________ reaction time involves selecting a specific and correct response from several choices when presented with several diffe
2 answers:
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Answer:
(a) vmax = 0.34m/s
(b) v = 0.13m/s
(c) v = 0.31m/s
(d) x = 0.039m
Explanation:
Given information about the spring-mass system:
m: mass of the object = 0.30kg
k: spring constant = 22.6 N/m
A: amplitude of the motion = 4.0cm = 0.04m
(a) The maximum speed of the object is given by the following formula:
(1)
w: angular frequency of the motion.
The angular frequency is calculated with the following relation:
(2)
You replace the expression (2) into the equation (1) and replace the values of the parameters:
The maximum speed of the object is 0.34 m/s
(b) If the object is compressed 1.5cm the amplitude of its motion is A = 0.015m, and the maximum speed is:
The speed is 0.13m/s
(c) To find the speed of the object when it passes the point x=1.5cm, you first take into account the equation of motion:
You solve the previous equation for t:
With this value of t, you can calculate the speed of the object with the following formula:
The speed of the object for x = 1.5cm is v = 0.31 m/s
(d) To calculate the values of x on which v is one-half the maximum speed, you first calculate the time t:
The position will be:
The position of the object on which its speed is one-half its maximum velocity is 0.039
Answer:
1) Time interval Blue Car Red Car
0 - 2 s Constant Velocity Increasing Velocity
2 - 3 s Constant Velocity Constant Velocity
3 - 5 s Constant Velocity Increasing Velocity
5 - 6 s Constant Velocity Decreasing Velocity
2) For Red and Blue car y₂ = 120 v = =
= 20 m/s
We get the same velocity for two cars because it is the average velocity of the car at the given interval of time. It is measured for initial and final position.
3) At t = 2s, the cars are the same position, and are moving at the same rate
Position - same
Velocity - same
The position-time graph shares the same spot for two cars.
Explanation:
It is given that,
Frequency of diagnostic ultrasound, f = 3.82 MHz = 3820 Hz
The speed of the sound in air, v = 343 m/s
(a) We need to find the wavelength in air of such a sound wave. Let it is given by λ₁
i.e.
(b) If the speed of sound in tissue is 1650 m/s .
Hence, this is the required solution.