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VikaD [51]
3 years ago
6

During the experiment if you could double the breakaway magnetic force with all other quantities left unchanged, what is the new

value for the critical velocity if it was vo initially?
Physics
1 answer:
sergiy2304 [10]3 years ago
6 0
There are some missing information in the question.
However, since you are talking about magnetic force, I think you refer to the Lorentz force. When a particle of charge q and velocity v is immersed in a magnetic field of intensity B, the force acting on the particle is:
F=qvBsin\theta
where \theta is the angle between the magnetic field and the direction of the particle.
Therefore, if force F is doubled, then also the velocity v must be double of its initial value:
v=2v_0
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The melting point will be lowered and broadened.

Explanation:

The melting point of a substance is the temperature at which the substance changes state from solid to liquid. The incomplete drying of a sample may result in the presence of impurities. There is presence of solid sample in the solution that fails to dry. The solid sample is not fully crystallized from the solvent or any other liquid with which it can form a true solution. The presence of these impurities in a sample for example, the incomplete removal of a recrystallization solvent cause the melting point to both lowered and broadened.

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A carpenter builds an exterior house wall with a layer of wood 3.0 cm thick on the outside and a layer of Styrofoam insulation 2
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Answer:

A. T=15.54 °C

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To solve this problem we need to use the Fourier's law for thermal conduction:

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\frac{Q}{A} =\frac{T_1-T_0}{d_w}k_w=\frac{T_2-T_1}{d_s}k_s\\T_1(\frac{k_w}{d_w}+\frac{k_s}{d_s})=T_2\frac{k_s}{d_s}+T_0\frac{k_w}{d_w}\\T_1=\frac{T_2\frac{k_s}{d_s}+T_0\frac{k_w}{d_w}}{\frac{k_w}{d_w}+\frac{k_s}{d_s}}\\T_1= 15.54 \°C

Then, to find the rate of heat flow per square meter, we have:

\frac{Q}{A}=\frac{T_1-T_0}{d_w}k_w=0.119 \frac{W}{m^2}\\\frac{Q}{A}=\frac{T_2-T_1}{d_s}k_s= 0.119 \frac{W}{m^2}

T_0: Temperature \ in \ the \ house\\T_1: Temperature \ at \ the \ plane \ between \ wood \ and \ styrofoam\\T_2: Temperature \ outside\\k_w: k \ for \ wood\\d_w: wood \ thickness\\k_s: k \ for \ styrofoam\\d_s: styrofoam \ thickness

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how does chemical energy cause a change?:

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