It’s definitely the 3rd one
Answer:
c. The coefficient of kinetic friction is less than the coefficient of static friction
Explanation:
When the box finally does break loose. Then the component of the box weight which is parallel to the board weight parallel component, is equal to the .
For the box to acce;erate thee must be non-zero net force acting on the box parallel to the board. Or we can say,
Therefore the force of kinetic friction must be less than the force of static friction. Thus,
Answer:
discrete lines are observed by the spectroscope, the emission of the lamp is of the ATOMIC source
Explanation:
Bulbs can emit light in several ways:
* When the emission is carried out by the heating of its filament, the bulb is called incandescent, in general its spectrum is similar to that of a black body, this is a continuous spectrum with a maximum dependent on the fourth power of the temperature of the filament.
* The emission can be by atomic transitions, in this case there is a discrete spectrum formed by the spectral lines of the material that forms the gas of the lamp, in general for the yellow emission the most used materials are mercury and sodium or a mixture of they.
Consequently, as discrete lines are observed by the spectroscope, the emission of the lamp is of the ATOMIC type
I think it’s C sorry if it’s wrong :(
Answer:
a₃ = -1.08 m/s², K = 1.42 J
Explanation:
The particle is in a periodic motion, so the general expression is
x = A cos (wt + Ф)
let's compare the terms with the expression they give us
x = 2 cos (t - 2)
the amplitude of motion is A = 2 m, the angular velocity w = 1 rad / s, and the phase is Ф = - 2.
to find the acceleration we use its definition
v = dx / dt
a = dv / dt
a =
let's perform the derivative
v = - A w sin (wt + Ф)
a = - A w² cos wt + Ф)
substituting the values
a = - 2 1² cos (t-2)
for t = 3 s
a₃ = 2 cos (3-2)
remember angles are in radians
a₃ = -1.08 m/s²
To calculate kinetic energy, let's find the velocity for t = 3 s
v = - 2 sin (t-2)
v = -2 sin (3-2)
v = - 1.683 m / s
body mass is m = 1 kg
we calculate
K = ½ m v²
K = ½ 1 (-1.683) ²
K = 1.42 J