Answer:
The voltage drop across the bulb is 115 V
Explanation:
The voltage drop equation is given by:

Where:
ΔW is the total work done (4.6kJ)
Δq is the total charge
We need to use the definition of electric current to find Δq

Where:
I is the current (2 A)
Δt is the time (20 s)


Then, we can put this value of charge in the voltage equation.

Therefore, the voltage drop across the bulb is 115 V.
I hope it helps you!
solution:
E\delta =\frac{R}{\epsilon0}(1-\frac{A}{\sqrt{4R^{2}}-ac}
=\frac{R}{\epsilon0}(1-\frac{1}{\sqrt{4r^{2}/^{_a{2}}+1}})
=\frac{R}{\epsilon0}(1-\frac{1}{\sqrt{4x^2+1}})
x=\frac{r}{a}
infinite case,
Ei=\frac{r}{\epsilon0}
\therefore e\delta =ei(1-\frac{1}{\sqrt{4x^{2}+1}})
we have to find x when,
ei-e\delta =1% ,y=ei=1/100 ei
or,ei-ei+\frac{ei}{\sqrt{4x^2+1}} = 1/100ei
\frac{1}{\sqrt{4x^2+1}}=\frac{1}{100}
4x^2+1 =10^4
x=\frac{\sqrt{\frac{10^4-1}{4}}}=49.99\approx 50
\therefore \frac{r}{a}\approx 50
<span>B is the correct answer. Neutrons have no charge. In contrast, protons are positively charged and electrons have negative charge. These charges are all relative to each other. The protons and neutrons can be found inside the atom of an element. Electrons float around the outside.</span>
Answer:
b
Explanation:
n = m(g +a)
n= normal force (N)
m=mass (kg)
g=acceleration of gravity
a= acceleration of elevator
rearrange:
a= n/m - g
a= (810 N/73 kg) - 9.8 m/s ^2
a= 1.3 m/s ^2 up
and the acceleration is upwards bc her weight is less than the scale reading