What is the hottest part of the Earth?

Two conductors are made of the same material and have the same length. Conductor A is a solid wire of diameter 1.0 mm.

pantera1 [17]

Answer:

R_a/R_b=3

Explanation:

The resistance in terms of the area and the length of the wire is given by:

R=pL/A

if we have two wires, the first one is a solid wire with a diameter of d_A = 1 x 10^-3 m, and the second one is a hollow wire with inner diameter of d_B,i = 1 x 10^-3 m and outer diameter of d_B,σ= 2 x 10^-3 m, so the cross sectional area of the first wire is:

A_a=πr^2_a

A_a=πd^2_A/4

hence the resistance is:

R_a=(4*p*L_a)/π*d^2_A (1)

the area of the second wire is:

A_b=π*r^2_B,σ-π*r^2_B,i

A_b=π/4(d^2_B,σ-d^2_B,i)

hence the resistance is:

R_b=(4*p*L_b)/π(d^2_B,σ-d^2_B,i) (2)

To find the ratio between the resistances R_a/R_b, we divide (1) over (2) to get:

R_a/R_b=(d^2_B,σ-d^2_B,i)*L_a/(d^2_a*L_b)

but the wires have the same length, therefore:

R_a/R_b=(d^2_B,σ-d^2_B,i)/(d^2_a)

substitute with the given values to get:

R_a/R_b=3

6 0

3 years ago

Read 2 more answers

Takumi works in his yard for 45 minutes each Saturday. He works in the morning, and he wears sunscreen and a hat each time he wo

MrRa [10]

Explanation :

Takumi wears sunscreen and a hat each time he works in the yard. This is to protect himself with the strong radiation coming from the sun. UV rays that are coming from the sun are the main cause of skin cancer.

Stochastic effects are the effects that are caused by chance. Cancer is one of the main stochastic effects.

So, the correct option is (b) "the severity of stochastic effects, such as cancer".

7 0

4 years ago

Read 2 more answers

The spreading of waves behind an aperture ismore for long wavelengths and less for short wavelengths.less for long wavelengths a

kompoz [17]

<h2>Answer: The spreading of waves behind an aperture ismore for long wavelengths and less for short wavelengths</h2>

Here we are talking about Diffraction and, in fact, waves diffract the most when their wavelength is about the same size of the gap or aperture.

Diffraction happens when a wave (mechanical or electromagnetic wave) meets an obstacle or a slit .When this occurs, <u>the wave bends around the corners of the obstacle or passes through the opening of the slit that acts as an obstacle, forming multiple patterns with the shape of the aperture of the slit. </u>

<u />

Note that the principal condition for the occurrence of this phenomena is that the obstacle must be comparable in size (similar size) to the size of the wavelength.

In other words, when the gap (or slit) size is larger than the wavelength, the wave passes through the gap and does not spread out much on the other side, but when the gap size is equal to the wavelength, maximum diffraction occurs and the waves spread out greatly.

This means the smaller the slit or obstacle (diffracting object), the wider the resulting diffraction pattern, and the greater the obstacle, the narrower de resulting patter.

6 0

3 years ago

A leaky 10-kg bucket is lifted from the ground to a height of 11 m at a constant speed with a rope that weighs 0.9 kg/m. Initial

nalin [4]

Answer:

the work done to lift the bucket = 3491 Joules

Explanation:

Given:

Mass of bucket = 10kg

distance the bucket is lifted = height = 11m

Weight of rope= 0.9kg/m

g= 9.8m/s²

initial mass of water = 33kg

x = height in meters above the ground

Let W = work

Using riemann sum:

the work done to lift the bucket =∑(W done by bucket, W done by rope and W done by water)

= $\int\limits^a_b {(Mass of Bucket + Mass of Rope + Mass of water)*g*d} \, dx$

Work done in lifting the bucket (W) = force × distance

Force (F) = mass × acceleration due to gravity

Force = 9.8 * 10 = 98N

W done by bucket = 98×11 = 1078 Joules

Work done to lift the rope:

At Height of x meters (0≤x≤11)

Mass of rope = weight of rope × change in distance

= 0.8kg/m × (11-x)m

W done = integral of (mass×g ×distance) with upper 11 and lower limit 0

W done = $\int\limits^1 _0 {9.8*0.8(11-x)} \, dx$

Note : upper limit is 11 not 1, problem with math editor

W done = 7.84 (11x-x²/2)upper limit 11 to lower limit 0

W done = 7.84 [(11×11-(11²/2)) - (11×0-(0²/2))]

=7.84(60.5 -0) = 474.32 Joules

Work done in lifting the water

At Height of x meters (0≤x≤11)

Rate of water leakage = 36kg ÷ 11m = $\frac{36}{11}$ kg/m

Mass of rope = weight of rope × change in distance

= $\frac{36}{11}$ kg/m × (11-x)m = 3.27kg/m × (11-x)m

W done = integral of (mass×g ×distance) with upper 11 and lower limit 0

W done = $\int\limits^1 _0 {9.8*3.27(11-x)} \, dx$

Note : upper limit is 11 not 1, problem with math editor

W done = 32.046 (11x-x²/2)upper limit 11 to lower limit 0

W done = 32.046 [(11×11-(11²/2)) - (11×0-(0²/2))]

= 32.046(60.5 -0) = 1938.783 Joules

the work done to lift the bucket =W done by bucket+ W done by rope +W done by water)

the work done to lift the bucket = 1078 +474.32+1938.783 = 3491.103

the work done to lift the bucket = 3491 Joules

8 0

4 years ago

A woman has a mass of 55 kg on earth. what would be the women's weight on the moon? (The moon’s gravity is 1.6 N).

Alex17521 [72]

Answer:

i dont know the answer sorry have a good day

Explanation:

hsjsjdkdkdkdkdkdk

8 0

3 years ago