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Svet_ta [14]
1 year ago
6

A 60-watt light bulb carries a current of 0. 5 ampere. the total charge passing through it in one hour is________.

Physics
1 answer:
tino4ka555 [31]1 year ago
8 0

<u>Answer:</u>

charge passed = 1800 C

<u>Explanation:</u>

• We can use the following formula to calculate charge:

\boxed{Q = It},

where:

Q = charge in Coulombs

I = current in amperes (0.5 A)

t = time in seconds (1 h = 60 min = 3600 s).

• Substituting the values into the formula:

charge = 0.5 × 3600

            = 1800 C

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Explanation:

Hope it helps

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3 years ago
Find the net work W done on the particle by the external forces during the motion of the particle in terms of the initial and fi
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Answer:

W=K_f-K_i

Explanation:

The work done on a particle by external forces is defined as:

W=\int\limits^{r_f}_{r_i} {F\cdot dr} \,

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W=\int\limits^{r_f}_{r_i}{ma\cdot dr} \,\\

Acceleration is defined as the derivative of the speed with respect to time:

W=m\int\limits^{r_f}_{r_i}{\frac{dv}{dt}\cdot dr} \,\\\\W=m\int\limits^{r_f}_{r_i}{dv \cdot \frac{dr}{dt}} \,

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W=m\int\limits^{v_f}_{v_i} v \cdot dv \,

Kinetic energy is defined as K=\frac{mv^2}{2}:

W=m\frac{v_f^2}{2}-m\frac{v_i^2}{2}\\W=K_f-K_i

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3 years ago
State the five important assumptionns of the kinetic theory of matter
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2 years ago
A capacitor is formed from two concentric spherical conducting shells separated by vacuum. The inner sphere has radius 11.0 cm ,
viktelen [127]
Part A)
First of all, let's convert the radii of the inner and the outer sphere:
r_A = 11.0 cm = 0.110 m
r_B = 16.5 cm=0.165 m
The capacitance of a spherical capacitor which consist of two shells with radius rA and rB is
C=4 \pi \epsilon _0  \frac{r_A r_B}{r_B- r_A}=4\pi(8.85 \cdot 10^{-12}C^2m^{-2}N^{-1}) \frac{(0.110m)(0.165m)}{0.165m-0.110m}=
=3.67\cdot 10^{-11}F

Then, from the usual relationship between capacitance and voltage, we can find the charge Q on each sphere of the capacitor:
Q=CV=(3.67\cdot 10^{-11}F)(100 V)=3.67\cdot 10^{-9}C

Now, we can find the electric field at any point r located between the two spheres, by using Gauss theorem:
E\cdot (4 \pi r^2) =  \frac{Q}{\epsilon _0}
from which
E(r) =  \frac{Q}{4 \pi \epsilon_0 r^2}
In part A of the problem, we want to find the electric field at r=11.1 cm=0.111 m. Substituting this number into the previous formula, we get
E(0.111m)=2680 N/C

And so, the energy density at r=0.111 m is
U= \frac{1}{2} \epsilon _0 E^2 =  \frac{1}{2} (8.85\cdot 10^{-12}C^2m^{-2}N^{-1})(2680 N/C)^2=3.17 \cdot 10^{-5}J/m^3

Part B) The solution of this part is the same as part A), since we already know the charge of the capacitor: Q=3.67 \cdot 10^{-9}C. We just need to calculate the electric field E at a different value of r: r=16.4 cm=0.164 m, so
E(0.164 m)= \frac{Q}{4 \pi \epsilon_0 r^2}=1228 N/C

And therefore, the energy density at this distance from the center is
U= \frac{1}{2}\epsilon_0 E^2 =  \frac{1}{2} (8.85\cdot 10^{-12}C^2m^{-2}N^{-1})(1228 N/C)^2=6.68 \cdot 10^{-6}J/m^3
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3 years ago
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I would assume that the magnesium would have to come into contact with an ion in order to become a magnesium ion. 
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