Refer to the diagram below.
Ignoring air resistance, use gravitational acceleration g = 9.8 m/s².
The pole vaulter drops with an initial vertical speed u = 0.
At impact with the pad, velocity v satisfies:
v² = 2 × (9.8 m/s²) × (4.2 m) = 82.32 (m/s)²
v = 9.037 m/s
As the pad compresses by 0.5 m to bring the vaulter to rest,
let the average acceleration (deceleration) be a m/s². Then:
0 = (9.037 m/s)² + 2 × a × 0.5 m
Solving for a gives:
a = - 82.32 / (2 × 0.5) = -82 m/s²
Thus, the deceleration magnitude is 82 m/s².
Answer:
The duration, t = 3.53 seconds
Explanation:
The following information is provided:
The equation to calculate the time t is expressed as:
...... (1)
Where
s denotes the distance in feet
We are to determine the duration taken by the stone to fall a distance of 200 feet, where s = 200 feet
Substituting the value of s into equation (1) yields:
t = 3.53 seconds
Thus, the time taken by the object is 3.53 seconds, which provides the required answer.
Answer:
The pen requires 7.2 mJ of energy to extend.
Explanation:
Provided:
Length = 1.8 cm
Spring constant = 300 N/m
Initial compression = 1.0 mm
Additional compression = 6.0 mm
Total compression = 1.0 + 6.0 = 7.0 mm
We need to determine the energy needed
This energy is equivalent to the variation in spring potential energy
Substitute the values into the formula
Therefore, a total of 7.2 mJ is needed to extend the pen.
Answer:
A rock weighing 50kg should be positioned at a distance of 0.5m from the pivot of the seesaw.
Explanation:
τchild=τrock
We will utilize the formula for torque:
(F)child(d)child)=(F)rock(d)rock)
The gravitational force acts equally on both objects.
(m)childg(d)child)=(m)rockg(d)rock)
We can eliminate gravity from both sides of the equation for simplification.
(m)child(d)child)=(m)rock(d)rock)
Now employing the given masses for the rock and child. The seesaw's total length is 2 meters, with the child sitting at one end, placing them 1 meter from the center of the seesaw.
(25kg)(1m)=(50kg)drock
Solve for the distance where the rock should be positioned in relation to the seesaw's center.
drock=25kg⋅m50kg
drock=0.5m
A. The horizontal component of velocity is
vx = dx/dt = π - 4πsin(4πt + π/2)
vx = π - 4πsin(0 + π/2)
vx = π - 4π(1)
vx = -3π
b. vy = 4πcos(4πt + π/2)
vy = 0
c. m = sin(4πt + π/2) / [πt + cos(4πt + π/2)]
d. m = sin(4π/6 + π/2) / [π/6 + cos(4π/6 + π/2)]
e. t = -1.0
f. t = -0.35
g. To find t, set
vx = π - 4πsin(4πt + π/2) = 0
Then use this to calculate vxmax
h. To determine t, set
vy = 4πcos(4πt + π/2) = 0
Then use this to find vymax
i. s(t) = [x(t)^2 + y(t)^2]^(1/2)
h. s'(t) = d[x(t)^2 + y(t)^2]^(1/2) / dt
k and l. Determine the values for t
d[x(t)^2 + y(t)^2]^(1/2) / dt = 0
And substitute to find both the maximum and minimum speeds.
