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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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Tamiku [17]

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3 years ago

In 1965, a group of students wore black arm bands to school in protest of American policies in Vietnam. Administrators banned th

Ede4ka [16]

Answer:

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Explanation:

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3 years ago

g A 1.45-kg block is pushed against a vertical wall by means of a spring (k = 860 N/m). The coefficient of static friction betwe

olga_2 [115]

Answer:

The minimum compression is $x= 0.046m$

Explanation:

From the question we are told that

The mass of the block is $m_b = 1.45 kg$

The spring constant is $k = 860 N/m$

The coefficient of static friction is $\mu = 0.36$

For the the block not slip it mean the sum of forces acting on the horizontal axis is equal to the forces acting on the vertical axis

Now the force acting on the vertical axis is the force due to gravity which is mathematically given as

$F_y = m_b*g$

And the force acting on the horizontal axis is force due to the spring which is mathematically represented as

$F_x = k *x * \mu$

where x is the minimum compression to keep the block from slipping

Now equating this two formulas and making x the subject

$x = \frac{m_b * g}{k * \mu}$

substituting values we have

$x = \frac{1.45 * 9.8}{860 *0.36}$

$x= 0.046m$

3 0

3 years ago

Two 10-cm-diameter charged rings face each other, 21.0 cm apart. Both rings are charged to +40.0 nC. What is the electric field

Katyanochek1 [597]

Complete question:

Two 10-cm-diameter charged rings face each other, 21.0 cm apart. Both rings are charged to +40.0 nC. What is the electric field strength at the midpoint between the two rings ?

Answer:

The electric field strength at the mid-point between the two rings is zero.

Explanation:

Given;

diameter of each ring, d = 10 cm = 0.1 m

distance between the rings, r = 21.0 cm = 0.21 m

charge of each ring, q = 40 nC = 40 x 10⁻⁹ C

let the midpoint between the two rings = x

The electric field strength at the midpoint between the two rings is given as;

$E_{mid} = E_{right} +E_{left}\\\\E_{right} = \frac{KQ}{(x^2 + r^2)^\frac{2}{3} } \\\\E_{leftt} = -\ \frac{KQ}{(x^2 + r^2)^\frac{2}{3} }\\\\E_{mid} = \frac{KQ}{(x^2 + r^2)^\frac{2}{3} } - \frac{KQ}{(x^2 + r^2)^\frac{2}{3} } = 0$

Therefore, the electric field strength at the mid-point between the two rings is zero.

7 0

3 years ago

An object accelerates from rest and travels 53 m west in 5.2 s. Determine the acceleration

zmey [24]

Answer:

20.4m/s²

Explanation:

Given parameters:

Initial velocity = 0m/s

Distance = 53m

Time = 5.2s

Unknown:

Acceleration = ?

Solution:

This is a linear motion and we use the right motion equation;

S = ut + $\frac{1}{2}$ at²

S is the distance

u is the initial velocity

a is the acceleration

t is the time

Insert the parameters and solve;

53 = (0x 5.2) + $\frac{1}{2}$ x a x 5.2

53 = 2.6a

a = $\frac{53}{2.6}$ = 20.4m/s²

6 0

3 years ago