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MariettaO [177]
3 years ago
5

Which freezes faster, hot water or cold water?

Physics
2 answers:
Digiron [165]3 years ago
8 0
Hot water<span> seems to </span>freeze faster<span> than </span>cold water<span>, known as the Mpemba effect. The effect was named after the Tanzanian student who in 1963 noticed that </span>hot<span> ice cream mix </span>freezes faster<span> than a </span>cold<span> one. hope this helped</span>
Marina CMI [18]3 years ago
5 0
Hot water freezes faster
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Rotor windings in wound rotor motors are connected to ? on the shaft of the motor. Brushes allow those windings to be connected
Mashutka [201]

Answer:

Slip Rings

Explanation:

The wound rotor motor has a three-phase winding with each one connected to seperate slip rings. These slip rings contain brushes which form a secondary circuit where resistance can be inserted and this will allow for the rotor current to run more in phase with the stator current which will result in increased torque that is created

6 0
4 years ago
Stars originate as large bodies of slowly rotating gas.Because of gravity, these clumps of gas slowly decrease in size.The angul
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Answer:

E.) conservation of angular momentum

Explanation:

The angular momentum is defined as:

L=r x mv

where r is the radius of the star, m is the mass and v the angular velocity.

and angular momentum is an amount that is conserved, so the angular momentum before the star is compressed must be equal to the angular momentum after the star was compressed:

r_{1} x mv_{1}=r_{2} x mv_{2}

the second radius is smaller than the first radius, since the star shrinked, the second angular velocity must be greater that the first.

In other words, the angular velicity increases as the star shrinks because of the conservation of angular momentum.

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3 years ago
For copper, ρ = 8.93 g/cm3 and M = 63.5 g/mol. Assuming one free electron per copper atom, what is the drift velocity of electro
viktelen [127]

Answer:

V_d = 1.75 × 10⁻⁴ m/s

Explanation:

Given:

Density of copper, ρ = 8.93 g/cm³

mass, M = 63.5 g/mol

Radius of wire = 0.625 mm

Current, I = 3A

Area of the wire, A = \frac{\pi d^2}{4} = A = \frac{\pi 0.625^2}{4}

Now,

The current density, J is given as

J=\frac{I}{A}=\frac{3}{ \frac{\pi 0.625^2}{4}}= 2444619.925 A/mm²

now, the electron density, n = \frac{\rho}{M}N_A

where,

N_A=Avogadro's Number

n = \frac{8.93}{63.5}(6.2\times 10^{23})=8.719\times 10^{28}\ electrons/m^3

Now,

the drift velocity, V_d

V_d=\frac{J}{ne}

where,

e = charge on electron = 1.6 × 10⁻¹⁹ C

thus,

V_d=\frac{2444619.925}{8.719\times 10^{28}\times (1.6\times 10^{-19})e} = 1.75 × 10⁻⁴ m/s

4 0
3 years ago
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Resistivity of nichrome is high.

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Answer:

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