Answer:a) 34.5 N; b) 24.5 N; c) 10 N; d) 1J
Explanation: In order to solve this problem we have to used the second Newton law given by:
∑F= m*a
F-f=m*a where f is the friction force (uk*Normal), from this we have
F= m*a+f=5 Kg*2 m/s^2+0.5*5Kg*9.8 m/s^2= 34.5 N
then f=uk*N=0.5*5Kg*9.8 m/s^2= 24.5N
the net Force = (34.5-24.5)N= 10 N
Finally the work done by the net force is equal to kinetic energy change so
W=∫Force net*dr= 10 N* 0.1 m= 1J
Answer:
D. The temperature does not change during a phase change because the average kinetic energy does not change. Therefore, the potential energy in the bonds between molecules must change.
Explanation:
When there is a change of state (for example, from solid into a liquid, as in this example), when energy is added to the system, the temperature of the substance does not change.
The reason for this is that the energy supplied is no longer used to increase the average kinetic energy of the particle, but instead it is used to break the bonds between the different particles/molecules. For instance, since in this case the substance is changing from solid to liquid, all the energy supplied during the phase change is used to break the bonds between the molecules of the solid: when the process is done, all the molecules will be free to slide past each other, and the substance has turned completely into a liquid.
The bonds between molecules store potential energy: therefore, this means that the energy supplied during the phase change is not used to change the kinetic energy, but to change the potential energy in the bonds between the molecules.
The question is incomplete.
The distance between the Moon and Earth influences: 1) the attractive gravitational force between them, 2) the tides, 3) the eclipses, 4) the period of each full turn of the moon around the Earth.
Assuming the question refers to the gravitational attraction, we must use the fact that, as per, Newton's Universal Gravitaional Law, the attractive force between the two bodies is inversely related to the square distance that separates them.
Then, if the Moon were twice as far, the gravitational pull would be one fourth (1/4) of actual pull.
Gravitational field exists in
the space surrounding a charged particle and exerts a force on other charged
particles. Gravitational waves are ripples of waves travelling outward from the
source. The more massive the orbit of two bodies, the more it emits
gravitational wave. And everything around it that is near within the wave
experiences a ‘pull’ toward the orbiting bodies.
Answer:
i. The radius 'r' of the electron's path is 4.23 × 
m.
ii. The frequency 'f' of the motion is 455.44 KHz.
Explanation:
The radius 'r' of the electron's path is called a gyroradius. Gyroradius is the radius of the circular motion of a charged particle in the presence of a uniform magnetic field.
r = 
Where: B is the strength magnetic field, q is the charge, v is its velocity and m is the mass of the particle.
From the question, B = 1.63 × T, v = 121 m/s, Θ = 
(since it enters perpendicularly to the field), q = e = 1.6 × 
C and m = 9.11 × 
Kg.
Thus,
r = ÷ sinΘ
But, sinΘ = sin = 1.
So that;
r =
= (9.11 × × 121) ÷ (1.6 × × 1.63 × )
= 1.10231 × 
÷ 2.608 × 
= 4.2266 ×
= 4.23 × m
The radius 'r' of the electron's path is 4.23 × m.
B. The frequency 'f' of the motion is called cyclotron frequency;
f = 
= (1.6 × × 1.63 × ) ÷ (2 ×
× 9.11 × )
= 2.608 × ÷ 5.7263 × 
= 455442.4323
f = 455.44 KHz
The frequency 'f' of the motion is 455.44 KHz.
