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2 years ago

Tony uses the device shown in the diagram to model how an electromagnet is used in his uncle’s scrap metal yard.

2 answers:

7 0

An electromagnet is one of the most fundamental experimental set-ups that is commonly used to describe the phenomenon of electromagnetic induction. In addition, there are a lot of ways in order to increase or decrease the strength of an electromagnet and one of which is by <span>spreading the wire loops farther apart.</span>

SpyIntel [72]2 years ago

3 0

Answer:

Option (1)

Explanation:

When a coil carrying current it behaves like an magnet. Such magnet is called electromagnet.

To vanish its magnetic field, the current should be stopped.

So, tony drops the staples in the scrap metal bin by removing the battery, so that the magnetism of nail becomes vanish.

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which has a greater kinetic energy, a bowling ball that has a mass of 5kg travelling at 6m/s, or a ship that has a mass of 12000

AVprozaik [17]

Answer:bowling ball has greater kinetic energy

Explanation:

Kinetic energy of bowling ball:

mass=m=5kg

Velocity=v=6m/s

Kinetic energy =ke

Ke=0.5 x m x v x v

Ke=0.5 x 5 x 6 x 6

Ke=90J

Kinetic energy of ship:

mass=m=120000kg

velocity=v=0.02m/s

Ke=0.5 x m x v x v

Ke=0.5 x 120000 x 0.02 x 0.02

Ke=24J

6 0

2 years ago

High‑speed ultracentrifuges are useful devices to sediment materials quickly or to separate materials. An ultracentrifuge spins

pantera1 [17]

Answer:

Fc = 7.14N

Explanation:

First of all, let's convert everything to the same unit system:

m = 0.0031kg R = 13.1cm * 1m / 100cm = 0.131m

ω = 50000 rev/min * 1rev /( 2π rad ) * 1min / 60s = 132.63 rad/s

Now we can calculate centripetal force as:

$Fc = m * \frac{V^2}{R} = m * \frac{(\omega*R)^2}{R}=m*R*\omega ^2$

Replacing the values we get the answer:

Fc = 7.14N

3 0

2 years ago

a rod of some material 0.20 m long elongates 0.20 mm on heating from 21 to 120°c. determine the value of the linear coefficient

Rufina [12.5K]

Answer:

The value of the linear coefficient of thermal expansion is : α=1.01 *10⁻⁵ (ºC)⁻¹

Explanation:

Li = 0.2m

ΔL = 0.2 mm = 0.0002m

T1 = 21ºC

T2 = 120ºC

ΔT =99ºC

α =ΔL/(Li*ΔT)

α =0.0002m /(0.2m * 99ºC)

α = 1.01 *10⁻⁵ (ºC)⁻¹

4 0

2 years ago

In Physics lab, Ellen has been given the task of constructing a simple motor. She must find common household items at home and b

Aleks [24]

Answer:

a) battery-->electrical current-->copper wire rotor -->magnet--> mechanical energy

Explanation:

6 0

2 years ago

Read 2 more answers

Eac of the two Straight Parallel Lines Each of two very long, straight, parallel lines carries a positive charge of 24.00 m C/m.

Cloud [144]

Answer:

The magnitude of the electric field at a point equidistant from the lines is $4.08\times10^{5}\ N/C$

Explanation:

Given that,

Positive charge = 24.00 μC/m

Distance = 4.10 m

We need to calculate the angle

Using formula of angle

$\theta=\sin^{-1}(\dfrac{\dfrac{d}{2}}{2d})$

$\theta=\sin^{-1}(\dfrac{1}{4})$

$\theta=14.47^{\circ}$

We need to calculate the magnitude of the electric field at a point equidistant from the lines

Using formula of electric field

$E=\dfrac{2k\lambda}{r}\times2\cos\theat$

Put the value into the formula

$E=\dfrac{2\times9\times10^{9}\times24.00\times2\times10^{-6}\cos14.47}{2.05}$

$E=408094.00\ N/C$

$E=4.08\times10^{5}\ N/C$

Hence, The magnitude of the electric field at a point equidistant from the lines is $4.08\times10^{5}\ N/C$

6 0

2 years ago