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.
<h3>How to solve the problem?</h3>
- 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:
Newtons first law states that:
<em>If</em><em> </em><em>a</em><em> </em><em>body</em><em> </em><em>i</em><em>s</em><em> </em><em>in</em><em> </em><em>rest</em><em> </em><em>or</em><em> </em><em>motion</em><em> </em><em>in</em><em> </em><em>a</em><em> </em><em>straight</em><em> </em><em>line</em><em>,</em><em> </em><em>it</em><em> </em><em>remains</em><em> </em><em>at</em><em> </em><em>rest</em><em> </em><em>or</em><em> </em><em>at</em><em> </em><em>motion</em><em> </em><em>in</em><em> </em><em>a</em><em> </em><em>straight</em><em> </em><em>line</em><em> </em><em>with</em><em> </em><em>constant</em><em> </em><em>speed</em><em> </em><em>until</em><em> </em><em>and</em><em> </em><em>unless</em><em> </em><em>and</em><em> </em><em>external</em><em> </em><em>unbalanced</em><em> </em><em>force</em><em> </em><em>acts</em><em> </em><em>on</em><em> </em><em>it</em><em>.</em>
<em>'</em><em>This</em><em> </em><em>law</em><em> </em><em>i</em><em>s</em><em> </em><em>also</em><em> </em><em>known</em><em> </em><em>as</em><em> </em><em>the</em><em> </em><em>law</em><em> </em><em>of</em><em> </em><em>Inertia</em><em>.</em><em>'</em>
Answer:
Here ball and rod will repel each other as they are of similar charges
Explanation:
As we know that the two charges attract or repel each other by electrostatic force
This force is given as

so we know if two charges are similar in nature then they will repel each other and if the two charges are opposite in nature then they will attract each other
So here when rod touch the ball then it transfer its charge to the ball and due to similar charges in ball and rod they both repel each other
To solve this problem we will apply the concepts related to Reyleigh's criteria. Here the resolution of the eye is defined as 1.22 times the wavelength over the diameter of the eye. Mathematically this is,

Here,
D is diameter of the eye


The angle that relates the distance between the lights and the distance to the lamp is given by,

For small angle, 
Here,
d = Distance between lights
L = Distance from eye to lamp
For small angle 
Therefore,



Therefore the distance is 5.367km.
Part A:
For this part we’re assuming all the kinetic energy of the moving bumper car is converted into elastic potential energy in the spring since the car is brought to rest. Therefore you can find the total kinetic energy to get your answer:
KE = ½ mv^2
KE = ½ (200)(8)^2
KE = 6400 J
Part B:
Now you can use Hooke’s law to find the force:
F = kx
F = (5000)(0.2)
F = 1000 N