Power in a wire where current is flowing can be calculated from the product of the square of the current and the resistance. Resistance is equal to the product of resistivity and length divided by the area of the wire. We do as follows:
Resistance = 2.44 × 10-8 ( 0.11) / (π)(0.0009)^2 = 1.055x10^-3 <span>Ω
P = I^2R = .170^2 (</span>1.055x10^-3 ) = 3.048x10^-5 W
Answer:112.82 m/s
Explanation:
Given
range of arrow=68 m
as the arrow travels it acquire a vertical velocity
-------1
Range is given by
R=ut
where u=initial velocity
substitute the value of t in eqn 1
--------2
and
substitute it in 2
u=112.82 m/s
Choices 'B'; and 'D' both begin with the correct words.
But they should end with the equation
R = V / I
Answer:
4.6×10^-7 m or 0.46nm
Explanation:
From
Wo= hc/λ
Where:
Wo= work function of the metal
h= planks constant
c= speed of light
λ= wavelength
λ= hc/Wo
λ= 6.6×10^-34 × 3×10^8/4.30×10^-19
λ= 4.6×10^-7 m
Answer:
First Quarter and Third Quarter.
Explanation:
Tides are formed as a consequence of the differentiation of gravity due to the Moon across to the Earth sphere.
Since gravity variates with the distance:
(1)
Where m1 and m2 are the masses of the two objects that are interacting and r is the distance between them.
For example, seeing the image below, point A is closer to the Moon than point b, and at the same time the center of mass of the Earth will feel more attracted to the Moon than point B. Therefore, that creates a tidal bulge in point A and point B.
When the Sun and the Moon are alight with respect to the Earth, then the Sun tidal force contributes to the tidal force of the Moon over the Earth. That makes the high tides even higher (spring tides).
However, when the Sun is not in the same line than the Moon (the Moon is at 90° with respect to the Sun), then the low tides are higher and the high tides are lower. That scenario is known as neap tides.
Therefore, that happens when the Moon is at First Quarter and Third Quarter.