Answer:
2.98 m/s^2
Explanation:
I have done this before and it was a question on my physics test
Answer:
Explanation:
A ) When gymnast is motionless , he is in equilibrium
T = mg
= 63 x 9.81
= 618.03 N
B )
When gymnast climbs up at a constant rate , he is still in equilibrium ie net force acting on it is zero as acceleration is zero.
T = mg
= 618.03 N
C ) If the gymnast climbs up the rope with an upward acceleration of magnitude 0.600 m/s2
Net force on it = T - mg , acting in upward direction
T - mg = m a
T = mg + m a
= m ( g + a )
= 63 ( 9.81 + .6)
= 655.83 N
D ) If the gymnast slides down the rope with a downward acceleration of magnitude 0.600 m/s2
Net force acting in downward direction
mg - T = ma
T = m ( g - a )
= 63 x ( 9.81 - .6 )
= 580.23 N
Answer:
a)
b)
c)
d)
e)
Explanation:
Given that:
- initial speed of turntable,

- full speed of rotation,

- time taken to reach full speed from rest,

- final speed after the change,

- no. of revolutions made to reach the new final speed,

(a)
∵ 1 rev = 2π radians
∴ angular speed ω:

where N = angular speed in rpm.
putting the respective values from case 1 we've


(c)
using the equation of motion:

here α is the angular acceleration



(b)
using the equation of motion:





(d)
using equation of motion:



(e)
using the equation of motion:


