Two vectors with magnitudes of 6 meters and 8 meters cannot have a resultant of:
48 meters
14 meters
10 meters
2 meters
Please explain. I do not understand how to figure this out!
The range of possible values goes from the sum of the two magnitudes (14) to the absolute value of the difference (2). Do you see why? Imagine all the possible ways that the two vectors can be added. The greatest result is when they are in the same direction, and the least result is when they are in opposite directions.
Well, let's put on our vector clown shoes and dive into this question!
To determine whether two vectors with magnitudes of 6 meters and 8 meters can have a certain resultant, we need to use the concept of vector addition. When you add two vectors, you combine their magnitudes and directions to find their resultant.
The magnitude of the resultant vector can be found using the Pythagorean theorem. It states that the square of the hypotenuse (resultant) is equal to the sum of the squares of the other two sides (vectors). In this case, we have a 6-meter vector and an 8-meter vector.
So, if we calculate the magnitude of the resultant vector, we should be able to determine whether it matches any of the given options.
Now, let's crunch some numbers:
- If we take a 6-meter vector and an 8-meter vector, and their magnitudes add up to 14 meters. Hey, look! That's one of the options! So, 14 meters is possible.
- If we calculate the magnitude of the resultant, using the Pythagorean theorem, we get √(6^2 + 8^2) = √(36 + 64) = √100 = 10 meters. Hey, there's another one of the options! So, 10 meters is possible too.
- Now, if we try the same process for the other options, we find that √(6^2 + 8^2) = √(36 + 64) = √100 = 10 meters and not 48 meters. Sorry, 48 meters is not possible.
- Lastly, √(6^2 + 8^2) = √(36 + 64) = √100 = 10 meters and not 2 meters. Darn it, 2 meters is also not possible.
So, to summarize, out of the four options, the magnitudes of 6 meters and 8 meters cannot have a resultant of 48 meters or 2 meters. But they can have a resultant of 14 meters and 10 meters. I hope this clears things up for you, and remember, sometimes understanding physics is a balancing act - just like a clown on a unicycle!
To determine if two vectors with magnitudes of 6 meters and 8 meters can have a resultant of a given magnitude, we can use the Triangle Inequality Theorem. According to this theorem, the sum of the magnitudes of any two sides of a triangle must be greater than or equal to the magnitude of the third side. In the case of vectors, the resultant of two vectors is represented by the magnitude of the third side of the triangle formed when the vectors are added together.
In this scenario, let's consider the vector with magnitude 6 meters as vector A and the vector with magnitude 8 meters as vector B.
Now, let's check the options one by one:
1. Can the resultant be 48 meters?
Using the Triangle Inequality Theorem, we add the magnitudes of the two vectors: 6 meters + 8 meters = 14 meters which is less than 48 meters. Therefore, a resultant of 48 meters is possible.
2. Can the resultant be 14 meters?
Following the same step, 6 meters + 8 meters = 14 meters. In this case, the magnitudes of the two vectors are equal to the given resultant magnitude. Therefore, a resultant of 14 meters is possible.
3. Can the resultant be 10 meters?
Again, using the same method, 6 meters + 8 meters = 14 meters. This is greater than 10 meters. Therefore, a resultant of 10 meters is possible.
4. Can the resultant be 2 meters?
Once again, applying the same approach, 6 meters + 8 meters = 14 meters. This magnitude is greater than 2 meters. Therefore, a resultant of 2 meters is possible.
Thus, based on these calculations, we conclude that none of the given options are correct. A resultant of 6 meters and 8 meters can have any of the mentioned magnitudes.
To find the resultant of two vectors, you can use the vector addition. The magnitude of the resultant vector is given by the formula:
|Resultant| = sqrt(|Vector1|^2 + |Vector2|^2 + 2 · |Vector1| · |Vector2| · cosθ)
Where |Vector1| and |Vector2| are the magnitudes of the vectors, and θ is the angle between them.
In this case, we have two vectors with magnitudes of 6 meters and 8 meters respectively. Let's see which of the given options is not possible for the resultant magnitude:
1. |Resultant| = 48 meters:
To have a resultant of 48 meters, we need to find an angle θ such that:
48^2 = 6^2 + 8^2 + 2 · 6 · 8 · cosθ
Solving this, we find that cosθ = -1. This means the two vectors are pointing in opposite directions (180 degrees apart), which is not possible since we are considering two physical vectors. Therefore, a resultant of 48 meters is not possible.
2. |Resultant| = 14 meters:
To have a resultant of 14 meters, we need to find an angle θ such that:
14^2 = 6^2 + 8^2 + 2 · 6 · 8 · cosθ
Solving this, we find that cosθ = 0. This means the two vectors are perpendicular to each other (90 degrees apart). It is possible for two vectors with magnitudes of 6 meters and 8 meters to have a resultant of 14 meters if they are perpendicular to each other.
3. |Resultant| = 10 meters:
To have a resultant of 10 meters, we need to find an angle θ such that:
10^2 = 6^2 + 8^2 + 2 · 6 · 8 · cosθ
Solving this, we find that cosθ ≈ -0.982. This means the angle θ is approximately 178 degrees. It is possible for two vectors with magnitudes of 6 meters and 8 meters to have a resultant of 10 meters if they are nearly pointing in opposite directions.
4. |Resultant| = 2 meters:
To have a resultant of 2 meters, we need to find an angle θ such that:
2^2 = 6^2 + 8^2 + 2 · 6 · 8 · cosθ
Solving this, we find that cosθ ≈ 0.998. This means the angle θ is approximately 3 degrees. It is possible for two vectors with magnitudes of 6 meters and 8 meters to have a resultant of 2 meters if they are nearly pointing in the same direction.
Therefore, the only option that is not possible is a resultant of 48 meters.