Answer:
A) The updated amplitude = 0.048 m
B) Period T = 0.6 seconds
Explanation: Please refer to the attached documents for the solution.
Megan rode the bus to school, which is located 8 kilometers from her home. If Megan's frame of reference is her house, and it to
1 answer:
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Answer:
130 m/s (to two significant figures)
Explanation:
In projectile motion, the launching velocity and launch angle help to determine both the horizontal and vertical velocity components.
u represents the initial projectile velocity = 150 m/s
uₓ = u cos θ = 150 cos 30° = 129.9 m/s
uᵧ = u sin θ = 150 sin 30° = 75.0 m/s
A projectile's motion can be viewed as made up of independent vertical and horizontal elements.
The vertical motion is affected by gravitational acceleration (which pulls down on the projectile), altering the vertical velocity component due to this acting force.
Conversely, there is no acting force in the horizontal direction, which means the horizontal component maintains a steady velocity throughout the projectile's flight.
Thus, at t = 4 s, the horizontal component of the projectile's speed remains equal to the initial horizontal velocity component.
At t = 4 s, the horizontal component of velocity is uₓ = u cos θ = 150 cos 30° = 129.9 m/s ≈ 130 m/s
Answer:
Explanation:
Definitions:
P is an upward acting force.
C represents the vertical contact force from the floor.
W is the crate's weight.
Since P cannot lift the crate, the normal reaction force remains active.
Thus, the forces are related as follows:
P - W + C = 0
Rearranged:
P = W - C
Starting speed of the coin (u)= 0 (Since it starts from rest)
Gravitational acceleration (a) = g = 9.81 m/s²
Duration of the fall (t) = 1.5 s
According to the motion equation we can state:
Substituting the known values into the equation yields:
v = 0 + 9.81 × 1.5
v = 14.715 m/s
The speed of the coin upon reaching the ground/Final speed of the coin = 14.715 m/s