Select the correct answer. Eli and Andy want to find out which of the two is stronger. Eli pushes a table with a force of 120 ne
2 answers:
Acceleration of the table: B. 0.50 meters/second2
Explanation:
The problem can be solved by using Newton's second law of motion, which states that the net force acting on an object is the product of its mass and its acceleration. Mathematically:
where
is the net force
m is the mass
a is the acceleration
For the table in this problem, we have:
is the net force on the table, because there are two forces of 125 N and 120 N acting in opposite directions
m = 10.0 kg is the mass of the table
Solving for a, we find the acceleration:
Learn more about Newton's second law:
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Answer: The electron number density (the number of electrons per unit volume) in the wire is .
Explanation:
Given: Current = 5.0 A
Area =
Density = 2.7 , Molar mass = 27 g
The electron density is calculated as follows.
where,
= density
M = molar mass
= Avogadro's number
Substitute the values into above formula as follows.
Thus, we can conclude that the electron number density (the number of electrons per unit volume) in the wire is .
Answer:
x(t)=0.337sin((5.929t)
Explanation:
A frictionless spring with a 3-kg mass can be held stretched 1.6 meters beyond its natural length by a force of 90 newtons. If the spring begins at its equilibrium position, but a push gives it an initial velocity of 2 m/sec, find the position of the mass after t seconds.
Solution. Let x(t) denote the position of the mass at time t. Then x satisfies the differential equation
Definition of parameters
m=mass 3kg
k=force constant
e=extension ,m
ω =angular frequency
k=90/1.6=56.25N/m
ω^2=k/m= 56.25/1.6
ω^2=35.15625
ω=5.929
General solution will be
differentiating x(t)
dx(t)=-5.929c1sin(5.929t)+5.929c2cos(5.929t)
when x(0)=0, gives c1=0
dx(t0)=2m/s gives c2=0.337
Therefore, the position of the mass after t seconds is
x(t)=0.337sin((5.929t)