A student walks 50 meters east, 40 meters north, 35 meters east, and then 20 m south. What is the magnitude and direction of the
1 answer:
Answer: A student walks 50 meters east, 40 meters north, 35 meters east, and then 20 m south. Then the magnitude and direction of the student's total displacement will be 87.32 m along the direction of AD or in east-south direction.
Explanation: To find the correct answer, we need to know about the Displacement of a body in motion.
<h3>What is displacement of a body in motion?</h3>- The displacement is the shortest distance between initial and final positions of a body.
- It's a vector quantity, and can positive, negative, or zero.
- The magnitude of displacement is less than or equal to the distance travelled.
- At first, we can draw a diagram showing the motion of the body.
- From the diagram, the displacement of the body will be equal to the distance between point A and D.
- To solve this, we can use Pythagoras theorem.
Thus, from the above calculations, we can conclude that, the displacement of the body will be equal to 87.32 m along the direction of AD or in east-south direction.
Learn more about the Displacement here:
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Answer:
0.82 mm
Explanation:
The formula for calculation an bright fringe from the central maxima is given as:
so for the distance of the second-order fringe when wavelength = 745-nm can be calculated as:
where;
n = 2
= 745-nm
D = 1.0 m
d = 0.54 mm
substituting the parameters in the above equation; we have:
= 0.00276 m
= 2.76 × 10 ⁻³ m
The distance of the second order fringe when the wavelength = 660-nm is as follows:
= 1.94 × 10 ⁻³ m
So, the distance apart the two fringe can now be calculated as:
= 2.76 × 10 ⁻³ m - 1.94 × 10 ⁻³ m
= 10 ⁻³ (2.76 - 1.94)
= 10 ⁻³ (0.82)
= 0.82 × 10 ⁻³ m
= 0.82 × 10 ⁻³ m
= 0.82 mm
Thus, the distance apart the second-order fringes for these two wavelengths = 0.82 mm