C because they are both going in a constant speed
Answer: hello below is the missing part of your question
A mass m = 10 kg rests on a frictionless table and accelerated by a spring with spring constant k = 5029 N/m. The floor is frictionless except for a rough patch. For this rough path, the coefficient of friction is μk = 0.49. The mass leaves the spring at a speed v = 3.4 m/s.
answer
x = 0.0962 m
Explanation:
<em>First step :</em>
Determine the length of the rough patch/spot
F = Uₓ (mg)
and w = F.d = Uₓ (mg) * d
hence;
d( length of rough patch) = w / Uₓ (mg) = 46.55 / (0.49 * 10 * 9.8) = 0.9694 m
<em>next : </em>
work done on unstretched spring length
Given that block travels halfway i.e. d = 0.9694 / 2 = 0.4847 m
w' = Uₓ (mg) * d
= 0.49 * 10 * 9.81 * 0.4847 = 23.27 J
also given that the Elastic energy of spring = work done ( w')
1/2 * kx^2 = 23.27 J
x = 
= 0.0962 m
Trimming the mold of and putting it on a weight scale
Answer:
39.2m/s
Explanation:
Given parameters:
Mass of book = 5kg
Time taken for fall = 4s
Unknown:
Final velocity of the book = ?
Solution:
Serena dropped the book from rest therefore, the initial velocity of the book is 0.
Let us find the appropriate motion equation to solve this problem;
V = U + gt
V is the final velocity
U is the initial velocity
g is the acceleration due to gravity = 9.8m/s²
t is the time taken
Insert the given parameters and solve;
V = 0 + 9.8 x 4 = 39.2m/s
Answer:
The gravitational force between them increases by a factor of 4
Explanation:
Gravitational force is a force of attraction between two objects with masses M and m which are separated by a distance R. It is given mathematically as:
Fg = GMm/R²
Where G = Gravitational constant.
If the distance between their centers, R, decreases by a factor of 2, then it means the new distance between their centers is:
r = R/2
Hence,the gravitational force becomes:
Fg = GMm/r²
Fg = GMm/(R/2)²
Fg = GMm/(R²/4)
Fg = 4GMm/R²
Hence,the gravitational force increases by a factor of 4.
