Answer:

Explanation:
When calculating Hall voltage, it is crucial to have the current, magnetic field strength, length, area, and number of charge carriers available. The Hall voltage can be expressed using the equation:

Where:
i= the current
B= the magnetic field strength
L = the length
n = the number of charge carriers
e= charge of an electron
We need to replace values and solve for n:



As a result, the charge carrier density is 
Answer:
x₂=2×1
Explanation:
According to the work-energy theorem, we can assume that the gravitational potential energy at the lowest point of compression is zero since the kinetic energy change is 0;
mgx-(kx)²/2 =0 where m refers to the object's mass, g indicates the acceleration due to gravity, k denotes spring constant, and x represents the spring's compression.
mgx=(kx)²/2
x=2mg/k----------------compression when the object is at rest
However, ΔK.E =-1/2mv²⇒kx²=mv² -----------where v symbolizes the object's velocity and K.E signifies kinetic energy
Thus, if kx²=mv² then
v=x *√(k/m) ----------------where v=0
<pDoubling v results in multiplying x *√(k/m) by 2, leading to x₂ being double x₁
Answer:
Maximum emf = 5.32 V
Explanation:
Provided data includes:
Number of turns, N = 10
Radius of loop, r = 3 cm = 0.03 m
It made 60 revolutions each second
Magnetic field, B = 0.5 T
We are tasked to determine the maximum emf produced in the loop, which is founded on Faraday's law. The induced emf can be calculated by:

For the maximum emf, 
Therefore,

Hence, the maximum emf generated in the loop is 5.32 V.
<span>A force of 110 N is applied at an angle of 30</span>°<span> to the horizontal. Because the force does not align directly either vertically or horizontally with the sled, it can be broken down into two components based on sine and cosine.
For the component parallel to the ground:
x = rcos</span>β
<span>x = 110cos30</span>°
<span>x = 95.26
For the component perpendicular to the ground:
y = rsin</span>β
<span>y = 110sin30</span>°
<span>y = 55</span>
Answer:
d_total = 12 m
Explanation:
In this kinematics scenario illustrated in the graph provided, we determine the distance traveled over a 24-second duration.
The comprehensive distance can be calculated as follows:
d_total = d₁ + d₂ + d₃
Given that d₂ on the graph is level (v=0), its distance equates to zero, hence d₂ = 0.
The distance for d₁ is calculated as:
d₁ = 12 - 6 = 6 m
For distance d₃:
d₃ = 6 - 0 = 6 m
Thus, the overall distance covered is:
d_total = 6 + 0 + 6
d_total = 12 m