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Ainat [17]
2 years ago
13

If you walk 1.2 km north and then 1.6 km east, what are the magnitude and direction of your resultant displacement?

Physics
1 answer:
vodomira [7]2 years ago
5 0

<u>Answer:</u>

 Option A is the correct answer.

<u>Explanation:</u>

  Let the east point towards positive X-axis and north point towards positive Y-axis.

 First walking 1.2 km north,  displacement = 1.2 j km

 Secondly 1.6 km east, displacement = 1.6 i km

 Total displacement = (1.6 i + 1.2 j) km

 Magnitude = \sqrt{1.2^2+1.6^2} = 2 km

 Angle of resultant with positive X - axis = tan^{-1}(1.2/1.6)=36.87^0 = 36.87⁰ east of north.

 

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*please refer to photo*
just olya [345]

Based on the calculations, the average velocity is equal to 360 m/s and the percent difference is equal to 4.72%.

<h3>What is average velocity?</h3>

An average velocity can be defined as the total distance covered by a physical object divided by the total time taken.

<h3>What is an average?</h3>

An average is also referred to as mean and it can be defined as a ratio of the sum of the total number in a data set to the frequency of the data set.

<h3>How to calculate the average velocity?</h3>

Mathematically, the average velocity for this data set would be calculated by using this formula:

Average = [F(v)]/n

Vavg = [v₁ + v₂ + v₃ + v₄ + v₅)/5

Since the values of the average velocity from the table are missing, we would assume the following values for the purpose of an explanation:

  • v₁ = 100 m/s
  • v₂ = 150 m/s
  • v₃ = 200 m/s
  • v₄ = 250 m/s
  • v₅ = 300 m/s

Substituting the parameters into the formula, we have:

Vavg = [300 + 450 + 500 + 250 + 300)/5

Vavg = 1800/5

Vavg = 360 m/s.

Next, we would calculate the percent difference by using this formula:

Percent \;difference = \frac{[V_{avg}\;-\;V_{sound}]}{V_{sound}} \times 100

Percent difference = [360 - 343]/360 × 100

Percent difference = 17/360 × 100

Percent difference = 0.0472 × 100

Percent difference = 4.72%.

Read more on average here: brainly.com/question/9550536

#SPJ1

3 0
8 months ago
What is the net force of a 25 g object with an acceleration of 3 m/s^2? 100PTS
gayaneshka [121]

Explanation:

Force=Mass×acceleration

force=25×3

force=75N

5 0
2 years ago
Read 2 more answers
Car A, moving in a straight line at a constant speed of 20. meters per second, is initially 200 meters behind car B, moving in t
MA_775_DIABLO [31]
First, let's express the movement of Car A and B in terms of their position over time (relative to car B)
For car A: y=20x-200   Car A moves 20 meters every second x, and starts 200 meters behind car B
For Car B: y= 15x      Car B moves 15 meters every second and starts at our basis point

Set the two equations equal to one another to find the time x at which they meet:
20x - 200 = 15x
200 = 5x 
x= 40
At time x=40 seconds, the cars meet. How far will Car A have traveled at this time? 
Car A moves 20 meters every second:
20 x 40 = 800 meters
8 0
3 years ago
Suppose a photon with an energy of 1.60 eV strikes a piece of metal. If the electron that it hits loses 0.800 eV leaving the met
JulsSmile [24]

To solve this problem it is necessary to apply the concepts related to the change of Energy in photons and the conservation of energy.

From the theory we could consider that the energy change is subject to

\Delta E = E_0 -W_f

Where

E_0 =Initial Energy

W_f = Energy loses

Replacing we have that

\Delta E = 1.6-0.8

\Delta E = 0.8eV

Therefore the Kinetic energy of the electron once it has broken free of the metal surface is 0.8eV

7 0
2 years ago
Which orbit has the highest energy?<br> n = 1<br> n = 2<br> n = 3
PIT_PIT [208]

Answer:

3

Explanation:

The closer an orbit is to the nucleus the fewer energy

4 0
3 years ago
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