What is thermal conductivity?

A straight, nonconducting plastic wire 8.00 cm long carries a charge density of 100 nC/m distributed uniformly along its length.

Sphinxa [80]

Answer:

(A) 16641 N/C (B) 72000 N/C

Explanation:

We have given the length of the wire = 8 cm =0.08 m

Charge density = 100 nC/m

So charge = 100 ×0.08 = 8 nC $=8\times 10^{-9}C$

(A) Electric field E that produce 6 cm directly above its midpoint will be

$E=\frac{KQ}{X\sqrt{X^2+a^2}}=\frac{9\times 10^9\times 8\times 10^{-9}}{0.06\sqrt{0.06^2+(\frac{0.08}{2})^2}}=16641N/C$

Here X is the distance where we have to find the electric field

(B) Now electric field due to flat ring will be

$E=\frac{KQX}{(X^2+a^2)^\frac{3}{2}}$

Here X is the distance where we have to find the electric field

So $E=\frac{9\times 10^9\times 8\times 10^{-9}}{(0.06^2+0.08^2)^\frac{3}{2}}=72000N/C$

7 0

4 years ago

You have two steel solid spheres. sphere 2 has twice the radius of sphere 1. part a by what factor does the moment of inertia i2

sertanlavr [38]

Answer:

moment of inertia of sphere 2 is 32 times the moment of inertia of sphere 1

Explanation:

The moment of inertia of a solid sphere about its axis is

$I=\frac{2}{5}MR^2$

where

M is the mass of the sphere

R is the radius of the sphere

The mass of the sphere can be rewritten as

$M=\rho V$

where

$\rho$ is the density

$V=\frac{4}{3}\pi R^3$ is the volume of the sphere

So the moment of inertia becomes

$I=\frac{2}{5}(\frac{4}{3}\pi \rho R^3)R^2 = \frac{8}{15}\pi \rho R^5$

Calling R the radius of sphere 1, the moment of inertia of sphere 1 is

$I_1=\frac{8}{15}\pi \rho R^5$

where $\rho$ is the density of steel, since the sphere is made of steel

Sphere 2 has twice the radius of sphere 1, so

R' = 2R

and so its moment of inertia is

$I_2=\frac{8}{15}\pi \rho R'^5=\frac{8}{15}\pi \rho (2R)^5=32(\frac{8}{15}\pi \rho R^5)=32I_1$

So, the moment of inertia of sphere 2 is 4 times the moment of inertia of sphere 1.

4 0

3 years ago

A bird flies directly into the windshield of a moving car. How does the FORCE exerted on the car compare to the FORCE exerted on

Anarel [89]

Answer:

b

Explanation:

7 0

3 years ago

Which of the following employs the use of an external combustion engine?

just olya [345]

The answer should be ''all the above''

7 0

4 years ago

What is the specific heat of the masses in this experiment? Infer the substance the masses are made of and explain your inferenc

Liono4ka [1.6K]

The metal whose specific heat capacity is close to the obtained value is aluminum.

<h3>What is specific heat capacity</h3>

The specific heat capacity of an object is the heat required to raise a unit mass of the substance by 1 kelvin.

$Q= mc\Delta \theta$

where;

c is the specific heat capacity
Δθ is change in temperature

Let the mass of the water = 50 g

mass of the metal for this first trial = 50 g

The heat gained by the water is calculated as follows

$Q = 50 \times 4.184 \times 8.4\\\\Q = 1757.28 \ J$

Specific heat capacity of the metal for the first trial is calculated as follows;

Heat gained by water = Heat lost by metal

$C = \frac{Q}{m\Delta T} = \frac{1757.28}{50\times 8.4} = 4.184 \ J/g^oC$

Specific heat capacity of the metal for the second trial;

mass of metal = 200 - mass of water = 150 g

$C_2 = \frac{1757.28}{150 \times 15.2} = 0.77 \ J/g^oC$

Specific heat capacity of the metal for the third trial;

$C_3 = \frac{1757.28}{250 \times 20.8} = 0.34\ J/g^oC$

Specific heat capacity of the metal for the fourth trial;

$C_4 = \frac{1757.28}{350 \times 25.4} = 0.19\ J/g^oC$

Specific heat capacity of the metal for the fifth trial;

$C_5 = \frac{1757.28}{450 \times 29.6} = 0.13\ J/g^oC$

Average specific heat capacity

$C = \frac{4.184 + 0.77 + 0.34+ 0.19 + 0.13 }{5} = 1.12 \ J/g^oC = 1120 J/kg^oC$

The metal whose specific heat capacity is close to the above value is aluminum.

Learn more about specific heat capacity here: brainly.com/question/16559442

8 0

2 years ago