Two 1.9 kg masses are 1.1 m apart (center to center) on a frictionless table. Each has + 9.6 μC of charge.
1 answer:
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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
Answer:
a) the maximum transverse speed of a point on the string at an antinode is 5.9899 m/s
b) the maximum transverse speed of a point on the string at x = 0.075 m is 4.2338 m/s
Explanation:
Given the data in the question;
as the equation of standing wave on a string is fixed at both ends
y = 2AsinKx cosωt
but k = 2π/λ and ω = 2πf
λ = 4 × 0.150 = 0.6 m
and f = v/λ = 260 / 0.6 = 433.33 Hz
ω = 2πf = 2π × 433.33 = 2722.69
given that A = 2.20 mm = 2.2×10⁻³
so = A × ω
= 2.2×10⁻³ × 2722.69 m/s
= 5.9899 m/s
therefore, the maximum transverse speed of a point on the string at an antinode is 5.9899 m/s
b)
A' = 2AsinKx
= 2.20sin( 2π/0.6 ( 0.075) rad )
= 2.20 sin( 0.7853 rad ) mm
= 2.20 × 0.706825 mm
A' = 1.555 mm = 1.555×10⁻³
so
= A' × ω
= 1.555×10⁻³ × 2722.69
= 4.2338 m/s
Therefore, the maximum transverse speed of a point on the string at x = 0.075 m is 4.2338 m/s