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3 years ago

What was your electric potential relative to a metal pipe if a spark jumped 1.1 cm through dry air from your finger to the pipe?

Physics

2 answers:

aleksley [76]3 years ago

7 0

Answer:

Electric potential = V = 33 KV

Explanation:

The breakdown is an electric stress phenomenon. The value of electric field at which breakdown of dry air occur is 3 × 10^6 V/m. So,

E = 3 × 10^6 V/m

d = 1.1 cm = 1.1 × 10^-2 m

We know that:

E = V/d

V = Ed

V = (3 × 10^6)( 1.1 × 10^-2)

V = 33000 V

V = 33 KV

romanna [79]3 years ago

6 0

The electric field is given by volts/distance: $E= \frac{V}{d}$ . The breakdown voltage of dry air is about 3x10^6V/m. So solving for V we get
$V=Ed$
or $V=(3e6V/m)(0.011m)=33,000V$

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Savatey [412]

Answer:

$41.81^{\circ}$

Explanation:

The tidal current flows to the east at 2.0 m/s and the speed of the kayaker is 3.0 m/s.

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In order to travel straight across the harbor, the vector addition of both the velocities (i.e the resultant velocity, $\vec {R}$ must be in the north direction.

Let $\overrightarrow{AB}$ is the speed of the kayaker having angle \theta measured north of east as shown in the figure.

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3 years ago

Two workers pull horizontally on a heavy box. but one pulls twice as hard as the other. The larger pull is directed at 21.0° wes

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Answer:

The magnitude of F1 is

$|F1|=358.74 \ N$

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$|F2|=179.37\ N$

And the direction of F2 is

$\alpha = 44.214^o$

Explanation:

<u>Net Force </u>

Forces are represented as vectors since they have magnitude and direction. The diagram of forces is shown in the figure below.

The larger pull F1 is directed 21° west of north and is represented with the blue arrow. The other pull F2 is directed to an unspecified direction (red arrow). Since the resultant Ft (black arrow) is pointed North, the second force must be in the first quadrant. We must find out the magnitude and angle of this force.

Following the diagram, the sum of the vector components in the x-axis of F1 and F2 must be zero:

$\displaystyle -2F\ sin21^o+F\ cos\alpha =0$