Answer:
28 cm and 32 cm
Explanation:
1. The spring pendulum hangs vertically, oscillates harmonic with amplitude 2cm and angular frequency 20 rad/s. The natural length of
a spring is 30cm. What is the minimum and maximum length of the spring during the oscillation? Take g = 10m/s2.
As the amplitude is 2 cm and the natural length is 30 cm. So, it oscillates between 30 -2 = 28 cm to 30 + 2 = 32 cm.
So, the minimum length is 28 cm and the maximum length is 32 cm.
Answer:
t = 0.657 s
Explanation:
First, let's use the appropiate equations to solve this:
V = √T/u
This expression gives us a relation between speed of a disturbance and the properties of the material, in this case, the rope.
Where:
V: Speed of the disturbance
T: Tension of the rope
u: linear density of the rope.
The density of the rope can be calculated using the following expression:
u = M/L
Where:
M: mass of the rope
L: Length of the rope.
We already have the mass and length, which is the distance of the rope with the supports. Replacing the data we have:
u = 2.31 / 10.4 = 0.222 kg/m
Now, replacing in the first equation:
V = √55.7/0.222 = √250.9
V = 15.84 m/s
Finally the time can be calculated with the following expression:
V = L/t ----> t = L/V
Replacing:
t = 10.4 / 15.84
t = 0.657 s
Answer:
The material with higher modulus will stretch less than
The material with lower modulus
Explanation:
A material with a higher modulus is stiffer and has better resistance to deformation. The modulus is defined as the force per unit area required to produce a deformation or in other words the ratio of stress to strain.
E= stress/stain
Hooks law states that provided the elastic limit is not exceeded the extension e of a spring is directly proportional to the load or force attached
F=ke
Where k is the constant which gives the measure of the spring under tension
Answer:
The pressure on the floor is 12 Pascal(Pa).
Explanation:
Given,
Force (F) = 240N
Area (A) = 20 cm^2
Pressure (P) = ?
we know that,
P = F/A
= 240/ 20
= 12 Pa
