(a) Find the size of the smallest detail observable in human tissue with 20.0-MHz ultrasound. (b) Is its effective penetration d
1 answer:
Answer:
(a) λ=77×10⁻⁶m
(b)Yes its enough to examine entire eye,
(c)λ=16.65μm.
Explanation:
The ultrasonic sound are used to detect problems in human body by using device consisting of transmitter and detector .
The speed of sound in human tissue is 1540m/s.
The wavelength of sound is 20MHz.
(a) let the velocity of sound in human tissue is v = 1540m/s.
v=fλ
λ= =
= 77×10⁻⁶m
λ=77×10⁻⁶m
(b) we can scan effectively upto the depth of about 500λ
500λ=500× (77×10⁻⁶)= 3.85cm
Yes its enough to examine entire eye,
(c) In air the velocity of sound is 331m/s.
λ= =
= 16.65μm
λ=16.65μm.
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The box has 3 forces acting on it:
• its own weight (magnitude <em>w</em>, pointing downward)
• the normal force of the incline on the box (mag. <em>n</em>, pointing upward perpendicular to the incline)
• friction (mag. <em>f</em>, opposing the box's slide down the incline and parallel to the incline)
Decompose each force into components acting parallel or perpendicular to the incline. (Consult the attached free body diagram.) The normal and friction forces are ready to be used, so that just leaves the weight. If we take the direction in which the box is sliding to be the positive parallel direction, then by Newton's second law, we have
• net parallel force:
∑ <em>F</em> = -<em>f</em> + <em>w</em> sin(35°) = <em>m a</em>
• net perpendicular force:
∑<em> F</em> = <em>n</em> - <em>w</em> cos(35°) = 0
Solve the net perpendicular force equation for the normal force:
<em>n</em> = <em>w</em> cos(35°)
<em>n</em> = (15 kg) (9.8 m/s²) cos(35°)
<em>n</em> ≈ 120 N
Solve for the mag. of friction:
<em>f</em> = <em>µ</em> <em>n</em>
<em>f</em> = 0.25 (120 N)
<em>f</em> ≈ 30 N
Solve the net parallel force equation for the acceleration:
-30 N + (15 kg) (9.8 m/s²) sin(35°) = (15 kg) <em>a</em>
<em>a</em> ≈ (54.3157 N) / (15 kg)
<em>a</em> ≈ 3.6 m/s²
Now solve for the block's speed <em>v</em> given that it starts at rest, with <em>v</em>₀ = 0, and slides down the incline a distance of ∆<em>x</em> = 3 m:
<em>v</em>² - <em>v</em>₀² = 2 <em>a</em> ∆<em>x</em>
<em>v</em>² = 2 (3.6 m/s²) (3 m)
<em>v</em> = √(21.7263 m²/s²)
<em>v</em> ≈ 4.7 m/s
Answer:
the force will increase by a factor 2.25
Explanation:
The gravitational force between the two stars is given by:
where
G is the gravitational constant
m1, m2 are the masses of the two stars
r is the distance between the stars
If the distance is decreased by one-third, it means that the new distance is 2/3 of the previous distance
So the new force will be
So, the force will be 2.25 times the previous value.