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Yuliya22 [10]
3 years ago
6

Which statement describes the relationship of resistance and current

Physics
1 answer:
slavikrds [6]3 years ago
6 0

This is picture describe as

More current mean less resistance

Less resistance mean more current

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Alex Ar [27]

Answer:

4- false

5- false (i think..)

Explanation:

6 0
3 years ago
How is the levitation of the SCMagLev train achieved? Just write a sentence.
Harman [31]

Answer:

<h2>The SCMaglev (superconducting maglev, formerly called the MLU) is a magnetic levitation (maglev) railway system developed by Central Japan Railway Company (JR Central) and the Railway Technical Research Institute</h2>

5 0
2 years ago
Read 2 more answers
A car of mass 500kg travelling at 60m/s has it speed reduced to 40m/s by a constant breaking force over a distance of 200m. find
uranmaximum [27]

Answer:

Ek1 = 900000 [J]

Ek1 = 400000 [J]

Explanation:

In order to solve this problem we must remember that kinetic energy is defined as the product of mass by velocity squared by a medium. Therefore using the following equation we have:

E_{k1}=\frac{1}{2}*m*v1^{2}

where:

m = mass = 500 [kg]

v1 = 60 [m/s]

So we have:

Ek1 = 0.5*500*(60^2)

Ek1 = 900000 [J]

and:

Ek2 = 0.5*500*(40^2)

Ek2 = 400000 [J]

6 0
3 years ago
An unlit match is held near (not touching) a Bunsen flame. The match doesn't get hot enough to light because
ohaa [14]
The answer is option D)
this is because the heat radiated by the flame is mostly absorbed by the air surrounding it, so the air becomes hot and its density decreases (because of expansion), therefore it goes up and it is replaced by cooler air. since all of the hot air flies up, non goes side ways to heat up the match stick, hence it remains cool and does not light up.

option A) also sounds correct, but it isn't. this is because the flame IS hot enough to burn the match stick, it's just that the match stick is positioned the wrong way
6 0
3 years ago
In the physics lab, a block of mass M slides down a frictionless incline from a height of 35cm. At the bottom of the incline it
bogdanovich [222]

Solution :

Given :

M = 0.35 kg

$m=\frac{M}{2}=0.175 \ kg$

Total mechanical energy = constant

or $K.E._{top}+P.E._{top} = K.E._{bottom}+P.E._{bottom}$

But $K.E._{top} = 0$ and $P.E._{bottom} = 0$

Therefore, potential energy at the top = kinetic energy at the bottom

$\Rightarrow mgh = \frac{1}{2}mv^2$

$\Rightarrow v = \sqrt{2gh}$

      $=\sqrt{2 \times 9.8 \times 0.35}$      (h = 35 cm = 0.35 m)

      = 2.62 m/s

It is the velocity of M just before collision of 'm' at the bottom.

We know that in elastic collision velocity after collision is given by :

$v_1=\frac{m_1-m_2}{m_1+m_2}v_1+ \frac{2m_2v_2}{m_1+m_2}$

here, $m_1=M, m_2 = m, v_1 = 2.62 m/s, v_2 = 0$

∴ $v_1=\frac{0.35-0.175}{0.5250}+\frac{2 \times 0.175 \times 0}{0.525}

      $=\frac{0.175}{0.525}+0$

     = 0.33 m/s

Therefore, velocity after the collision of mass M = 0.33 m/s

 

3 0
3 years ago
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