Answer:
19.1 deg 
Explanation:
v = speed of the proton = 8 x 10⁶ m/s 
B = magnitude of the magnetic field = 1.72 T 
q = magnitude of charge on the proton = 1.6 x 10⁻¹⁹ C 
F = magnitude of magnetic force on the proton = 7.20 x 10⁻¹³ N 
θ = Angle between proton's velocity and magnetic field 
magnitude of magnetic force on the proton is given as 
F = q v B Sinθ
7.20 x 10⁻¹³ = (1.6 x 10⁻¹⁹) (8 x 10⁶) (1.72) Sinθ
Sinθ = 0.327 
θ = 19.1 deg 
 
        
             
        
        
        
Answer:
<em>the phase relationship between two waves.</em>
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Explanation:
Coherence describes all properties of the correlation between physical quantities between waves. It is an ideal property of waves that determines their interference. In a situation in which there is a correlation or phase relationship between two waves. If the properties of one of the waves can be measure directly, then, some of the properties of the other wave can be calculated.
 
        
             
        
        
        
Answer:
The average induced emf in the coil is 0.0286 V
Explanation:
Given;
diameter of the wire, d = 11.2 cm = 0.112 m
initial magnetic field, B₁ = 0.53 T
final magnetic field, B₂ = 0.24 T
time of change in magnetic field, t = 0.1 s
The induced emf in the coil is calculated as;
E = A(dB)/dt
where;
A is area of the coil = πr²
r is the radius of the wire coil = 0.112m / 2 = 0.056 m
A = π(0.056)²
A = 0.00985 m²
E = -0.00985(B₂-B₁)/t
E = 0.00985(B₁-B₂)/t
E = 0.00985(0.53 - 0.24)/0.1
E = 0.00985 (0.29)/ 0.1
E = 0.0286 V
Therefore, the average induced emf in the coil is 0.0286 V
 
        
             
        
        
        
Answer:
72.22 N
Explanation:
F = weight
m = mass of body
M = mass of earth
R = radius of earth
G = universal constant of gravitation
F_1= 650 N
 F_1 = GMm/R^2
two earth radius above the surface of the earth:
F_2=  GMm/(3R)^2=  GMm/9R^2= F_1/9= 650/9
=72.22 N
 
        
             
        
        
        
Answer:
the magnitude of the total angular momentum of the blades is <em>743.71 kg·m²</em>
Explanation:
Converting the angular speed into radians per second:
ω = 334 rpm · (2π rad / 1 rev) · (1 min / 60 s)
ω = 34.98 rad/s
The rotational kinetic energy of the blades is given by:
EK = 1/2 I ω²
where
- I is the moment of inertia 
- ω is the angular speed
Therefore, rearranging the above equation, we get:
1/2 I ω² = EK
I ω² = 2 EK
I = 2(EK) / ω²
I = 2(4.55 × 10⁵ J) / (34.98 rad/s)²
<em>I = 743.71 kg·m²</em>
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Therefore, the magnitude of the total angular momentum of the blades is <em>743.71 kg·m²</em>.