Displacement is a vector quantity. So, you incorporate the vector calculations when you try to determine the resultant vector. This is the shortest path from the starting point to the endpoint. If they are moving on one axis only, you use sign conventions. For motions moving to the left, use the negative sign. If it's moving to the right, then use the positive sign. Now, it the object moves 2 km to the left, and 2 km also to the right, the displacement is zero.
Displacement = 2 km - 2km = 0
Generally, the equation is:
<span>Displacement = Distance of motion to the right - Distance of motion to the left</span>
According to Newton's second law, the force applied to an object is equal to the product between the mass of the object and its acceleration:

where F is the magnitude of the force, m is the mass of the object and a its acceleration.
In this problem, the object is the insect, with mass

. The acceleration of the insect is

, therefore we can calculate the force exerted by the car on the insect:

How do we find the force exerted by the insect on the car?
According to Newton's third law (known as action-reaction law), when an object A exerts a force on an object B, object B also exerts a force equal and opposite on object A. Therefore, the force exerted by the insect on the car is equal to the force exerted by the car on the object, so it is 0.01 N.
<span>ultraviolet
Have a great day!</span>
Answer:
d = 4180.3m
wavelengt of sound is 0.251m
Explanation:
Given that
frequency of the sound is 5920 Hz
v=1485m/s
t=5.63s
let d represent distance from the vessel to the ocean bottom.
an echo travels a distance equivalent to 2d, that is to and fro after it reflects from the obstacle.


wavelengt of sound is
= v/f
= (1485)/(5920)
= 0.251 m
Answer:
A. The time taken for the car to stop is 3.14 secs
B. The initial velocity is 81.64 ft/s
Explanation:
Data obtained from the question include:
Acceleration (a) = 26ft/s2
Distance (s) = 256ft
Final velocity (V) = 0
Time (t) =?
Initial velocity (U) =?
A. Determination of the time taken for the car to stop.
Let us obtain an express for time (t)
Acceleration (a) = Velocity (V)/time(t)
a = V/t
Velocity (V) = distance (s) /time (t)
V = s/t
a = s/t^2
Cross multiply
a x t^2 = s
Divide both side by a
t^2 = s/a
Take the square root of both side
t = √(s/a)
Now we can obtain the time as follow
Acceleration (a) = 26ft/s2
Distance (s) = 256ft
Time (t) =..?
t = √(s/a)
t = √(256/26)
t = 3.14 secs
Therefore, the time taken for the car to stop is 3.14 secs
B. Determination of the initial speed of the car.
V = U + at
Final velocity (V) = 0
Deceleration (a) = –26ft/s2
Time (t) = 3.14 sec
Initial velocity (U) =.?
0 = U – 26x3.14
0 = U – 81.64
Collect like terms
U = 81.64 ft/s
Therefore, the initial velocity is 81.64 ft/s