Conservation of linear momentum

An aluminum wire with a diameter of 0.115 mm has a uniform electric field of 0.235 V/m imposed along its entire length. The temp

tekilochka [14]

Answer with Explanation:

We are given that

Diameter of coil=d=0.115mm

Radius, r= $\frac{d}{2}=\frac{0.115}{2}=0.0575mm=0.0575\times 10^{-3} m$

Using $1mm=10^{-3} m$

Electric field=E=0.235V/m

T=55 degree C

$T_0=20^{\circ} C$

$\rho_0=2.82\times 10^{-8}\Omega m$

$\alpha=3.9\times 10^{-3}/C$

(a).We know that

$\rho=\rho_0(1+\alpha(T-T_0))$

Substitute the values

$\rho=2.82\times 10^{-8}(1+3.9\times 10^{-3}(55-20))$

$\rho=3.2\times 10^{-8}\Omega m$

(b).Current density, $J=\frac{E}{\rho}$

Using the formula

$J=\frac{0.235}{3.2\times 10^{-8}}=7.3\times 10^6A/m^2$

c.Total current,I=JA

Where $A=\pi r^2$

$\pi=3.14$

Using the formula

$I=7.3\times 10^6\times 3.14\times (0.0575\times 10^{-3})^2$

I=0.076A

d.Length of wire=l=2m

$V=El$

Substitute the values

$V=0.235\times 2=0.47 V$

5 0

3 years ago

Newton’s first law relates motion to balanced and unbalanced forces.

BlackZzzverrR [31]

True, an object at rest stays and rest and an object in motion stays in motion

8 0

3 years ago

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what is the focalength of the combination of two thin renses of power -5D and -2D placeci in contact with each other?

Ghella [55]

Answer:

f = - 0.143 m = - 14.3 cm

Explanation:

First, we will calculate the power of the combination of lenses:

P = P₁ + P₂

where,

P = Power of Combination = ?

P₁ = Power of first lens = - 5D

P₂ = Power of second lens = - 2D

Therefore,

P = - 5D - 2D

P = - 7D

Now, the focal length can be given as:

$f = \frac{1}{P} \\\\f = \frac{1}{-\ 7\ D} \\\\$

Negative focal length indicates that combination will act as diverging lens.

6 0

3 years ago

When blueshift occurs,the preceived frequency of the wave would be?

LiRa [457]

Answer:

When blueshift happens, the perceived frequency of the wave would be higher than the actual frequency.

Explanation:

As the name suggests, when blueshift happens to electromagnetic waves, the frequency of the observed wave would shift towards the blue (high-frequency) end of the visible spectrum. Hence, there would be an increase to the apparent frequency of the wave.

Blueshifts happens when the source of the wave and the observer are moving closer towards one another.

Assume that the wave is of frequency $f\; {\rm Hz}$ at the source. In other words, the source of the wave sends out a peak after every $(1/f)\; {\text{seconds}}$ .

Assume that the distance between the observer and the source of the wave is fixed. It would then take a fixed amount of time for each peak from the source to reach the observer.

The source of this wave sends out a peak after each period of $(1/f)\; {\text{seconds}}$ . It would appear to the observer that consecutive peaks arrive every $(1/f)\; {\text{seconds}}\!$ . That would correspond to a frequency of $f\; {\rm Hz}$ .

On the other hand, for a blueshift to be observed, the source of the wave needs to move towards the observer. Assume that the two are moving towards one another at a constant speed of $v \; {\rm m \cdot s^{-1}}$ .

Again, the source of this wave would send out a peak after each period of $(1/f)\; {\text{seconds}}$ . However, by the time the source sends out the second peak, the source would have been $v \cdot (1 / f) \; { \rm m}= (v / f)\; {\rm m}$ closer to the observer then when the source sent out the first peak.

When compared to the first peak, the second peak would need to travel a slightly shorter distance before it reach the observer. Hence, from the perspective of the observer, the time difference between the first and the second peak would be shorter than $(1/f)\; {\text{seconds}}$ . The observed frequency of this wave would be larger than the original $f\; {\rm Hz}$ .

6 0

3 years ago

After pouring water into a cup with holes in the bottom what happened to the water

Nesterboy [21]

It goes through the holes in the cup .

3 0

3 years ago

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