Answer:
Battery voltage will be equal to 9.65 volt
Explanation:
We have given capacitance 
Resistance 
Time constant of RC circuit is 

Time is given t = 0.15 sec
Current i = 0.46 mA
Current in RC circuit is given by



V = 9.65 volt
So battery emf will be equal to 9.65 volt
 
        
             
        
        
        
Answer:
q = 8.57 10⁻⁵ mC
Explanation:
For this exercise let's use Newton's second law
          F = ma
where force is magnetic force
          F = q v x B
the bold are vectors, if we write the module of this expression we have
          F = qv B sin θ
as the particle moves perpendicular to the field, the angle is θ= 90º
         F = q vB
the acceleration of the particle is centripetal
         a = v² / r
we substitute
         qvB = m v² / r
          qBr = m v
           q = 
The exercise indicates the time it takes in the route that is carried out with constant speed, therefore we can use
           v = d / t
the distance is ¼ of the circle,
           d =  
            d = 
we substitute
            v =   
            r =  
            
let's calculate
            r = 2 2.2 10-3 88 /πpi
 2 2.2 10-3 88 /πpi
            r = 123.25 m
          
let's substitute the values
            q =  7.2 10-8 88 / 0.6 123.25
7.2 10-8 88 / 0.6 123.25
             q = 8.57 10⁻⁸ C
Let's reduce to mC
            q = 8.57 10⁻⁸ C (10³ mC / 1C)
            q = 8.57 10⁻⁵ mC
 
        
             
        
        
        
A. All natural radiation is at a level low enough to be safe
 
        
             
        
        
        
Velocity<span> is a</span>vector<span> quantity; it is direction-aware.</span>
        
                    
             
        
        
        
Answer:
v=12.5 i + 12.5 j m/s
Explanation:
Given that
m₁=m₂ = m
m₃ = 2 m
Given that speed of the two pieces 
u₁=- 25 j m/s
u₂ =- 25 i m/s
Lets take the speed of the third mass = v m/s
From linear momentum conservation
Pi= Pf
0 = m₁u₁+m₂u₂ + m₃ v
0 = -25 j m  - 25 i m + 2 m v
2 v=25 j   + 25 i m/s
v=12.5 i + 12.5 j m/s
Therefore the speed of the third mass will be v=12.5 i + 12.5 j m/s