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fiasKO [112]
3 years ago
11

Suppose that instead of a long straight wire, a shortstraight wire was used. The distance from the wire to thepoint that the mag

netic field is measured would then be comparableto the length of the short wire. How would you expect themeasured value of the magnetic field to change if the field weresampled at the same distance from each of the two wires?
Physics
1 answer:
vichka [17]3 years ago
4 0

Answer:

Thus, if field were sampled at same distance, the field due to short wire is greater than field due to long wire.

Explanation:

The magnetic field, B of long straight wire can be obtained by applying ampere's law

B= \frac{\mu_0 I}{2\pi r}

I is here current, and r's the distance from the wire to the field of measurement.

The magnetic field is obviously directly proportional to the current wire. From this expression.

As the resistance of the long cable is proportional to the cable length, the short cable becomes less resilient than the long cable, so going through the short cable (where filled with the same material) is a bigger amount of currents. If the field is measured at the same time, the field is therefore larger than the long wire because of the short wire.

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A train starts from rest and travels for 5.0 s with a uniform acceleration of 1.5 m/s2. What is the final velocity of the train?
alexandr1967 [171]

Answer:

Final speed of the train is 7.5 m/s

Explanation:

It is given that,

Uniform acceleration of the train is, a = 1.5 m/s²

It starts from rest and travels for 5.0 s. We have to find the final velocity of the train. By using first equation of motion as :

v=u+at

Here, train starts from rest so, u = 0

v=0+1.5\ m/s^2\times 5\ s  

v = 7.5 m/s

So, the final velocity of the train is 7.5 m/s. Hence, this is the required solution.

7 0
3 years ago
Read 2 more answers
. A vertical electric field is set up in space to compensate for the gravitational force on a point charge. What is the required
Delicious77 [7]

(a) The required magnitude of the electric field when the point charge is an electron is 5.57 x 10⁻¹¹ N/C.

(b) The required magnitude of the electric field when the point charge is an proton is 1.02 x 10⁻⁷ N/C.

<h3>Magnitude of electric field </h3>

The magnitude of electric field is given by the following equation.

F = qE

But F = mg

mg = qE

E = mg/q

where;

  • E is the electric field
  • m is mass of the particle
  • g is acceleration due to gravity
  • q is charge of the particle
<h3>For an electron</h3>

E = (9.11 x 10⁻³¹ x 9.8)/(1.602 x 10⁻¹⁹)

E = 5.57 x 10⁻¹¹ N/C

<h3>For proton</h3>

E = (1.67 x 10⁻²⁷ x 9.8)/(1.602 x 10⁻¹⁹)

E = 1.02 x 10⁻⁷ N/C

Thus, the required vertical electric field is greater when the charge is proton.

Learn more about electric field here: brainly.com/question/14372859

#SPJ1

4 0
2 years ago
A 50 kg object traveling at 100 m/s collides (perfectly elastic) with a 50 kg object initially at rest.
tino4ka555 [31]

Answer:

a

Explanation:

8 0
3 years ago
Two identical cars, one on the moon and one on the earth, have the same speed and are rounding banked turns that have the same r
Naily [24]

Answer:

The value of the centripetal forces are same.

Explanation:

Given:

The masses of the cars are same. The radii of the banked paths are same. The weight of an object on the moon is about one sixth of its weight on earth.

The expression for centripetal force is given by,

F_{c} = \dfrac{mv^{2}}{r}

where, m is the mass of the object, v is the velocity of the object and r is the radius of the path.

The value of the centripetal force depends on the mass of the object, not on its weight.

As both on moon and earth the velocity of the cars and the radii of the paths are same, so the centripetal forces are the same.

3 0
3 years ago
A small logo is embedded in a thick block of crown glass (n = 1.52), 4.70 cm beneath the top surface of the glass. The block is
harkovskaia [24]

The concept required to solve this problem is the optical relationship that exists between the apparent depth and actual or actual depth. This is mathematically expressed under the equations.

d'w = d_w (\frac{n_{air}}{n_w})+d_g (\frac{n_{air}}{n_g})

Where,

d_g = Depth of glass

n_w = Refraction index of water

n_g = Refraction index of glass

n_{air} = Refraction index of air

d_w = Depth of water

I enclose a diagram for a better understanding of the problem, in this way we can determine that the apparent depth in the water of the logo would be subject to

d'w = d_w (\frac{n_{air}}{n_w})+d_g (\frac{n_{air}}{n_g})

d'w = (1.7cm) (\frac{1}{1.33})+(4.2cm)(\frac{1}{1.52})

d'w = 4.041cm

Therefore the distance below the upper surface of the water that appears to be the logo is 4.041cm

3 0
3 years ago
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