Answer:
40 m/s
Explanation:
given,
height of the fall, h = 82 m
time taken to fall, t = 1.3 s
rock velocity, v = ?
acceleration due to gravity, g = 9.8 m/s²
rock is released initial velocity, u = 0 m/s
using equation of motion
v² = u² + 2 a s
v² = 0 + 2 x 9.8 x 82
v² = 1607.2
v = 40 m/s
hence, rock's velocity is equal to 40 m/s
Answer:a) 34.5 N; b) 24.5 N; c) 10 N; d) 1J
Explanation: In order to solve this problem we have to used the second Newton law given by:
∑F= m*a
F-f=m*a where f is the friction force (uk*Normal), from this we have
F= m*a+f=5 Kg*2 m/s^2+0.5*5Kg*9.8 m/s^2= 34.5 N
then f=uk*N=0.5*5Kg*9.8 m/s^2= 24.5N
the net Force = (34.5-24.5)N= 10 N
Finally the work done by the net force is equal to kinetic energy change so
W=∫Force net*dr= 10 N* 0.1 m= 1J
B) Acceleration is directly proportional to the mass of the object
Answer:
so 9/3=3 current is 3 amperes
Explanation:
