La respuesta es "Un avion que vuela al norte con rapidez constante y altitud constante".
Para que la fuerza neta sea 0, la aceleracion debe ser 0, para esto la velocidad debe ser constante.
Para que la velocidad sea constante el objecto debe estar moviendo con rapidez (magnitud de la velocidad) constante y sin cambiar direccion; ya que la velocidad es un vector asi es que depende en magnitud y direccion.
En las demas opciones la magnitud de la velocidad (rapidez) cambia y/o la direccion.
You can write an hypothesis such as this:
The weight of an object has effects on the operating frictional force, the greater the weight, the higher the operating frictional force.
The father is the one with the higher weight while the son has the lower weight. The operating frictional force is the friction that their weights exert.
Answer:
e) electrons are transferred from the rod to the silk.
Explanation:
An initially neutral glass rod contains equal number of electron and proton.
If the rod becomes positively charged after being rubbed with silk, then the rod must have lost some its electron to the silk since electrons are more mobile than protons, leaving the rod with excess positive charge (protons), and the silk will be negatively charged (excess electron).
Thus, the rod becomes positively charged by transfer of electrons from rod to the silk.
e) electrons are transferred from the rod to the silk.
Density is mass divides by volume, so
89.6g / 10cm^3 =8.96g /cm^3
*cm^3 is a standard unit of volume*
Answer:
a) v = √ g x
, b) W = 2 m g d
, c) a = ½ g
Explanation:
a) For this exercise we use Newton's second law, suppose that the block of mass m moves up
T-W₁ = m a
W₃ - T = M a
w₃ - w₁ = (m + M) a
a = (3m - m) / (m + 3m) g
a = 2/4 g
a = ½ g
the speed of the blocks is
v² = v₀² + 2 ½ g x
v = √ g x
b) Work is a scalar, therefore an additive quantity
light block s
W₁ = -W d = - mg d
3m heavy block
W₂ = W d = 3m g d
the total work is
W = W₁ + W₂
W = 2 m g d
c) in the center of mass all external forces are applied, they relate it is
a = ½ g
