A ball is thrown at a 60.0° angle above the horizontal across level ground. It is thrown from a height of 2.00 m above the groun
d with a speed of 20.0 m/s and experiences no appreciable air resistance. The time the ball remains in the air before striking the ground is closest to_____________.a. 16.2 s.b. 3.07 s.c. 3.32 s.d. 3.53 s.e. 3.64 s.
1 answer:
Answer:
Option E is correct.
Time the ball remains in the air before striking the ground is closest to 3.64 s
Explanation:
yբ = yᵢ + uᵧt + gt²/2
yբ = 0
yᵢ = 2 m
uᵧ = u sinθ = 20 sin 60 = 17.32 m/s
g = -9.8 m/s², t = ?
0 = 2 + 17.32t - 4.9t²
4.9t² - 17.32t - 2 = 0
Solving the quadratic equation,
t = 3.647 s or t = -0.1112 s
time is a positive variable, hence, t = 3.647 s. Option E.
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Answer:
18.7842493212 W
Explanation:
T = Tension = 1871 N
= Linear density = 3.9 g/m
y = Amplitude = 3.1 mm
= Angular frequency = 1203 rad/s
Average rate of energy transfer is given by
The average rate at which energy is transported by the wave to the opposite end of the cord is 18.7842493212 W
Answer:
a) see attached, a = g sin θ
b)
c) v = √(2gL (1-cos θ))
Explanation:
In the attached we can see the forces on the sphere, which are the attention of the bar that is perpendicular to the movement and the weight of the sphere that is vertical at all times. To solve this problem, a reference system is created with one axis parallel to the bar and the other perpendicular to the rod, the weight of decomposing in this reference system and the linear acceleration is given by
Wₓ = m a
W sin θ = m a
a = g sin θ
b) The diagram is the same, the only thing that changes is the angle that is less
θ' = 9/2 θ
c) At this point the weight and the force of the bar are in the same line of action, so that at linear acceleration it is zero, even when the pendulum has velocity v, so it follows its path.
The easiest way to find linear speed is to use conservation of energy
Highest point
Em₀ = mg h = mg L (1-cos tea)
Lowest point
Emf = K = ½ m v²
Em₀ = Emf
g L (1-cos θ) = v² / 2
v = √(2gL (1-cos θ))
