<h2>
Speed with which it return to its initial level is 100 m/s</h2>
Explanation:
We have equation of motion v² = u² + 2as
Initial velocity, u = 100 m/s
Acceleration, a = -9.81 m/s²
Final velocity, v = ?
Displacement, s = 0 m
Substituting
v² = u² + 2as
v² = 100² + 2 x -9.81 x 0
v² = 100²
v = ±100 m/s
+100 m/s is initial velocity and -100 m/s is final velocity.
Speed with which it return to its initial level is 100 m/s
Answer:
the correct answer is D
Explanation:
In this exercise, the vectors are in the same west-east direction, so we can assume that the positive direction is east and perform the algebraic sum.
R = δ + ε
where
δ = 2.0 m
ε = 7.0 m
the positive sign indicates that it is heading east
R = 2.0 + 7.0
R = 9.0 m
the direction is east
the correct answer is D
Answer:
16 km
Explanation:
Drawing a right triangle to model the problem helps. I started by drawing the lines of the triangle to model the hiker's journey- a vertical straight line for 11 km north and then a horizontal line connected to the top of it for 11 km east; I then drew the hypothenuse to connect the two lines.
The hypothenuse is what we have to solve for, so we will use the Pythagorean Theorem, a^2 + b^2 = c^2. Since both distances are 11 km both a and b in the equation are 11.
11^2 + 11^2 = c^2
121 + 121 = c^2
242 = c^2
c = 15.56
Rounding the answer makes it 16 km for the hiker's magnitude of displacement.
The layers form from sand dunes
Explanation:
t = usin©/g
Where t is the time to reach the maximum height
Time spent in air is T = 2t
Hence, T = 2usin©/g
T = 2 x 20 x sin 65°/ 9.8
T = 3.69s
