A=mgh
m=300g=0.3kg
g=9,81 m/s^2
h=10m
A=29.43J
<h2>Acceleration due to gravity in moon is 1.5 m/s²</h2>
Explanation:
We have equation of motion s = ut + 0.5 at²
Here the ball travels 3 m less distance in fifth second compared to third second.
That is
s₃ = s₅ + 3
Now we have
Distance traveled in third second, s₃ = u x 3 - 0.5 x g x 3² - u x 2 - 0.5 x g x 2²
s₃ = u - 2.5 g
Also
Distance traveled in fifth second, s₅ = u x 5 - 0.5 x g x 5² - u x 4 - 0.5 x g x 4²
s₅ = u - 4.5 g
That is
u - 2.5 g = u - 4.5 g + 3
2 g = 3
g = 1.5 m/s²
Acceleration due to gravity in moon = 1.5 m/s²
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
Answer:
it is 27 degree.
Explanation:
because angle of reflection is equal to angle of Inc dence
Answer:
Explanation:
KE = ½mv² = ½(6.8)8² = 217.6 J
round as appropriate because that result is way too much precision for the inputs provided. Arguably should be 200 J based on the single significant digit of the velocity.
