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Ostrovityanka [42]

2 years ago

8

How would a small bar magnet be oriented when placed at position X?

1 answer:

Illusion [34]2 years ago

7 0

Based on the document attached, the small bar magnet will be oriented when placed at position X is option d:

<h3>What is the magnet position about?</h3>

The orientation of the magnets is known to be one that do affect its magnetic force. Atoms atoms, such as iron, are known to often gives a stronger magnetic force when compared to other atoms.

These atoms can be seen as tiny magnets, that has the north and south of each pole. If the poles is known to be oriented in all directions, the material will not bring about a net magnetic force.

Therefore, Based on the document attached, the small bar magnet will be oriented when placed at position X is option d because the shapes of the magnetic field lines in regards to the bar magnet are said to be closed.

Learn more about magnet from

brainly.com/question/12839896

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a freight train travels at v=60(1-e^-t) ft/s, where t is the elapsed time in seconds. Determine the distance traeled in tree sec

lesantik [10]

Answer

given,

$v = 60(1-e^{-t})\ ft/s$

t = 3 s

we know,

$v = \dfrac{dx}{dt}$

$a = \dfrac{dv}{dt}$

position of the particle

$dx = v dt$

integrating both side

$\int dx =\int 60(1-e^{-t}) dt$

$x = 60 t + 60 e^{-t}$

Position of the particle at t= 3 s

$x = 60\times 3+ 60 e^{-3}$

x = 182.98 ft

Distance traveled by the particle in 3 s is equal to 182.98 ft

now, particle’s acceleration

$a = \dfrac{dv}{dt}$

$a = \dfrac{d}{dt}60(1-e^{-t})$

$a = 60 e^{-t}$

at t= 3 s

$a = 60 e^{-3}$

a = 2.98 ft/s²

acceleration of the particle is equal to 2.98 ft/s²

7 0

4 years ago

A wave has a wavelength of 10mm and a frequency of 5 hz what is the speed?

geniusboy [140]

V=fλ
v=5*0.01
Therefore v=0.05

3 0

4 years ago

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baherus [9]

Answer:

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3 years ago

Consider two parallel plate capacitors. The plates on Capacitor B have half the area as the plates on Capacitor A, and the plate

vichka [17]

Answer:

CB = 4.45 x 10⁻⁹ F = 4.45 nF

Explanation:

The capacitance of a parallel plate capacitor is given by the following formula:

C = ε₀A/d

where,

C = Capacitance

ε₀ = Permeability of free space

A = Area of plates

d = Distance between plates

FOR CAPACITOR A:

C = CA = 17.8 nF = 17.8 x 10⁻⁹ F

A = A₁

d = d₁

Therefore,

CA = ε₀A₁/d₁ = 17.8 x 10⁻⁹ F ----------------- equation 1

FOR CAPACITOR B:

C = CB = ?

A = A₁/2

d = 2 d₁

Therefore,

CB = ε₀(A₁/2)/2d₁

CB = (1/4)(ε₀A₁/d₁)

using equation 1:

CB = (1/4)(17.8 X 10⁻⁹ F)

<u>CB = 4.45 x 10⁻⁹ F = 4.45 nF</u>

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3 years ago

HELP URGENT!!!!!!!!!!!!!!!!!!!!!!!!

nirvana33 [79]

D
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3 years ago