To solve this problem we will apply the two concepts mentioned. To find the constant we will apply Hooke's law, and to find the period we will apply the relationship between the mass and the spring constant. Let us begin,
PART A) For this section we will use Hooke's law. In turn, since the force applied is equivalent to weight, we will use Newton's law for which weight is defined as the product between mass and gravity. This weight is equal to the Spring Force.
Here,
k = Spring constant
= Displacement
F = Force, the same as the Weight (mg)
Then we have that
Therefore the spring constant is 13.23N/m
PART B) To find the period of oscillation, the relationship that allows us to find is given by the following mathematical function,
Here
m = mass
k = Spring constant
Replacing,
Therefore the period of the oscillation is 0.491s
Answer:
T= 8.061N*m
Explanation:
The first thing to do is assume that the force is tangential to the square, so the torque is calculated as:
T = Fr
where F is the force, r the radius.
if we need the maximum torque we need the maximum radius, it means tha the radius is going to be the edge of the square.
Then, r is the distance between the edge and the center, so using the pythagorean theorem, r i equal to:
r = 
r = 0.5374m
Finally, replacing the value of r and F, we get that the maximun torque is:
T = 15N(0.5374m)
T= 8.061N*m
C is the answer I think good luck
Answer:
22 N upward
Explanation:
From the question,
Applying newton's second law of motion
F = m(v-u)/t....................... Equation 1
Where F = Average force exerted by the ground on the ball, m = mass of the baseball, v = final velocity, u = initial velocity, t = time of contact
Note: Let upward be negative and downward be positive
Given: m = 0.14 kg, v = -1.00 m/s, u = 1.2 m/s, t = 0.014 s
Substitute into equation 1
F = 0.14(-1-1.2)/0.014
F = 0.14(-2.2)/0.014
F = 10(-2.2)
F = -22 N
Note the negative sign shows that the force act upward
The average it the constant speed.
