Answer:
Turns of the primary coil: 500
Current in the primary coil: Ip= 0.01168A
Explanation:
Considering an ideal transformer I can propose the following equations:
Vp×Ip=Vs×Is
Vp= primary voltaje
Ip= primary current
Vs= secondary voltaje
Is= secondary current
Np×Vs=Ns×Vp
Np= turns of primary coil
Ns= turns of secondary coil
From these equations I can clear the number of turns of the primary coil:
Np= (Ns×Vp)/Vp = (20×120V)/4.8V = 500 turns
To determine the current in the secondary coil I use the following equation:
Is= (1.4W)/4.8V = 0.292A
Therefore I can determine the current in the primary coil with the following equation:
Ip= (Vs×Is)/Vp = (4.8V×0.292A)/120V = 0.01168A
The answer will be C based on what I had researched
Force = mass*acceleration
acceleration = Force/mass
If you double the mass then acceleration will be halved in order for the equation to still equal.
New acceleration = Force/(2)mass
New acceleration = (1/2)* Force/mass------->if you compare this to original acceleration equation above it is 1/2
Answer:
distance between the two second-order minima is 2.8 cm
Explanation:
Given data
distance = 1.60 m
central maximum = 1.40 cm
first-order diffraction minima = 1.40 cm
to find out
distance between the two second-order minima
solution
we know that fringe width = first-order diffraction minima /2
fringe width = 1.40 /2 = 0.7 cm
and
we know fringe width of first order we calculate slit d
β1 = m1λD/d
d = m1λD/β1
and
fringe width of second order
β2 = m2λD/d
β2 = m2λD / ( m1λD/β1 )
β2 = ( m2 / m1 ) β1
we know the two first-order diffraction minima are separated by 1.40 cm
so
y = 2β2 = 2 ( m2 / m1 ) β1
put here value
y = 2 ( 2 / 1 ) 0.7
y = 2.8 cm
so distance between the two second-order minima is 2.8 cm
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