Answer:
Half life of S = 3.76secs
Explanation:
The concept of half life in radioactivity is applied. Half life is the time taken for a radioactive material to decay to half of its initial size.
For part 1 - How much signal will be degraded in 1secs = 1/3.9 = 0.2564
for part 2 - How much signal will be degraded in 1secs = 1/104 = 0.009615
Simply say = 1/3.9 + 1/104 = 0.266015
So both part 1 and part 2 took 1/0.266015 = 3.76secs is the half life of S when both pathways are active
Answer: Dalton’s model
Explanation:
In the attached image we can see four atomic models labeled with four letters:
W represents the current and accepeted atomic model: a nucleus with an electron cloud, where the orbit and position of the electrons around the nucleus is defined by specific regions (associated with specific energy levels) where there is a greater probability of finding the electron at any given moment. It is important to note this model was improved by the works in quantum physics done by Louis de Broglie and Erwin Schrodinger.
X represents Rutherford's model (This model was proposed after Thomson's model). Ernest Rutherford conducted a series of experiments in order to corroborate Thomson's atomic model. However the results of the experiment led him to find out there is a concentration of charge in the atom's core (which was later called nucleus) surrounded by electrons. This lead to a new atomic model, in which the atom has a positive charged nucleus surrounded by negative charged particles that move similar to the orbit of the planet around the Sun.
Y represents Thomson's model, also called the <em>plum pudding</em> model. This scientific found out that atoms contain small subatomic particles with a negative charge (later called electrons). However, taking into consideration that at that time there was still no evidence of the atom nucleus, Thomson thought the electrons were immersed in the atom of positive charge that counteracted the negative charge of the electrons. Just like the raisins embedded in a pudding or bread.
Z represents Bohr's model. This model was proposed by the danish physicist Niels Bohr after Rutherford's model. In fact, this model was Rutherford's model with the following addition: electrons orbit the nucleus (like planets around the sun) in specific orbits at different energy levels around the nucleus.
So, the only missing model is <u>Dalton's model</u>, which was the first atomic model: the atom represented as a solid, indestructible and indivisible mass. An idea that was already accepted by that time since the ancient Greeks.
Question 1:
Answer:
The moment of inertia of Alex's rolling hoop is 0.197 
Explanation:
<u>Given</u>:
Mass of the hoop = 0.350 g
Radius of the hoop = 75.0 cm
<u>To Find:</u>
The moment of inertia of Alex's rolling hoop = ?
<u>Solution</u><u>:</u>
The moment of inertia = 
where
m is the mass
r is the radius
Converting cm to m, we get
75.0 cm = 0.75 m
Now substituting the values,
=> moment of inertia = 
=> moment of inertia = 
=> moment of inertia = 
Question 2:
Answer:
The combined angular momentum of the masses is 1.76 
If she pulls her arms in to 0.12 m, her new linear speed is 
Explanation:
Given:
Mass = 2.0 kg
Radius = 0.8 m
Velocity = 1.2 m/s
a.The combined angular momentum of the masses:

Substituting the values,

L= 1.76 
b. If she pulls her arms in to 0.12 m, what is her new linear speed




Answer:
295.42 N
Explanation:
From Newton's law of universal gravitation.
F = Gmm'/r².................. Equation 1
Where F = Gravitational force, G = Universal constant, m = mass of the human, m' = mass of mass, r = radius of mass.
Given: m = 80 kg, m' = 6.4×10²³ kg, r = 3.4×10⁶ m.
Constant: G = 6.67×10⁻¹¹ Nm²/Kg²
Substitute into equation 1
F = 6.67×10⁻¹¹(80)(6.4×10²³ )/( 3.4×10⁶)²
F = 3415.04×10¹²/(11.56×10¹²)
F = 3415.04/11.56
F = 295.42 N
Hence the gravitational force = 295.42 N
Answer:
d) 1.2 mT
Explanation:
Here we want to find the magnitude of the magnetic field at a distance of 2.5 mm from the axis of the coaxial cable.
First of all, we observe that:
- The internal cylindrical conductor of radius 2 mm can be treated as a conductive wire placed at the axis of the cable, since here we are analyzing the field outside the radius of the conductor. The current flowing in this conductor is
I = 15 A
- The external conductor, of radius between 3 mm and 3.5 mm, does not contribute to the field at r = 2.5 mm, since 2.5 mm is situated before the inner shell of the conductor (at 3 mm).
Therefore, the net magnetic field is just given by the internal conductor. The magnetic field produced by a wire is given by

where
is the vacuum permeability
I = 15 A is the current in the conductor
r = 2.5 mm = 0.0025 m is the distance from the axis at which we want to calculate the field
Substituting, we find:
