high speed through a station whose platform measures 900 m, and the station ma
TiliK225 [7]

Answer:

0.66c

Explanation:

Use length contraction equation:

L = L₀ √(1 − (v²/c²))

where L is the contracted length,

L₀ is the length at 0 velocity,

v is the velocity,

and c is the speed of light.

900 = 1200 √(1 − (v²/c²))

3/4 = √(1 − (v²/c²))

9/16 = 1 − (v²/c²)

v²/c² = 7/16

v = ¼√7 c

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3 years ago
A cyclist going downhill is accelerating at 1.2 m/s2. If the final velocity of the cyclist is 16 m/s after 10 seconds, what is t
Blababa [14]

Answer:

Initial Velocity is 4 m/s

Explanation:

What is acceleration?

It is the change in velocity with respect to time, or the rate of change of velocity.

We can write this as:

a=\frac{\Delta v}{t}

Where

a is the acceleration

v is velocity

t is time

\Delta  is "change in"

For this problem , we are given

a = 1.2

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Putting into formula, we get:

a=\frac{\Delta v}{t}\\1.2=\frac{\Delta v}{10}\\\Delta v = 1.2*10\\\Delta v = 12

So, the change in velocity is 12 m/s

The change in velocity can also be written as:

\Delta v = Final  \ Velocity - Initial \ Velocity

It is given Final Velocity = 16, so we put it into formula and find Initial Velocity. Shown Below:

\Delta v = Final  \ Velocity - Initial \ Velocity\\12=16-Initial \ Velocity\\Initial \ Velocity = 16 - 12 = 4

hence,

Initial Velocity is 4 m/s

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4 years ago
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This type of energy can be transferred in three different ways: 1) direction contact through collisions (also called conduction)
MA_775_DIABLO [31]
I would say nuclear.
4 0
3 years ago
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