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3 years ago

When a wire within a closed circuit is coiled upon a nail, the nail will _______.

Physics

2 answers:

lara31 [8.8K]3 years ago

7 0

Answer:

will spark

Explanation:

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Natali5045456 [20]3 years ago

5 0

when wire is coiled upon a nail then it will posses magnetic characteristics

This is due to the coiled shape it will behave like a solenoid and the magnetic field of solenoid is given as

$B = \mu_o ni$

so here when wire is coiled up on a nail then due to the magnetic field of the coil the nail will attain magnetic characteristic.

so correct answer will be

<em>Possess magnetic properties </em>

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Current can be made to flow in a stationary conductor by ............................

lara31 [8.8K]

Answer:

A is the answer. Im only 12 and i hope this explanation helps you.

Explanation:

Lenz's Law of Electromagnetic Induction. Faraday's Law tells us that inducing a voltage into a conductor can be done by either passing it through a magnetic field, or by moving the magnetic field past the conductor and that if this conductor is part of a closed circuit, an electric current will flow.

3 0

3 years ago

#3: a container has the dimensions of 30 cm x 50 mm x 0.2 m. the density of its contents is 2.5 g/cm3. what is the mass of the s

WARRIOR [948]

Good afternoon!

We calculate the volume of the container in cm³. To do that, we must put the units in cm:

30 cm → 30 cm
50 mm → 5 cm
0.2 m → 20 cm

The volume is:

V = 30 . 5 . 20

V = 3000 cm³

Now, we calculate the mas with the formula:

m = dV

m = 2.5 · 3000

m = 7500 g

Dividing by 1000, we have the mass in kg:

m = 7.5 kg

4 0

4 years ago

Read 2 more answers

2. What do pitch and loudness have in common?

DanielleElmas [232]

Answer:

Both are subject to a persons interpretation

Explanation:

We hear people describe this when somebody is making an irresistible sound. usually people say the baby has a pitch scream.

5 0

3 years ago

A 30-km, 34.5-kV, 60-Hz, three-phase line has a positive-sequence series impedance z 5 0.19 1 j0.34 V/km. The load at the receiv

zmey [24]

Answer:

(a) With a short line, the A,B,C,D parameters are:

A = 1pu B = 1.685∠60.8°Ω C = 0 S D = 1 pu

(b) The sending-end voltage for 0.9 lagging power factor is 35.96 $KV_{LL}$

Explanation:

(a)

Considering the short transition line diagram.

Apply kirchoff's voltage law to the short transmission line.

Write the equation showing the relations between the sending end and the receiving end quantities.

Compare the line equations with the A,B,C,D parameter equations.

(b)

Determine the receiving-end current for 0.9 lagging power factor.

Determine the line-to-neutral receiving end voltage.

Determine the sending end voltage of the short transition line.

Determine the line-to-line sending end voltage which is the sending end voltage.

(c)

Determine the receiving-end current for 0.9 leading power factor.

Determine the sending-end voltage of the short transition line.

Determine the line-to-line sending end voltage which is the sending end voltage.

8 0

3 years ago

A nonconducting spherical shell, with an inner radius of 4 cm and an outer radius of 6 cm, has charge spread nonuniformly throug

Vaselesa [24]

Answer:

$1.57 * 10^{3} Q$

Explanation:

The volume charge density is defined by ρ = $\frac{Q}{V}$ (Equation A), where Q is the charge and V, the volume.

The units in the S.I. are $\frac{Coulombs}{m^{3} }$ , so we have to express the radius in meters:

inner radius = $4 cm * \frac{1 m}{100 cm} = 0.04m$

outer radius = $6 cm * \frac{1m}{100cm} = 0.06m$

Now, we know that the volume of the sphere is calculated by the formula:

$V = \frac{4}{3}\pi r^{3}$ , and as we have an spherical shell, the volume is calculated by the difference between the outher and inner spheres:

V = $\frac{4}{3}\pi (r_{o} ^{3} - r_{i} ^{3})$ , where $r_{o}$ is the outer radius and $r_{i}$ is the inner radius.

Replacing the volume formula in the Equation A:

ρ = $\frac{Q}{\frac{4}{3}\pi(r^{3} _{o}-r_{i} ^{3})}$

ρ = $\frac{3Q}{4\pi (r_{o} ^{3}-r_{i} ^{3} ) }$

Replacing the values of the outer and inner radius whe have:

ρ = $\frac{3Q}{4\pi (1.52 * 10^{-4})}$

ρ = $1.57 * 10^{3} Q$

4 0

4 years ago