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
<span> Use the Law of Cosines, where you have a triangle with included angle of 145 degrees and sides of 16 and 18. You are then solving the equation: </span>
<span>d^2 = 16^2 + 18^2 - 2(16)(18)cos(145) </span>
Answer:
mass is lifted 1.8 m. What is the potential energy of the mass 4. A 100 kg
Answer: Examples of conductors include metals, aqueous solutions of salts (i.e., ionic compounds dissolved in water), graphite, and the human body. Examples of insulators include plastics, Styrofoam, paper, rubber, glass and dry air.
Answer:
N = 337.96 N
Explanation:
∅ = 32º
F = 249 N
m = 21 Kg
N = ?
We can apply:
∑ F = 0 (↑)
- Fy - W + N = 0 ⇒ N = Fy + W
⇒ F*Sin ∅ + m*g = N
⇒ N = (249 N*Sin32º) + (21 Kg*9.81 m/s²)
⇒ N = 337.96 N (↑)
