All categories

3 years ago

What is the main difference between a maglev train and a normal train?

1 answer:

5 0

Regular trains follow train tracks and travel on rails; Maglev trains are Magnetically Levitated using powerful electromagnets, so that they actually hover slightly above the ground. This massively reduces friction and allows them to travel much faster (kind of like how a puck can go really far with only a little push on an air hockey table).

You might be interested in

Write the equation that links current, potential difference, and resistance

mrs_skeptik [129]

You can write the equation in 3 different ways, depending on which quantity you want to be the dependent variable. Any one of the three forms can be derived from either of the other two with a simple algebra operation. They're all the same relationship, described by "Ohm's Law".

==> Current = (potential difference) / (resistance)

==> Potential difference = (current) x (resistance)

==> Resistance = (potential difference) / (resistance)

4 0

3 years ago

How can I make the atomic model of boron?

zloy xaker [14]

This should help

there always needs to be 2 electrons in the first shell then 8 electrons in each shell and continue till you not have any more number of electron left

8 0

4 years ago

What force do scientist think is responsible for matter clumping together

Stells [14]

It MUST be either glue or gravity.

3 0

3 years ago

A space probe is fired as a projectile from the Earth's surface with an initial speed of 2.05 104 m/s. What will its speed be wh

Elanso [62]

Answer:

The value is $v = 2.3359 *10^{4} \ m/s$

Explanation:

From the question we are told that

The initial speed is $u = 2.05 *10^{4} \ m/s$

Generally the total energy possessed by the space probe when on earth is mathematically represented as

$T__{E}} = KE__{i}} + KE__{e}}$

Here $KE_i$ is the kinetic energy of the space probe due to its initial speed which is mathematically represented as

$KE_i = \frac{1}{2} * m * u^2$

=> $KE_i = \frac{1}{2} * m * (2.05 *10^{4})^2$

=> $KE_i = 2.101 *10^{8} \ \ m \ \ J$

And $KE_e$ is the kinetic energy that the space probe requires to escape the Earth's gravitational pull , this is mathematically represented as

$KE_e = \frac{1}{2} * m * v_e^2$

Here $v_e$ is the escape velocity from earth which has a value $v_e = 11.2 *10^{3} \ m/s$

=> $KE_e = \frac{1}{2} * m * (11.3 *10^{3})^2$

=> $KE_e = 6.272 *10^{7} \ \ m \ \ J$

Generally given that at a position that is very far from the earth that the is Zero, the kinetic energy at that position is mathematically represented as

$KE_p = \frac{1}{2} * m * v^2$

Generally from the law energy conservation we have that

$T__{E}} = KE_p$

$2.101 *10^{8} m + 6.272 *10^{7} m = \frac{1}{2} * m * v^2$

=> $5.4564 *10^{8} = v^2$

=> $v = \sqrt{5.4564 *10^{8}}$

=> $v = 2.3359 *10^{4} \ m/s$

4 0

3 years ago

A steel pipe of 12-in. outer diameter is fabricated from 1 4 -in.-thick plate by welding along a helix that forms an angle of 25

Varvara68 [4.7K]

Answer:

Normal stress = 66/62.84 = 1.05kips/in²

shearing stress = T/2 = 0.952/2 = 0.476 kips/in²

Explanation:

A steel pipe of 12-in. outer diameter d₂ =12in d₁= 12 -4in = 8in

4 -in.-thick

angle of 25°

Axial force P = 66 kip axial force

determine the normal and shearing stresses

Normal stress б = force/area = P/A

= 66/ (П* (d₂²-d₁²)/4

=66/ (3.142* (12²-8²)/4

= 66/62.84 = 1.05kips/in²

Tangential stress T = force* cos ∅/area = P/A

= 66* cos 25/ (П* (d₂²-d₁²)/4

=59.82/ (3.142* (12²-8²)/4

= 59.82/62.84 = 0.952kips/in²

shearing stress = tangential stress /2

= T/2 = 0.952/2 = 0.476 kips/in²

8 0

3 years ago