The compass doesn’t give you the value of the net magnetic field, just the direction. So, how do you get the magnitude of a particular field from this? The trick is to assume the value of the Earth’s magnetic field and the direction of the compass. Let’s assume that at this location on the Earth, the magnetic field is pointing directly North with a horizontal component of about 2 x 10-5 T.
Now suppose that I do something to create a magnetic field in a known direction and perpendicular to the horizontal component of the Earth’s magnetic field. Here is an example where I put a current carrying wire right over the compass needle. Since the compass is underneath the wire, the magnetic field due to the wire will be 90° to the Earth’s magnetic field.
Answer:
A. Geocentric: This model is Earth Centered
. Retrograde motion is explained by epicycles
.
B. Heliocentric: This model is Sun centered. Retrograde motion is explained by the orbital speeds of planets
C. Both geocentric and heliocentric: Epicycles and deferents help explain planetary motion
. Planets move in circular orbits and with uniform motion
. The brightness of a planet increases when the planet is closest to Earth.
Explanation:
The principle of the Ptolemy's geocentric model was developed on the assumption that the center of the universe is the Earth. On the other hand, the principle of the Copernicus' heliocentric model was based on the assumption that the center of the universe is the sun. However, both models have a common ideology on uniform circular motion and epicycles.
We have: a = v/t
Here, t = 2 s [ Given ]
a = 9.8 m/s² [constant value for earth system ]
Substitute their values into the expression:
9.8 = v/2
v = 9.8 × 2
v = 19.6 m/s
In short, Your Answer would be Option B
Hope this helps!
The reactants are on the left and the products are on the right of the equation
The work W done by the electric field in moving the proton is equal to the difference in electric potential energy of the proton between its initial location and its final location, therefore:
where q is the charge of the proton,
, with
being the elementary charge, and
and
are the initial and final voltage.
Substituting, we get (in electronvolts):
and in Joule:
