Answer:
a) 5.851× 10¹⁰m/s²
b) 2.411×10⁻¹¹s
c) 1.70×10⁻¹¹m
d) 1.661×10⁻²⁷KJ
Explanation:
A proton in the field experience a downward force of magnitude,
F = eE. The force of gravity on the proton will be negligible compared to the electric force
F = eE
a= eE/m
= 1.602×10⁻¹⁹ × 610/1.67×10⁻²⁷
= 5.851× 10¹⁰m/s²
b)
V = u + at
u= 0
v= 1.4106m/s
v= (0)t + at
t= v/a
= 1.4106m/s/5.851 ×10¹⁰
= 2.411×10⁻¹¹s
c)
S = ut + at²
= (o)t + 5.851×10¹⁰×(2.411×10⁻¹¹)²
= 1.70×10⁻¹¹m
d)
Ke = 1/2mv²
= (1.67×10⁻²⁷×)(1.4106)²/2
= 1.661×10⁻²⁷KJ
To solve this problem it is necessary to apply the definition given in Faraday's law in a solenoid for which it is noted that


Where,
N = Number of loops
A = Cross sectional Area
B = Magnetic Field



Therefore the correct answer is A.
Answer:
No work is performed or required in moving the positive charge from point A to point B.
Explanation:
Lets take
Q= Positive charge which move from point A to point B along
Voltage difference,ΔV =V₁ - V₂
The work done
W = Q . ΔV
Given that charge is moved from point A to point B along an equipotential surface.It means that voltage difference is zero.
ΔV = 0
So
W = Q . ΔV
W = Q x 0
W= 0 J
So work is zero.
Answer:
a) ![(Qa*g*Vb)-(Qh*Vb*g)=(Qh*Vb*a)\\where \\g=gravity [m/s^2]\\a=acceleration [m/s^2]](https://tex.z-dn.net/?f=%28Qa%2Ag%2AVb%29-%28Qh%2AVb%2Ag%29%3D%28Qh%2AVb%2Aa%29%5C%5Cwhere%20%5C%5Cg%3Dgravity%20%5Bm%2Fs%5E2%5D%5C%5Ca%3Dacceleration%20%5Bm%2Fs%5E2%5D)
b) a = 19.61[m/s^2]
Explanation:
The total mass of the balloon is:
![massball=densityheli*volumeheli\\\\massball=0.41 [kg/m^3]*0.048[m^3]\\massball=0.01968[kg]\\\\](https://tex.z-dn.net/?f=massball%3Ddensityheli%2Avolumeheli%5C%5C%5C%5Cmassball%3D0.41%20%5Bkg%2Fm%5E3%5D%2A0.048%5Bm%5E3%5D%5C%5Cmassball%3D0.01968%5Bkg%5D%5C%5C%5C%5C)
The buoyancy force acting on the balloon is:
![Fb=densityair*gravity*volumeball\\Fb=1.23[kg/m^3]*9.81[m/s^2]*0.048[m^3]\\Fb=0.579[N]](https://tex.z-dn.net/?f=Fb%3Ddensityair%2Agravity%2Avolumeball%5C%5CFb%3D1.23%5Bkg%2Fm%5E3%5D%2A9.81%5Bm%2Fs%5E2%5D%2A0.048%5Bm%5E3%5D%5C%5CFb%3D0.579%5BN%5D)
Now we need to make a free body diagram where we can see the forces that are acting over the balloon and determinate the acceleration.
In the attached image we can see the free body diagram and the equation deducted by Newton's second law