Answer:
The west axis component of walk is, A = 5 miles
The east axis component of walk is, B = 8 miles
Explanation:
Given data,
The displacement of hiker south of west, R = 9.4 miles
The angle formed south of west, Ф = 60°
The vector components along,
the west axis is
A = R cosФ
A = 9.4 x cos 60
A = 4.7 miles
A = 5 miles
The west axis component of walk is, A = 5 miles
the south axis is
B = R sinФ
B = 9.4 sin 60
B = 8.14 miles
B = 8 miles
The east axis component of walk is, B = 8 miles
Answer:
The ball and the wall experience the same force.
Explanation:
According to the third law of Newton, which states that "for every action, there is an equal and opposite reaction", this means that when an object 1 acts on object 2 with a certain force, object 2 also acts on object 1 with the same magnitude of force but in an opposite direction.
According to this question, a ball hits a wall with a certain force. This means that the wall will react on the ball with the same force magnitude, but in an opposite manner. Hence, the ball and the wall experience the same force.
The reason for that is that P-waves (primary waves) travel faster than S-waves (secondary waves).
If we call

the speed of the primary waves and

the speed of the secondary waves, and we call

the distance of the seismogram from the epicenter, we can write the time the two waves take to reach the seismogram as


So the lag time between the arrival of the P-waves and of the S-waves is

We see that this lag time is proportional to the distance S, therefore the larger the distance, the greater the lag time.
The force between the molecules involved in the bond is 6. 426 *10^-11 Newton
<h3>How to determine the force</h3>
Using the formula:
F = K[q1 x q2]/D^2
where K is coulombs constant =9 *10 ^9 Nm^2/C^2.
q1 and q2 = charges = 1.60x10 -20C
d = distance between the charges = 2x10 -10 m
Substitute the values into the formula
F =
F = 
F = 
F =
Newton
Thus, the force between the molecules involved in the bond is 6. 426 *10^-11 Newton
Learn more about electrostatic force here:
brainly.com/question/8424563
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