(a) Differentiate the position vector to get the velocity vector:
<em>r</em><em>(t)</em> = (3.00 m/s) <em>t</em> <em>i</em> - (4.00 m/s²) <em>t</em>² <em>j</em> + (2.00 m) <em>k</em>
<em>v</em><em>(t)</em> = d<em>r</em>/d<em>t</em> = (3.00 m/s) <em>i</em> - (8.00 m/s²) <em>t</em> <em>j</em>
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(b) The velocity at <em>t</em> = 2.00 s is
<em>v</em> (2.00 s) = (3.00 m/s) <em>i</em> - (16.0 m/s) <em>j</em>
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(c) Compute the electron's position at <em>t</em> = 2.00 s:
<em>r</em> (2.00 s) = (6.00 m) <em>i</em> - (16.0 m) <em>j</em> + (2.00 m) <em>k</em>
The electron's distance from the origin at <em>t</em> = 2.00 is the magnitude of this vector:
||<em>r</em> (2.00 s)|| = √((6.00 m)² + (-16.0 m)² + (2.00 m)²) = 2 √74 m ≈ 17.2 m
(d) In the <em>x</em>-<em>y</em> plane, the velocity vector at <em>t</em> = 2.00 s makes an angle <em>θ</em> with the positive <em>x</em>-axis such that
tan(<em>θ</em>) = (-16.0 m/s) / (3.00 m/s) ==> <em>θ</em> ≈ -79.4º
or an angle of about 360º + <em>θ</em> ≈ 281º in the counter-clockwise direction.
The correct answer to the question above is The third Option: C; ultrasound imaging of the liver. The ultrasound imaging of the liver is definitely not an application of Doppler technology.
Hope this helps! :)
Answer: option B
Explanation: when a neutral atom loses an electron or gains a positive charge electron, it becomes a positive ion (positively charged) and when an neutral atom gains an electronic charge or losses a positive charge electron, it becomes a negative ion (negatively charged).
Answer:
354 m/s
Explanation:
For the second overtune (Third harmonic) of an open pipe,
λ = 2L/3................................ Equation 1
Where L = Length of the open pipe, λ = Wave length.
Given: L = 1.75 m.
Substitute into equation 1
λ = 2(1.75)/3
λ = 1.17 m.
From the question,
V = λf.......................... Equation 2
V = speed of sound in the room, f = frequency
Given: f = 303 Hz.
Substitute into equation 2
V = 1.17(303)
V = 353.5
V ≈ 354 m/s
Hence the right answer is 354 m/s
Explanation:
Given that,
Length of gold wire, l = 4 m
Voltage of battery, V = 1.5 V
Current, I = 4 mA
The resistivity of gold, 
Resistance in terms of resistivity is given by :
Also, V = IR
So,
A is area of wire,
, r is radius, r = d/2 (diameter=d)
Out of four option, near option is (C) 17 μm.
