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umka21 [38]
3 years ago
14

If the area of an iron rod is 10 cm by 0.5 cm and length is 35 cm. Find the value of resistance, if 11x10^-8 ohm.m be the resist

ivity of iron.
Physics
1 answer:
malfutka [58]3 years ago
6 0

Answer:

Resistance of the iron rod, R = 0.000077 ohms    

Explanation:

It is given that,

Area of iron rod, A=10\ cm\times 0.5\ cm=5\ cm^2 = 0.0005\ m^2

Length of the rod, L = 35 cm = 0.35 m

Resistivity of Iron, \rho=11\times 10^{-8}\ \Omega-m

We need to find the resistance of the iron rod. It is given by :

R=\rho\dfrac{L}{A}

R=11\times 10^{-8}\times \dfrac{0.35\ m}{0.0005\ m^2}

R=0.000077 \Omega

So, the resistance of the rod is 0.000077 ohms. Hence, this is the required solution.

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Answer:

Yes

Explanation:

The spring force is given as:

          F  = kd

F is the spring force

K is the spring constant

d is the magnitude of the stretch

 Since k is a constant, therefore, doubling the stretch distance will double the force.

Both stretch distance and force applied can be said to be directly proportional to one another.

3 0
3 years ago
By what factor is the self-inductance of an air solenoid changed if its length and number of coil turns are both tripled
fredd [130]

Answer:

The new self inductance is 3 times of the initial self inductance.

Explanation:

The self inductance of a solenoid is given by :

L=\dfrac{\mu_oN^2 A}{L}

Where

N is number of turns per unit length

A is area of cross section

l is length of solenoid

If length and number of coil turns are both tripled,

l' = 3l and N' = 3N

New self inductance is given by :

L'=\dfrac{\mu_oN'^2 A}{L'}\\\\=\dfrac{\mu_o(3N)^2 A}{3L}\\\\=3\dfrac{\mu_oN^2 A}{L}\\\\=3L

So, the new self inductance is 3 times of the initial self inductance.

4 0
3 years ago
The inner conductor of a coaxial cable has a radius of 0.800 mm, and the outer conductor’s inside radius is 3.00 mm. The space b
ZanzabumX [31]

Answer:

The maximum potential difference is 186.02 x 10¹⁵ V

Explanation:

formula for calculating maximum potential difference

V = \frac{2K_e \lambda}{k}ln(\frac{b}{a})

where;

Ke is coulomb's constant = 8.99 x 10⁹ Nm²/c²

k is the dielectric constant = 2.3

b is the outer radius of the conductor = 3 mm

a is the inner radius of the conductor = 0.8 mm

λ is the linear charge density = 18 x 10⁶ V/m

Substitute in these values in the above equation;

V = \frac{2K_e \lambda}{k}ln(\frac{b}{a}) =  \frac{2*8.99*10^9*18*10^6 }{2.3}ln(\frac{3}{0.8}) =140.71 *10^{15} *1.322 \\\\V= 186.02 *10^{15} \ V

Therefore, the maximum potential difference this cable can withstand is 186.02 x 10¹⁵ V

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3 years ago
Which of the following best describes wind?
Afina-wow [57]

Answer:

The answer is D the rising of warm air pushing down cool air.

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A hockey ball accelerates from 0.m/s to 25m/s in 0.05 seconds what is the acceleration of the ball ?
S_A_V [24]

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500

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I think this ?

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