Answer:
It is always less than 1 because the load arm is always longer than effort arm.
Explanation:
In the formula, MA= effort divided by load which makes it less than 1.
This helps by reducing the applied force(effort). It is a speed multiplier.
Hope it helps.
Answer:
To find the acceleration of the object we have to apply Newton second law of motion that is F = mass × acceleration.
Explanation:
Given ,
F = 130N
M = 24kg
A = ?
F = m× a
then ,
130N = 24kg ×a
a = 130/24 = 5 m/s.
Answer:
0.103 m/s to the south
Explanation:
The total momentum of the launcher+ball system must be conserved before and after the launch, so we can write:
![p_i = p_f\\0 = m_L v_L + m_B v_B](https://tex.z-dn.net/?f=p_i%20%3D%20p_f%5C%5C0%20%3D%20m_L%20v_L%20%2B%20m_B%20v_B)
where
is the total initial momentum (before the launch)
is the mass of the launcher
is the velocity of the launcher after the launch
is the mass of the ball
is the velocity of the ball after the launch (we take the north direction as positive)
Solving for
, we find
![v_L = -\frac{m_B v_B}{m_L}=-\frac{(0.057 kg)(+36 m/s)}{20 kg}=-0.103 m/s](https://tex.z-dn.net/?f=v_L%20%3D%20-%5Cfrac%7Bm_B%20v_B%7D%7Bm_L%7D%3D-%5Cfrac%7B%280.057%20kg%29%28%2B36%20m%2Fs%29%7D%7B20%20kg%7D%3D-0.103%20m%2Fs)
and the negative sign means that the direction is south.
Answer:
the static charge is not always distributed on the surface of the conductor, there are also charges in the volume but of lesser magnitude
Explanation:
In this hypothetical system the electric force is of type
F =
in this case the force decays to zero much faster,
if we call Fo the force of Coulomb's law
F₀ = ![k \frac{q_1 q_2 }{r^2}](https://tex.z-dn.net/?f=k%20%5Cfrac%7Bq_1%20q_2%20%7D%7Br%5E2%7D)
assuming the constant k is the same
the relationship between the two forces is
F / F₀ = 1 / r
F = F₀ / r
when analyzing this expression the force decays much faster to zero.
In an electric conductor, charges of the same sign may not feel any repulsive force from other charges that are at a medium distance, so there is a probability that some charges are distributed in the volume of the material, this does not happen with coulomb's law
Consequently, the static charge is not always distributed on the surface of the conductor, there are also charges in the volume but of lesser magnitude
D. because they aren't mixed up evenly