The position of a 50 g oscillating mass is given by x(t)=(2.0cm)cos(10t−π/4), where t is in s. If necessary, round your answers
1 answer:
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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:
Explanation:
As we know that the magnetic field near the center of solenoid is given as
now we know that initially the length of the solenoid is L = 18 cm and N number of turns are wounded on it
So the magnetic field at the center of the solenoid is 2 mT
now we pulled the coils apart and the length of solenoid is increased as L = 21 cm
so we have
now plug in all values in it
Answer:
T = 60 s
Explanation:
There are 6 poles on the track which are equally spaced
so the angular separation between the poles is given as
so the angular speed of the train is given as
now we have time period of the train given as