Easy !
Take any musical instrument with strings ... a violin, a guitar, etc.
The length of the vibrating part of the strings doesn't change ...
it's the distance from the 'bridge' to the 'nut'.
Pluck any string. Then, slightly twist the tuning peg for that string,
and pluck the string again.
Twisting the peg only changed the string's tension; the length
couldn't change.
-- If you twisted the peg in the direction that made the string slightly
tighter, then your second pluck had a higher pitch than your first one.
-- If you twisted the peg in the direction that made the string slightly
looser, then your second pluck had a lower pitch than the first one.
Recall that work is the amount of energy transferred to an object when it experiences a displacement and is acted upon by an external force. It is given a symbol of W and is measured in joules (J).
W=\vec{F}\cdot \Delta \vec{d}
We can use this formula to determine the work done by very specific forces, generating specific types of energy. We will examine three types of energy in this activity: gravitational potential, kinetic, and thermal. Before we start deriving equations for gravitational potential energy and kinetic energy, we should note that since work is the transfer and/or transformation of energy, we can also write its symbol as \Delta E.
The horizontal component of the magnetic field is 12.6 μT.
The magnetic influence on moving electric currents, electric charges, and magnetic materials is described by a magnetic field, which is a vector field. When a charge moves through a magnetic field, a force that is perpendicular to both its own velocity and the magnetic field operates on it.
The horizontal component of the Earth's magnetic field is perpendicular to the axis of a circular coil with five turns and a diameter of D = 30.0 cm that is vertically orientated.
A coil current of I = 0.600 A causes a horizontal compass to deflect 45.0° from magnetic north when it is positioned in the coil's center.
Let B be the magnetic field and R be the radius of the circular coil.
Then the horizontal component of the Earth's magnetic field is given as:
B(h) = B(coil) = μ₀ NI / 2R
B(h) = (4π × 10⁻⁷ ) (5)(0.6) / 0.3
B(h) = 12.6 μT
Learn more about magnetic field here:
brainly.com/question/14411049
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Answer:
109.385m
Explanation:
In 1 day, the hour hand travels 2 circles, or 4π rad in angular. The distance it travels is its angle times the radius
7.4 * 4π = 93 mm
In 1 day, the minute hand travels 24*60 = 1440 circles, or 1440 * 2π = 2880π rad in angular. The distance it travels is
12.1 * 2880π = 109478 mm
So the distance traveled by the tip of the minute hand that exceed the distance traveled by the tip of the hour hand is
109478 - 93 = 109385 mm or 109.385 m
