The guitarist shortens the oscillating length of the properly tuned D-string by 0.15 m by pressing on the string with a finger.
1 answer:
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Answer:
all the details are in the attached picture, the answers are marked with colour.
The value of maximum velocity will be 0.3464 m/s². Energy or work is equal to the product of force and displacement.
<h3>What is velocity?</h3>The change of distance with respect to time is defined as speed. Speed is a scalar quantity. It is a time-based component. Its unit is m/sec.
The given data in the problem is
Mass of the body is,(m= 200g)
Amplitude is,(A =20 mm)
Maximum force is,F= 0.6 N
To find;
Maximum velocity
Energy or work is equal to the product of force and displacement.
Hence,the value of maximum velocity will be 0.3464 m/s².
To learn more about the velocity, refer to the link;
#SPJ2
Answer:
The shell hit at a distance of 1.9 x 10² km
The time of flight of the shell was 5.3 x 10² s
Explanation:
The position of the shell is given by the vector "r":
r = (x0 + v0 * t * cos α ; y0 + v0 * t * sin α + 1/2 g t²)
where:
x0 = initial horizontal position
v0 = magnitude of the initial velocity
t = time
α = launching angle
y0 = initial vertical position
g = acceleration of gravity
When the shell hit, the vertical component (ry) of the vector position r is 0. See figure.
Then:
ry = 0 = y0 + v0 * t * sin α + 1/2 g t²
Since the gun is at the center of our system of reference, y0 and x0 = 0
0 = t (v0 sin α + 1/2 g t)
t= 0 is discarded as solution
v0 sin α + 1/2 g t = 0
t = -2v0 sin α / g
t = (-2 * 2610 m/s * sin 81.9°)/ (-9.8 m/s²) = 5.3 x 10² s. This is the time of flight of the shell until it hit.
Then, the distance at which the shell hit is:
Distance = Module of r = ( x0 + v0 * t * cos α; 0) = x0 + v0 * t * cos α
Distance = 2.61 km/s * 5.3 x 10² s * cos 81.9 = 1.9 x 10² km
