Answer: The spring constant is K=392.4N/m
Explanation:
According to hook's law the applied force F will be directly proportional to the extension e produced provided the spring is not distorted
The force F=ke
Where k=spring constant
e= Extention produced
h=2m
Given that
e=20cm to meter 20/100= 0.2m
m=100g to kg m=100/1000= 0.1kg
But F=mg
Ignoring air resistance
assuming g=9.81m/s²
Since the compression causes the plastic ball to poses potential energy hence energy stored in the spring
E=1/2ke²=mgh
Substituting our values to find k
First we make k subject of formula
k=2mgh/e²
k=2*0.1*9.81*2/0.1²
K=3.921/0.01
K=392.4N/m
Answer:
the velocity of the bullet-wood system after the collision is 2.48 m/s
Explanation:
Given;
mass of the bullet, m₀ = 20 g = 0.02 kg
velocity of the bullet, v₀ = 250 m/s
mass of the wood, m₁ = 2 kg
velocity of the wood, v₁ = 0
Let the velocity of the bullet-wood system after collision = v
Apply the principle of conservation of linear momentum to calculate the final velocity of the system;
Initial momentum = final momentum
m₀v₀ + m₁v₁ = v(m₀ + m₁)
0.02 x 250 + 2 x 0 = v(2 + 0.02)
5 + 0 = v(2.02)
5 = 2.02v
v = 5/2.02
v = 2.48 m/s
Therefore, the velocity of the bullet-wood system after the collision is 2.48 m/s
Answer:
The average atomic mass is 79.91 amu.
Explanation:
Since
Atomic mass can be find by Multiplying the relative abundance of each isotope by its atomic mass, then add them together to get the atomic mass of the element.
so
Atomic mass = (0.5069)(78.92 amu) + (0.4931)(80.92 amu)
=79.91 amu
So the Atomic mass of the bromine is 79.91amu.
The type of energy that depends on position is called
kinetic energy
Answer:
0.130
Explanation:
From the given data, the coefficient of static friction for each trial are:
1. 0.053
2. 0.081
3. 0.118
4. 0.149
5. 0.180
6. 0.198
The sum of the coefficient of static friction = 0.053 + 0.081 + 0.118 + 0.149 + 0.180 + 0.198
= 0.779
So that;
the average coefficient of static friction = 
= 
= 0.12983
The average coefficient of static friction is 0.130
