Up to a point, the elongation of a spring is directly proportional to the force applied to it. Once you extend the spring more t
2 answers:
You might be interested in
Refer to the diagram shown below.
Neglect wind resistance, and use g = 9.8 m/s².
The pole vaulter falls with an initial vertical velocity of u = 0.
If the velocity upon hitting the pad is v, then
v² = 2*(9.8 m/s²)*(4.2 m) = 82.32 (m/s)²
v = 9.037 m/s
The pole vaulter comes to res after the pad compresses by 50 cm (or 0.5 m).
If the average acceleration (actually deceleration) is (a m/s²), then
0 = (9.037 m/s)² + 2*(a m/s²)*(0.5 m)
a = - 82.32/(2*0.5) = - 82 m/s²
Answer: - 82 m/s² (or a deceleration of 82 m/s²)
Neglect wind resistance, and use g = 9.8 m/s².
The pole vaulter falls with an initial vertical velocity of u = 0.
If the velocity upon hitting the pad is v, then
v² = 2*(9.8 m/s²)*(4.2 m) = 82.32 (m/s)²
v = 9.037 m/s
The pole vaulter comes to res after the pad compresses by 50 cm (or 0.5 m).
If the average acceleration (actually deceleration) is (a m/s²), then
0 = (9.037 m/s)² + 2*(a m/s²)*(0.5 m)
a = - 82.32/(2*0.5) = - 82 m/s²
Answer: - 82 m/s² (or a deceleration of 82 m/s²)
Answer:
I = 2 kgm^2
Explanation:
In order to calculate the moment of inertia of the door, about the hinges, you use the following formula:
(1)
I: moment of inertia of the door
α: angular acceleration of the door = 2.00 rad/s^2
τ: torque exerted on the door
You can calculate the torque by using the information about the Force exerted on the door, and the distance to the hinges. You use the following formula:
(2)
F: force = 5.00 N
d: distance to the hinges = 0.800 m
You replace the equation (2) into the equation (1), and you solve for α:
Finally, you replace the values of all parameters in the previous equation for I:
The moment of inertia of the door around the hinges is 2 kgm^2