When jumping down from a table to the ground, it is helpful to bend your knees when you hit the ground because...

The paper dielectric in a paper-and-foil capacitor is 0.0800 mm thick. Its dielectric constant is 2.50, and its dielectric stren

jeka94

Answer:

a) 0.723 m²

b) 2000V

Explanation:

Given that

Thickness of the capacitor, d = 0.08*10^-3 m

Dielectric constant of the capacitor, k = 2.5

Dielectric strength of the capacitor, E = 50*10^6

Capacitance of the capacitor, C = 0.2*10^-6

Permittivity of free space, E• = 8.85*10^-12

a)

The area, A is given by the formula

A = (C * d) / (k * E•)

A = (0.2*10^-6 * 0.08*10^-3) / (2.5 * 8.85*10^-12)

A = 1.6*10^-11 / 2.213*10^-11

A = 0.723 m²

b)

Potencial difference, V is given by the formula

V = E * d

V = 1/2 * 50*10^6 * 0.08*10^-3

V = 1/2 * 4000

V = 2000 V

8 0

4 years ago

Our Sun is considered a(n) _____ star.<br> A. small <br> B. large <br> C. giant <br> D. average

faltersainse [42]

Checking a table of star sizes, we'll see that our Sun is a medium sized star, so the answer is D, average.

5 0

3 years ago

Suppose Gabor, a scuba diver, is at a depth of 15m. Assume that: The air pressure in his air tract is the same as the net water

s2008m [1.1K]

Complete Question

The complete question is shown on the first and second uploaded image

Answer:

a

Now the ratio of the gases in Gabor's lungs at the depth of 15m to that at

the surface is $\frac{(n/V)_{15\ m}}{(n/V)_{surface}} = 2.5$

b

The number of moles of gas that must be released is $n= 0.3538\ mols$

Explanation:

We are told from the question that the pressure at the surface is 1 atm and for each depth of 10m below the surface the pressure increase by 1 atm

This means that the pressure at the depth of the surface would be

$P_d = [\frac{15m}{10m} ] (1 atm) + 1 atm$

$= 2.5 atm$

The ideal gas equation is mathematically represented as

$PV = nRT$

Where P is pressure at the surface

V is the volume

R is the gas constant = 8.314 J/mol. K

making n the subject we have

$n = \frac{PV}{RT}$

Considering at the surface of the water the number of moles at the surface would be

$n_s = \frac{P_sV}{RT}$

Substituting $1 atm = 101325 N/m^2$ for $P_s$ , $6L = 6*10^{-3}m^3$ for volume , 8.314 J/mol. K for R , (37° +273) K for T into the equation

$n_s = \frac{(1atm)(6*10^{-3} m^3)}{(8.314J/mol \cdot K)(37 +273)K}$

$= 0.2359 mol$

To obtain the number of moles at the depth of the water we use

$n_d = \frac{P_d V}{RT}$

Where $P_d \ and \ n_d \$ are pressure and no of moles at the depth of the water

Substituting values we have

$n_d = \frac{(2.5)(101325 N/m^2)(6*10^{-3}m^3)}{(8.314 J/mol \cdot K)(37 + 273)K}$

$= 0.5897 mol$

Now to obtain the number of moles released we have

$n = n_d - n_s$

$= 0.5897mol - 0.2359mol$

$=0.3538 \ mol$

The molar concentration at the surface of water is

$[\frac{n}{V} ]_{surface} = \frac{0.2359mol}{6*10^-3m^3}$

$=39.31mol/m^3$

The molar concentration at the depth of water is

$[\frac{n}{V} ]_{15m} = \frac{0.5897}{6*10^{-3}}$

$= 98.28 mol/m^3$

Now the ratio of the gases in Gabor's lungs at the depth of 15m to that at the surface is

$\frac{(n/V)_{15\ m}}{(n/V)_{surface}} = \frac{98.28}{39.31} =2.5$

6 0

3 years ago

Which copper wire would have the highest resistance?

Dafna11 [192]

Answer:

Just consider the resistance as water flowing through a pipe. If the pipe is too small in radius (consider thin wire as small pipe) water can’t flow easly. If the pipe is big in radius (consider thick wire as big pipe) water can flow easly. So flowing water through a small pipe for a long distance will inversly affect the normal flow where as flowing water through a big pipe for a small distance will not affect too much the normal flow.

So long & thin wire has high resistance. Short & thick wire has low resistance.

B. A wire that is 2 m long and has a cross-sectional area of 0.066

7 0

3 years ago

A bullet of mass 0.016 kg traveling horizontally at a speed of 280 m/s embeds itself in a block of mass 3 kg that is sitting at

ivanzaharov [21]

(a) 1.49 m/s

The conservation of momentum states that the total initial momentum is equal to the total final momentum:

$p_i = p_f\\m u_b + M u_B = (m+M)v$

where

m = 0.016 kg is the mass of the bullet

$u_b = 280 m/s$ is the initial velocity of the bullet

M = 3 kg is the mass of the block

$u_B = 0$ is the initial velocity of the block

v = ? is the final velocity of the block and the bullet

Solving the equation for v, we find

$v=\frac{m u_b}{m+M}=\frac{(0.016 kg)(280 m/s)}{0.016 kg+3 kg}=1.49 m/s$

(b) Before: 627.2 J, after: 3.3 J

The initial kinetic energy is (it is just the one of the bullet, since the block is at rest):

$K_i = \frac{1}{2}mu_b^2 = \frac{1}{2}(0.016 kg)(280 m/s)^2=627.2 J$

The final kinetic energy is the kinetic energy of the bullet+block system after the collision:

$K_f = \frac{1}{2}(m+M)v^2=\frac{1}{2}(0.016 kg+3 kg)(1.49 m/s)^2=3.3 J$

(c) The Energy Principle isn't valid for an inelastic collision.

In fact, during an inelastic collision, the total momentum of the system is conserved, while the total kinetic energy is not: this means that part of the kinetic energy of the system is losted in the collision. The principle of conservation of energy, however, is still valid: in fact, the energy has not been simply lost, but it has been converted into other forms of energy (thermal energy).

4 0

4 years ago