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Likurg_2 [28]
3 years ago
12

Mark and Sofia walk together down a long, straight road. They walk without stopping for 4 miles. At this point Sofia says their

displacement during the trip must have been 4 miles; Mark says their current position must be 4 miles. Who, if either, is correct?a)Markb)Mark, only if their starting point is the origin of a coordinate systemc)Sofia, only if their starting point is the origin of a coordinate systemd)Sofia
Physics
1 answer:
kap26 [50]3 years ago
7 0

Answer:

d) Sofia.

Explanation:

If they walked down a straight road, their displacement will be 4 miles. However, their position depends on the origin of a coordinate system which could be located anywhere. If the origin of the coordinate system was at the starting point of their walk, then and only then, you can say that their position is also 4 miles.

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14*3(36/2(1*3)/6<br><img src="https://tex.z-dn.net/?f=14%20%5Ctimes%203%2836%20%5Cdiv%202%281%20%5Ctimes%203%29%20%5Cdiv%206" id
GarryVolchara [31]
<h2>10.5</h2><h3>remember pemdas</h3>
  • parentheses
  • exponents <em>excluded for this problem</em>
  • multiplication
  • division
  • addition
  • subtraction
<h3>step 1. start with what is at the top of the list.</h3>
  • parentheses
<h3>step 2. do 1 times 3 since it comes before division, and is in parentheses.</h3>

1 × 3 = 3

<h3>step 3. find 36 divided by 2</h3>

36 ÷ 2 = 18

<h3>step 4. add the values together</h3>

18 + 3 = 21

<h3>step 5. find 14 times 3</h3>

14 × 3 = 42 <em>you can also do 7 × 6 and will get the same result because 7 is half of 14 and 3 is half of 6</em>

<h3>step 6. add what is outside the parentheses</h3>

21 + 42 = 63

<h3>step 7. divide by 6</h3>

63 ÷ 6 = 10.5 -- > 10 remainder of 3 <em>remainder means left over</em>

<em />

7 0
2 years ago
A uniform bar has two small balls glued to its ends. The bar has length L and mass M, while the balls each have mass m and can b
vazorg [7]

Answer:

Explanation:

Length of bar = L

mass of bar = M

mass of each ball = m

Moment of inertia of the bar about its centre perpendicular to its plane is

I_{1}=\frac{ML^{2}}{12}

Moment of inertia of the two small balls about the centre of the bar perpendicular to its plane is

I_{2}=2\times m\times \frac{L^{2}}{4}

I_{2}=\frac{mL^{2}}{2}

Total moment of inertia of the system about the centre of the bar perpendicular to its plane is

I = I1 + I2

I=\frac{ML^{2}}{12}+\frac{mL^{2}}{2}

I=\frac{(M +6m)L^{2}}{12}

8 0
3 years ago
You can not exert force on a wall unless
ludmilkaskok [199]

unless...the wall simulaneiously exerts the same amount of force on you.

5 0
4 years ago
What is the electric potential energy of an electron at the negative end of the cable, relative to the positive end of the cable
VashaNatasha [74]

Answer:

Electric potential energy at the negative terminal: 1.92\cdot 10^{-18}J

Explanation:

When a particle with charge q travels across a potential difference \Delta V, then its change in electric potential energy is

\Delta U = q \Delta V

In this problem, we know that:

The particle is an electron, so its charge is

q=-1.60\cdot 10^{-19}C

We also know that the positive terminal is at potential

V_+=0V

While the negative terminal is at potential

V_-=-12 V

Therefore, the potential difference (final minus initial) is

\Delta V = -12-0 = -12 V

So, the change in potential energy of the electron is

\Delta U = (-1.6\cdot 10^{-19})(-12)=1.92\cdot 10^{-18}J

This means that the electron when it is at the negative terminal has 1.92\cdot 10^{-18}J of energy more than when it is at the positive terminal.

Since the potential at the positive terminal is 0, this means that the electric potential energy of the electron at the negative end is

1.92\cdot 10^{-18}J

3 0
4 years ago
Which objects have the most similar eccentricities?
mafiozo [28]
It depends on the shape of ellipse of that object, if their shapes are quite similar then there eccentricities would be similar.

Real life example. - <span> Neptune, </span>Venus<span>, and </span>Earth<span> are the planets in our </span>solar system<span> with the least eccentric orbits (They all have approx. orbits with same eccentricities)

Hope this helps!</span>
4 0
4 years ago
Read 2 more answers
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