Reasons why inductors opposes charges passing through it<br>

A car’s bumper is designed to withstand a 4.0-km/h (1.1-m/s) collision with an immovable object without damage to the body of th

Alex Ar [27]

Answer:

avriage force F = 2722.5 N

Explanation:

For this problem we can use Newton's second law, to calculate the average force and acceleration we can find it by kinematics.

vf² = v₀² - 2 ax

The final carriage speed is zero (vf = 0)

0 = v₀² - 2ax

a = v₀² / 2x

a = 1.1²/(2 0.200)

a = 3.025 m / s²

a = 3.0 m/s²

We calculate the average force

F = ma

F = 900 3,025

F = 2722.5 N

3 0

2 years ago

A spherical capacitor contains a charge of 3.00 nC when connected to a potential difference of 230 V. If its plates are separate

Assoli18 [71]

Answer:

Part(a): the capacitance is 0.013 nF.

Part(b): the radius of the inner sphere is 3.1 cm.

Part(c): the electric field just outside the surface of inner sphere is $\bf{2.81 \times 10^{4}~n~C^{-1}}$ .

Explanation:

We know that if 'a' and 'b' are the inner and outer radii of the shell respectively, 'Q' is the total charge contains by the capacitor subjected to a potential difference of 'V' and ' $\epsilon_{0}$ ' be the permittivity of free space, then the capacitance (C) of the spherical shell can be written as

$C = \dfrac{4 \pi \epsilon_{0}}{(\dfrac{1}{a} - \dfrac{1}{b})}~~~~~~~~~~~~~~~~~~~~~~~~~~~(1)$

Part(a):

Given, charge contained by the capacitor Q = 3.00 nC and potential to which it is subjected to is V = 230V.

So the capacitance (C) of the shell is

$C &=& \dfrac{Q}{V} = \dfrac{3 \times 10^{-90}~C}{230~V} = 1.3 \times 10^{-11}~F = 0.013~nF$

Part(b):

Given the inner radius of the outer shell b = 4.3 cm = 0.043 m. Therefore, from equation (1), rearranging the terms,

$&& \dfrac{1}{a} = \dfrac{1}{b} + \dfrac{1}{C/4 \pi \epsilon_{0}} = \dfrac{1}{0.043} + \dfrac{1}{1.3 \times 10^{-11} \times 9 \times 10^{9}} = 31.79\\&or,& a = \dfrac{1}{31.79}~m = 0.031~m = 3.1~cm$

Part(c):

If we apply Gauss' law of electrostatics, then

$&& E~4 \pi a^{2} = \dfrac{Q}{\epsilon_{0}}\\&or,& E = \dfrac{Q}{4 \pi \epsilon_{0}a^{2}}\\&or,& E = \dfrac{3 \times 10^{-9} \times 9 \times 10^{9}}{0.031^{2}}~N~C^{-1}\\&or,& E = 2.81 \times 10^{4}~N~C^{-1}$

3 0

2 years ago

Help with Order of Magnitude question?

solong [7]

The rotation of Earth is equivalent to one day which is comprised of 24 hours. To determine the number of miles in Earth's circumference, one simply have to multiply the given rate by the appropriate conversion factor and dimensional analysis. This is shown below.

C = (1038 mi/h)(24 h/1 day)
C = 24,912 miles

From the given choices, the nearest value would have to be 20,000 mile. The answer is the second choice.

7 0

2 years ago

Tim made tomato soup from scratch. When he cleaned up, he noticed that the aluminum pan was damaged. Analyze what happened to th

mr_godi [17]

Answer:

D) The acidic tomato juice reacted unfavorably with the aluminum pan

5 0

3 years ago

Read 2 more answers

The membrane that surrounds a certain type of living cell has a surface area of 4.7 x 10-9 m2 and a thickness of 1.3 x 10-8 m. A

vagabundo [1.1K]

Answer:

Q = 1.2*10⁻¹² C

Explanation:

For any capacitor, by definition the capacitance C is equal to the relationship between the charge on one of the conductors and the potential difference between them, as follows:

$C = \frac{Q}{V} (1)$

For the special case of a parallel plate capacitor, just by application of Gauss' law to a rectangular surface half out of the outer surface, and half inside it, it can be showed that the value of the capacitance C is a parameter defined only by geometric constants, as follows:

$C = \frac{\epsilon_{0}*\epsilon _{r} * A}{d} (2)$

So, due to the left sides in (1) and (2) are equal each other, right sides must be equal too.
Replacing ε₀, εr (dielectric constant), A, d and V by their values, we can solve for Q, as follows:

$Q =\frac{\epsilon_{0} * \epsilon_{r} *A* V}{d} = \frac{(8.85*(4.7)^{2}*79.5)e-24 (F/m*m2*V)}{1.3e-8m} = 1.2e-12 C = 1.2 pC (3)$

6 0

2 years ago

Reasons why inductors opposes charges passing through it​

Reasons why inductors opposes charges passing through it