Answer:
P₂ = 1.22 kPa
Explanation:
This problem can be solved using the equation of state:
where,
P₁ = initial pressure = 1 KPa
P₂ = final pressure = ?
V₁ = initial Volume = 1 liter
V₂ = final volume = 1.1 liter
T₁ = initial temperature = 290 k
T₂ = final temperature = 390 k
Therefore,
<u>P₂ = 1.22 kPa</u>
The spring is initially stretched, and the mass released from rest (v=0). The next time the speed becomes zero again is when the spring is fully compressed, and the mass is on the opposite side of the spring with respect to its equilibrium position, after a time t=0.100 s. This corresponds to half oscillation of the system. Therefore, the period of a full oscillation of the system is
Which means that the frequency is
and the angular frequency is
In a spring-mass system, the maximum velocity of the object is given by
where A is the amplitude of the oscillation. In our problem, the amplitude of the motion corresponds to the initial displacement of the object (A=0.500 m), therefore the maximum velocity is
Answer
given,
mass of copper rod = 1 kg
horizontal rails = 1 m
Current (I) = 50 A
coefficient of static friction = 0.6
magnetic force acting on a current carrying wire is
F = B i L
Rod is not necessarily vertical
the normal reaction N = mg-F y
static friction f = μ_s (mg-F y )
horizontal acceleration is zero
B_w = B sinθ
B_d = B cosθ
iLB cosθ= μ_s (mg- iLB sinθ)
B = 0.1 T