The plum pudding model, which has been abandoned since the discovery of the nucleus, stated that electrons were embedded in a "mush" of positive material. The nuclear model says they are placed around a central nucleus.
Answer:
271.862 N/m
Explanation:
From Hook's Law,
mgh = 1/2ke²............... Equation 1
Where
m = mass of the ball, g = acceleration due to gravity, k = spring constant, e = extension, h = height fro which the ball was dropped.
Making k the subject of the equation,
k =2mgh/k²....................... Equation 2
Note: The potential energy of the ball is equal to the elastic potential energy of the spring.
Given: m = 60.3 g = 0.0603 kg, g = 9.8 m/s², e = 4.68317 cm = 0.0468317 m, h = 53.7 cm = 0.537 m
Substitute into equation 2
k = 2(0.0603)(9.8)(0.537)/0.048317²
k = 0.6346696/0.0023345
k = 271.862 N/m
Answer:
a) 
b) 
c) 
Explanation:
From the question we are told that:
Given Frequencies
a. 100 Hz,
b. 1 kHz,
c. 100 kHz.
Generally the equation for Waveform Period is mathematically given by
Therefore
a)
For
b)
For
c)
For
<span>No, because the truck applies more pressure than the bridge can support.</span>
Thank you for your question, what you say is true, the gravitational force exerted by the Earth on the Moon has to be equal to the centripetal force.
An interesting application of this principle is that it allows you to determine a relation between the period of an orbit and its size. Let us assume for simplicity the Moon's orbit as circular (it is not, but this is a good approximation for our purposes).
The gravitational acceleration that the Moon experience due to the gravitational attraction from the Earth is given by:
ag=G(MEarth+MMoon)/r2
Where G is the gravitational constant, M stands for mass, and r is the radius of the orbit. The centripetal acceleration is given by:
acentr=(4 pi2 r)/T2
Where T is the period. Since the two accelerations have to be equal, we obtain:
(4 pi2 r) /T2=G(MEarth+MMoon)/r2
Which implies:
r3/T2=G(MEarth+MMoon)/4 pi2=const.
This is the so-called third Kepler law, that states that the cube of the radius of the orbit is proportional to the square of the period.
This has interesting applications. In the Solar System, for example, if you know the period and the radius of one planet orbit, by knowing another planet's period you can determine its orbit radius. I hope that this answers your question.
