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Tomtit [17]
3 years ago
14

How can a conductor be made superconductor?​

Physics
1 answer:
omeli [17]3 years ago
3 0

Answer:

a superconductor in a magnetic field and you'll make electric currents flow through its surface. These currents create a magnetic field that exactly cancels the original field trying to get inside the superconductor and repelling the magnetic field outside.

Explanation:

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An object travels with velocity v = 4.0 meters/second and it makes an angle of 60.0° with the positive direction of the y-axis.
Galina-37 [17]
Vx=cos60(4)
x-component of velocity
If you think about it, it makes a right triangle when you combine all the different types of forces together such as v, vx and vy. Then, you can use trigonometry and soh cah toa in order to figure out vx.
8 0
3 years ago
What's the frequency of a wave with a wavelength of 10 and velocity of 200m/s?
uranmaximum [27]

Answer:

\boxed {\boxed {\sf 20 \ Hz}}

Explanation:

The frequency of a wave can be found using the following formula.

f=\frac{v}{\lambda}

where <em>f</em> is the frequency, <em>v</em> is the velocity/wave speed, and λ is the wavelength.

The wavelength is 10 meters and the velocity is 200 meters per second.

  • 1 m/s can also be written as 1 m*s^-1

Therefore:

v= 200 \ m*s^{-1} \\\lambda = 10 \ m

Substitute the values into the formula.

f=\frac{200 \ m*s^{-1}}{10 \ m}

Divide and note that the meters (m) will cancel each other out.

f=\frac{200 \ s^{-1}}{10 \ }

f=20 \ s^{-1}

  • 1 s^-1 is equal to Hertz
  • Therefore, our answer of 20 s^-1 is equal to 20 Hz

f= 20 \ Hz

The frequency of the wave is <u>20 Hertz</u>

7 0
3 years ago
Jessica is running a 10K. She alternates between running and walking each kilometer. She runs at a rate of 1 kilometer every 5 m
saveliy_v [14]

Answer:

c.100 minutes

Explanation:

Total distance = 10 km

Runs for 1 km every 5 minutes

walks 1 km every 15 min

She alternates between walking and running  so,  Jessica will walk 5 km and run 5 Km

Time taken by Jessica for walking : 5 km

Time taken to walk 1 km=5 minutes

Time taken to walk 5 km

=> 5 X 5

=>25 minutes

Time taken by Jessica for Running : 5km

Time taken to run 1 km = 15 minutes

 

=> 5 X 15

=>75 minutes

Total time taken   = Time taken by Jessica for walking + Time taken by Jessica for Running

=>25 minutes +75 minutes

=> 100 minutes

8 0
3 years ago
Read 2 more answers
Consider the motion of a 4.00-kg particle that moves with potential energy given by U(x) = + a) Suppose the particle is moving w
gtnhenbr [62]

Correct question:

Consider the motion of a 4.00-kg particle that moves with potential energy given by

U(x) = \frac{(2.0 Jm)}{x}+ \frac{(4.0 Jm^2)}{x^2}

a) Suppose the particle is moving with a speed of 3.00 m/s when it is located at x = 1.00 m. What is the speed of the object when it is located at x = 5.00 m?

b) What is the magnitude of the force on the 4.00-kg particle when it is located at x = 5.00 m?

Answer:

a) 3.33 m/s

b) 0.016 N

Explanation:

a) given:

V = 3.00 m/s

x1 = 1.00 m

x = 5.00

u(x) = \frac{-2}{x} + \frac{4}{x^2}

At x = 1.00 m

u(1) = \frac{-2}{1} + \frac{4}{1^2}

= 4J

Kinetic energy = (1/2)mv²

= \frac{1}{2} * 4(3)^2

= 18J

Total energy will be =

4J + 18J = 22J

At x = 5

u(5) = \frac{-2}{5} + \frac{4}{5^2}

= \frac{4-10}{25} = \frac{-6}{25} J

= -0.24J

Kinetic energy =

\frac{1}{2} * 4Vf^2

= 2Vf²

Total energy =

2Vf² - 0.024

Using conservation of energy,

Initial total energy = final total energy

22 = 2Vf² - 0.24

Vf² = (22+0.24) / 2

Vf = \sqrt{frac{22.4}{2}

= 3.33 m/s

b) magnitude of force when x = 5.0m

u(x) = \frac{-2}{x} + \frac{4}{x^2}

\frac{-du(x)}{dx} = \frac{-d}{dx} [\frac{-2}{x}+ \frac{4}{x^2}

= \frac{2}{x^2} - \frac{8}{x^3}

At x = 5.0 m

\frac{2}{5^2} - \frac{8}{5^3}

F = \frac{2}{25} - \frac{8}{125}

= 0.016N

8 0
4 years ago
Say you want to make a sling by swinging a mass M of 1.9 kg in a horizontal circle of radius 0.042 m, using a string of length 0
padilas [110]

Answer: T= 715 N

Explanation:

The only external force (neglecting gravity) acting on the swinging mass, is the centripetal force, which. in this case, is represented by the tension in the string, so we can say:

T = mv² / r

At the moment that the mass be released, it wil continue moving in a straight line at the same tangential speed that it had just an instant before, which is the same speed included in the centripetal force expression.

So the kinetic energy will be the following:

K = 1/2 m v² = 15. 0 J

Solving for v², and replacing in the expression for T:

T = 1.9 Kg (3.97)² m²/s² / 0.042 m = 715 N

3 0
4 years ago
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