The magnitude of your displacement can be equal to the distance you covered, or it can be less than the distance you covered. But it can never be greater than the distance you covered.
This is because displacement is a straight line, whereas distance can be a straight line, a squiggly line, a zig-zag line, a line with loops in it, a line with a bunch of back-and-forths in it, or any other kind of line.
The straight line is always the shortest path between two points.
An unconformity, or abnormality in the layers of rocks, signifies that the process of layering was somehow interrupted. The only feasible choice for this would be (B), where erosion would have interrupted the layering process.
Answer:
a. Acceleration, a = 1.88 m/s²
b. Time, t = 7.87 seconds.
Explanation:
Given the following data;
Initial velocity, U = 14.5m/s
Final velocity, V = 29.3m/s
Distance, S = 172m
a. To find the acceleration of the speedboat;
We would use the third equation of motion;
V² = U² + 2aS
Substituting into the formula
29.3² = 14.5² + 2a*172
858.49 = 210.25 + 344a
344a = 858.49 - 210.25
344a = 648.24
a = 648.24/344
Acceleration, a = 1.88 m/s²
b. To find the time;
We would use the first equation of motion;
V = U + at
29.3 = 14.5 + 1.88t
1.88t = 29.3 - 14.5
1.88t = 14.8
Time, t = 14.8/1.88
Time, t = 7.87 seconds.
Answer:
The value of acceleration that accomplishes this is 8.61 ft/s² .
Explanation:
Given;
maximum distance to be traveled by the car when the brake is applied, d = 450 ft
initial velocity of the car, u = 60 mph = (1.467 x 60) = 88.02 ft/s
final velocity of the car when it stops, v = 0
Apply the following kinematic equation to solve for the deceleration of the car.
v² = u² + 2as
0 = 88.02² + (2 x 450)a
-900a = 7747.5204
a = -7747.5204 / 900
a = -8.61 ft/s²
|a| = 8.61 ft/s²
Therefore, the value of acceleration that accomplishes this is 8.61 ft/s² .
This problem involves Newton's universal law of gravitation and the equation to follow would be.
F = GM₁M₂/r²
Given: M₁ = 0.890 Kg; M₂ = 0.890 Kg; F = 8.06 x 10⁻¹¹ N; G = 6.673 X 10⁻¹¹ N m²/Kg²
Solving for distance r = ?
r = √GM₁M₂/F
r = √(6.673 x 10⁻¹¹ N m₂/Kg²)(0.890 Kg)(0.890 Kg)/ 8.06 x 10⁻¹¹ N
r = 0.81 m