Susan’s monthly energy usage is shown in the table.

The melting point of pure water is _____. 32°C 100°C 0°C 212°C

BabaBlast [244]

Answer:

0°C and 32°F

Explanation:

ihavetotypemoresouhyeahig

3 0

3 years ago

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At a separation distance of 0.500-m, two like-charged balloons experience a repulsive force of 0.320 N. If the distance is is de

soldi70 [24.7K]

Answer:

2.9 N

Explanation:

When the separation distance, r, is 0.5 m, the electrostatic force is 0.32 N. Electrostatic force is given as:

F = (k * q1 * q2) / r²

Where F = force acting on the balloons

k = Coulombs constant

Therefore:

0.32 = (k * q1 * q2) / 0.5²

=> k * q1 * q2 = 0.32 * 0.5² ------------(1)

When the distance is decreased by 3, that is r = r/3 = 0.5/3

F = (k * q1 * q2) / (0.5/3)² ------------(2)

Putting (1) into (2):

=> F = (0.32 * 0.5²) / (0.5/3)²

F = (0.32 * 0.5² * 3²) / 0.5²

F = 2.9 N

Therefore, the force would be 2.9 N

3 0

3 years ago

A wire is oriented along the x-axis. It is connected to two batteries, and a conventional current of 2.4 A runs through the wire

Deffense [45]

Answer:

$\vec{F}=0.40176 N \hat{k}$

Explanation:

To calculate the force we need to use this equation

$\vec{F}=\int\limits^L_0 {i(\vec{dl}\times\vec{B})}$

where L is the total length of the wire

So in this case the small element of current is

$\vec{dl} = dx \hat{i}$

Because x is the direction of the current flow.

As is said in the problem B is such that

$\vec{B} = B \hat{j} = 0.62\hat{j} [ T]$

so to use the equation above we first calculate the following cross product:

$\vec{dl}\times\vec{B}=dx \hat{i}\times B \hat{j} = Bdx\hat{k}$

so the force:

$F = \int\limits^L_0 {i(\vec{dl}\times\vec{B})}=\int\limits^L_0{iBdx\hat{k}}$

So here we use the fact that B=0 in any point of the x axis that is not $x^{'}=0.27 [m]$ , that means that we only need to do the integration between a very short distant behind the point $x^{'}=0.27 [m]$ and a very short distant after that point, meaning: