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3 years ago

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Physics

1 answer:

grin007 [14]3 years ago

6 0

Answer:

You have to Find x, y, and z such that x³+y³+z³=k, for each k from 1 to 100.

Explanation: And here's the answer in Binary for some fun: 01000110 01101001 01101110 01100100 00100000 01111000 00101100 00100000 01111001 00101100 00100000 01100001 01101110 01100100 00100000 01111010 00100000 01110011 01110101 01100011 01101000 00100000 01110100 01101000 01100001 01110100 00100000 01111000 11000010 10110011 00101011 01111001 11000010 10110011 00101011 01111010 11000010 10110011 00111101 01101011 00101100 00100000 01100110 01101111 01110010 00100000 01100101 01100001 01100011 01101000 00100000 01101011 00100000 01100110 01110010 01101111 01101101 00100000 00110001 00100000 01110100 01101111 00100000 00110001 00110000 00110000 00101110

Separate Answer: 01001000 01100101 01101100 01101100 01101111 00100000 01110100 01101000 01100101 01110010 01100101

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A test tube of length L and cross-sectional area A is submerged in water with the open end down so that the edge of the tube is

Margaret [11]

Answer:

if we measure the change in height of the gas within the had and obtain a straight line in relation to the depth we can conclude that the air complies with Boye's law.

Explanation:

The air in the tube can be considered an ideal gas,

P V = nR T

In that case we have the tube in the air where the pressure is P1 = P_atm, then we introduce the tube to the water to a depth H

For pressure the open end of the tube is

P₂ = P_atm + ρ g H

Let's write the gas equation for the colon

P₁ V₁ = P₂ V₂

P_atm V₁ = (P_atm + ρ g H) V₂

V₂ = V₁ P_atm / (P_atm + ρ g h)

If the air obeys Boyle's law e; volume within the had must decrease due to the increase in pressure, if we measure the change in height of the gas within the had and obtain a straight line in relation to the depth we can conclude that the air complies with Boye's law.

The main assumption is that the temperature during the experiment does not change

6 0

3 years ago

Suppose a straight 1.00-mm-diameter copper wire could just "float" horizontally in air because of the force due to the Earth’s m

insens350 [35]

To solve this problem it is necessary to apply the concepts related to the Force since Newton's second law, as well as the concept of Electromagnetic Force. The relationship of the two equations will allow us to find the magnetic field through the geometric relations of density and volume.

$F_{mag}= BIL$

Where,

B = Magnetic Field

I = Current

L = Length

<em>Note: $F_{mag}$ is a direct adaptation of the vector relation $F=q \times V \times B$ </em>

From Newton's second law we know that the relation of Strength and weight is determined as

$F_g = mg$

Where,

m = Mass

g = Gravitational Acceleration

For there to be balance the two forces must be equal therefore

$F_{mag} = F_g$

$BIL = mg$

Our values are given as,

Diameter $(d) = 1.0mm = 1*10^{-3}m$

Radius $(r) = \frac{d}{2} = \frac{1*10^{-3}}{2} = 0.5*10^{-3}m$

Magnetic Field $(B) = 5.0*10^{-5} T$

From the relationship of density another way of expressing mass would be

$\rho = \frac{m}{V} \rightarrow m = \rho V$

At the same time the volume ratio for a cylinder (the shape of the wire) would be

$V = \pi r^2 L \rightarrow L =Length, r= Radius$

Replacing this two expression at our first equation we have that:

$BIL = mg$

$BIL = ( \rho V)g$

$BIL = ( \rho \pi r^2 L)g$

Re-arrange to find I

$I = \frac{( \rho \pi r^2 L)g}{BL}$

$I = \frac{( \rho \pi r^2 )g}{B}$

We have for definition that the Density of copper is $8.9*10^3 Kg/m^3$ , gravity acceleration is $9.8m/s^2$ and the values of magnetic field (B) and the radius were previously given, then: