B. By vibrations in wires or strings
Answer:
Ptolemy proposed a model, he reference system is centered on the Earth
Copernicus, proposed a deferent system, this system is centered on the Sun, where it is at the origin of the system
Explanation:
Thousands of years ago, Ptolemy proposed a model to explain the movement of the planets and stars in the sky, in this model the reference system is centered on the Earth, so each body is orbiting in different spheres around the Earth as its center, this system had very complicated calculations and curves to be able to explain the orbits of the planets.
More recently Copernicus, proposed a deferent system, this system is centered on the Sun, where it is at the origin of the system, in this system the movement of the planets are ellipses, which is a much simpler explanation and has been widely accepted, in current systems the reference system is fixed in the bodies more massive, since this simplifies the explanation of the movements.
Answer:
31302 Volts and 55/111 Amps (≈0.5)
Explanation:
Secondary voltage / 141 = 1110 / 5
Secondary voltage = (1110*141) / 5
Secondary voltage = 31302
Amperage = 110/ (31302/141) = 55/111
Answer:
C. g/cm³
Explanation:
The slope is measured by calculating the variation of the Y values over the X values between two points on a line.
So, the formula is: Slope = Δy/Δx
That means that we also take the units.
In this case, the Y-axis unit is in g, while the X-axis unit is in cm³.
Dividing a Y-variation over an X-variation will give you g/cm³.
In this case, let's assume the line passes through (10,100) (not exactly, but close enough for the example), and it passes through (0,0)
So the slope would be: (100-0) g / (10-0) cm³ = 10 g/cm³
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)
