Answer:
<em>The force that would be applied on the rope just to start moving the wagon is 122 N</em>
Explanation:
Frictional force opposes motion between two surfaces in contact. It is the force that must be applied before a body starts to move. Static friction opposes the motion of two bodies that are in contact but are not moving. The magnitude of static friction to overcome for the body to move can be calculated using equation 1.
F = μ x mg .............................. 1
where F is the frictional force;
μ is the coefficient of friction ( μs, in this case, static friction);
m is mass of the object and;
g is the acceleration due to gravity( a constant equal to 9.81 m/
)
from the equation we are provide with;
μs = 0.25
m = 50 kg
g = 9.81 m/
F =?
Using equation 1
F = 0.25 x 50 kg x 9.81 m/
F = 122.63 N
<em>Therefore a force of 122 N must be applied to the rope just to start the wagon.</em>
The mass number of a nucleus (except Hydrogen) is: number of protons + number of neutrons.
A=Z+N
A=mass number=protons + neutrons.
Z=atomic number=number of protons.
N=number of neutrons.
In the case of Hydrogen it depends of isotope of Hydrogen .
the hydrogen has three isotopes,
protium : A=Z, because N=0
deuterium: A=Z+N; N=1
tritium: A=Z+N; N=2
Answer:
(a) The total energy of the object at any point in its motion is 0.0416 J
(b) The amplitude of the motion is 0.0167 m
(c) The maximum speed attained by the object during its motion is 0.577 m/s
Explanation:
Given;
mass of the toy, m = 0.25 kg
force constant of the spring, k = 300 N/m
displacement of the toy, x = 0.012 m
speed of the toy, v = 0.4 m/s
(a) The total energy of the object at any point in its motion
E = ¹/₂mv² + ¹/₂kx²
E = ¹/₂ (0.25)(0.4)² + ¹/₂ (300)(0.012)²
E = 0.0416 J
(b) the amplitude of the motion
E = ¹/₂KA²

(c) the maximum speed attained by the object during its motion
