PV=nRT
(P)(.010)=(n)(.08201)(0)
(v1/t1)=(v2/t2)
(.010/t1)=(v2/0)
The volume would be zero
Answer:
Angular velocity is same as frequency of oscillation in this case.
ω = 
x ![[\frac{L^{2}}{mK}]^{3/14}](https://tex.z-dn.net/?f=%5B%5Cfrac%7BL%5E%7B2%7D%7D%7BmK%7D%5D%5E%7B3%2F14%7D)
Explanation:
- write the equation F(r) = -K
with angular momentum <em>L</em>
- Get the necessary centripetal acceleration with radius r₀ and make r₀ the subject.
- Write the energy of the orbit in relative to r = 0, and solve for "E".
- Find the second derivative of effective potential to calculate the frequency of small radial oscillations. This is the effective spring constant.
- Solve for effective potential
- ω = x
Answer:
0.31 m
Explanation:
m = mass of the block = 1.5 kg
H = height from which the block is released on ramp = 0.81 m
k = spring constant of the spring = 250 N/m
x = maximum compression of the spring
using conservation of energy
Spring potential energy gained by spring = Potential energy lost by block
(0.5) k x² = mgH
(0.5) (250) x² = (1.5) (9.8) (0.81)
x = 0.31 m
Answer:
86605.08 N
Explanation:
The equation to calculate the force is:
Force = mass * acceleration
The force and the acceleration does not have the same direction in this case, so we need to decompose the force into its horizontal component, which is the force that will generate the horizontal acceleration:
Force_x = Force * cos(30)
Then, we have that:
Force_x = mass * acceleration
Force * cos(30) = 25000 * 3
Force * 0.866 = 75000
Force = 75000 / 0.866 = 86605.08 N
<h2>
Answer: 13.61 N/m</h2>
Hooke's law establishes that the elongation of a spring is directly proportional to the modulus of the force 
applied to it, <u>as long as the spring is not permanently deformed</u>:
(1)
Where:
is the elastic constant of the spring. The higher its value, the more work it will cost to stretch the spring.
is the length of the spring without applying force.
is the length of the spring with the force applied.
According to this, we have a spring where only the force due gravity is applied.
In other words, the force applied is the weigth 
of the block:
(2)
Where 
is the mass of the block and 
is the gravity acceleration.
(3)
(4)
Knowing the force applied and 
and 
, we can substitute the values in equation (1) and find :
(5)
(6)
<u>Finally:</u>
