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
Gradpoint? The answer is B
Answer:
A car moves up a hill at a constant velocity
Explanation:
Since the velocity is constant, the speed is also constant and so is the kinetic energy. However, total mechanical energy is sum of gravitational potential energy and kinetic energy, and the car is moving up the hill so its potential energy rises.
Thus, in the circumstances described the mechanical energy cannot be conserved.
The correct answer is A car moving up the hill with constant velocity.
North: 1m
South: 0,8m
Direction:
1m>0,8m
so mouse moved north
Distance:
1m-0,8m=0,2m
so mouse traveled 0,2m
Answer: The mouse moved 0,2m north.
"Non nobis Domine, non nobis, sed Nomini tuo da gloriam."
Regards M.Y.
Answer:
Va = 5000 m / 3600 s = 1.39 m/s
(Va - Vb) 60 = 10
Vb = Va - .167 = 1.22 m/s
(Va - Vb) T = 4200 Π where T is time for A to complete 1 more lap
.17 T = 4200 Π
T = 24700 Π time for A to again catch B
N = 1.39 * 24700 Answer:
Va = 5000 m / 3600 s = 1.39 m/s
(Va - Vb) 60 = 10
Vb = Va - .167 = 1.22 m/s
(Va - Vb) T = 4200 Π where T is time for A to complete 1 more lap
.17 T = 4200 Π
T = 24700 Π time for A to again catch B
N = 1.39 * 24700 Π / (4200 Π) = 8.2 laps
A will make 8 but not 9 rounds before catching B
