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Dima020 [189]
2 years ago
10

Which observational tool helped astronomers Arno Penzias and Robert Wilson discover the existence of the cosmic microwave backgr

ound (CMB)?
Physics
1 answer:
Softa [21]2 years ago
4 0
<span>In 1964, while experimenting with the Holmdel Horn Antenna, which was used as a radio telescope, Penzias and Wilson accidentally discovered the microwave background radiation that exists universally. The Holmdel Horn Antenna was used to support the "Big Bang Theory" as opposed to the "Steady State Theory".</span>
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Resolving vectors question. See diagram enclosed.

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A 20 cm-long wire carrying a current of 6 A is immersed in a uniform magnetic field of 3 T. If the magnetic field is oriented at
SOVA2 [1]

Answer:

the  magnitude of the force that the wire will experience = 1.8 N

Explanation:

The force on a current carrying wire placed in a magnetic field is :

F = Idl × B

where:

I = current flowing through the wire

dl = length of the wire

B = magnetic field

We can equally say that :

|F| = IdlBsin \theta

where : sin θ is the angle at which the orientation from the magnetic field  to the wire occurs = 30°

Then;

|F| = B\ I \ L \ sin \theta

Given that:

L = 20 cm = 0.2 m

I = 6 A

B = 3 T

θ = 30°

Then:

F = 3 × 6 × 0.2 sin 30°

F = 1.8 N

Therefore, the  magnitude of the force that the wire will experience = 1.8 N

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3 years ago
The table represent the thickness, top density, and bottom density of the different layers of the Earth. In most of the layers,
Nostrana [21]

Answer:inner core?

Explanation:

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3 years ago
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If you know what a JW is LET ME KNOW and it is a type of ppl
lisabon 2012 [21]

Answer:

only thing I think of when I see that is 'Just Wondering'

Explanation:

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2 years ago
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A permanent magnet has a magnetic dipole moment of 0.160 A · m^2. The magnet is in the presence of an external uniform magnetic
Elena L [17]

Answer:

the magnitude of the torque  on the permanent magnet = 7.34×10⁻³ Nm

the potential energy (in J) of the system consisting of the permanent magnet and the magnetic field provided by the coils =  -1.0485 ×10⁻² J

Explanation:

The torque is given by :

\bar {N} = \bar {m} * \bar {B}

where ;

m = 0.160 A.m²

B = 0.0800 T

θ = 35°

So the magnitude of the torque N = mBsinθ

N = (0.160)(0.0800)(sin 35°)

N = 0.007341

N = 7.34×10⁻³ Nm

Hence, the magnitude of the torque  on the permanent magnet = 7.34×10⁻³ Nm

b) The potential energy \bar{U} = \bar{-m} * \bar{B}

U = -mBcosθ

U = (- 0.160)(0.0800)(cos 45)

U = -0.010485

U = -1.0485 ×10⁻² J

Thus, the potential energy (in J) of the system consisting of the permanent magnet and the magnetic field provided by the coils =  -1.0485 ×10⁻² J

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