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3 years ago

15

Which one of the following statements concerning superconductors is false? A constant current can be maintained in a superconduc

ting ring for several years without an emf. Superconductors are perfect conductors. Critical temperatures for some superconductors exceed 100 K. All materials are superconducting at temperatures near absolute zero kelvin. Below its critical temperature, the resistivity of a superconductor is zero Ω-m

1 answer:

julsineya [31]3 years ago

8 0

Answer:

All materials are superconducting at temperatures near absolute zero kelvin.

Explanation:

All materials are superconducting at temperatures near absolute zero kelvin is false concerning superconductors.

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1. Si comprimes un globo hasta reducir su volumen a un tercio de su valor original, ¿cuánto aumenta la presión en su interior?

harina [27]

Answer:

See step by step sexplanation

Explanation:

1.-Sabemos que la relación:

P₁ * V₁ = P₂ * V₂

Para una temperatura constante debe mantenerse entonces si el globo se comprime hasta llevarlo a 1/3 de su valor inicial, entonces necesariamente para cumplir con la relación mencionada, la presión aumenta tres veces su valor original

2.-La definición de presión es fuerza por unidad de superficie, entonces la fuerza es determinada por la altura de la columna de liquido en el recipiente y no por la cantidad total de liquido, de acuerdo a esto habrá más presión en la base del florero, ya que la columna de agua tiene más altura.

3.-No se puede estar de acuerdo con el criterio del plomero. En su solución no plantea el aumento de la altura del tanque, para el logro del aumento de la presión que es realmente lo que hay que hacer

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3 years ago

An airplane starts from rest at the end of a runway and accelerates at a constant rate. In the first second, the airplane travel

Licemer1 [7]

Answer:

$v=4.44\frac{m}{s}$

Explanation:

Given that the airplane starts from the rest (this is initial velocity equals to zero) and accelerates at a constant rate, position can be described like this: $x=v_{0}t +\frac{1}{2} at^{2}$ where x is the position, t is the time a is the acceleration and $v_{0}$ is initial velocity. In this way acceleration can be found. $a=\frac{2(x-v_{0}t) }{t^{2} } =\frac{2(1.11m-0)}{1s^{2} } =2.22\frac{m}{s^{2} }$ .

Now we are able to found velocity at any time with the formula: $v=v_{0} +at = 0\frac{m}{s} +(2.22\frac{m}{s^{2}}.2s)=4.44\frac{m}{s}$

3 0

3 years ago

B) ¿Con qué experimentos o instrumentos se calibrarán los patrones distribuidos

Setler [38]

English please !!!!!!!

8 0

3 years ago

A certain lightning bolt moves 10.0 C of charge. How many charges have been moved if the fundamental unit of charge is 1.6 x 10-

Andrej [43]

To solve this problem, use the ratio given by the total number of electrons or protons that exist as a function of the total charge, and inversely proportional to the value of the fundamental charge. The number of fundamental unit $|q_e|$ that constitutes a charge of 40.0C can be calculated as

$N = \frac{|Q|}{|q_e|}$

Here,

$q_e$ = Value of charge and it is the fundamental charge

Q = Total Charge

N = Total number of electron or protons

The number of fundamental units is calculated as follows

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$N = 6.25*10^{19}$

Therefore the number of fundamental charge units moved by lightning bolt is $6.25*10^{19}$

8 0

3 years ago

If the swimmer starts at rest, slides without friction, and descends through a vertical height of 2.41 m

AveGali [126]

Answer:

6.88 m/s

Explanation:

The Conservation of Energy states that:

Initial Kinetic Energy + Initial Potential Energy = Final Kinetic Energy + Final Potential Energy

So we can write

$mgh_{i}+\frac{1}{2}mv_{i} ^{2}=mgh_{f}+\frac{1}{2}mv_{f} ^{2}$

We can cancel the common factor of $m$ which leaves us with

$gh_{i}+\frac{1}{2}v_{i} ^{2}=gh_{f}+\frac{1}{2}v_{f} ^{2}$

Lets solve for $v_f$

$gh_{i}+\frac{v_{i} ^{2}}{2}=gh_{f}+\frac{v_{f} ^{2}}{2}$

Subtract $gh_f$ from both sides of the equation.

$gh_{i}+\frac{v_{i} ^{2}}{2}-gh_{f}=\frac{v_{f} ^{2}}{2}$

Multiply both sides of the equation by 2.

$2(gh_{i}+\frac{v_{i} ^{2}}{2}-gh_{f})={v_{f} ^{2}$

Simplify the left side.

Apply the distributive property.

$2(gh_{i})+2\frac{v_{i} ^{2}}{2}+2(-gh_{f})={v_{f} ^{2}$

Cancel the common factor of 2.

$2gh_{i}+v_{i} ^{2}-2gh_{f}={v_{f} ^{2}$

Take the square root of both sides of the equation to eliminate the exponent on the right side.

${v_{f}=\sqrt{2gh_{i}+v_{i} ^{2}-2gh_{f}}$

We are given $g,v_{i},h_{i},h_{f}$ .

We can now solve for the final velocity.

${v_{f}=\sqrt{(2*9.81*2.41)+(0^{2})-(2*9.81*0)$

Anything multiplied by 0 is 0.

${v_{f}=\sqrt{2*9.81*2.41$

${v_{f}=\sqrt{47.2842$

$v_f=6.88$

7 0

1 year ago