Certain insects can achieve seemingly impossible accelerations while jumping. the click beetle accelerates at an astonishing 400
1 answer:
(a) The beetle's launch speed is 6.4 m/s.
(b) The duration to reach the launch speed is 1.63 ms.
(c) The maximum altitude attained by the beetle is 2.1 m.
(a) Starting from rest, the beetle accelerates upward at 400 g, reaching the launch speed over a distance of 0.53 cm, where g represents gravitational acceleration.
Using the motion equation,
Here, u indicates the initial velocity of the beetle, v the final velocity, a the beetle's acceleration, and s the distance covered.
The values used are 0 m/s for u, 400 g for a, 9.8 m/s² for g, and 0.52 × 10⁻² m for s.
The resulting launch speed of the beetle is 6.4 m/s.
(b) To calculate the time t needed for the beetle's upward launch, use the motion equation:
Substituting 0 m/s for u, 400 g for a, 9.8 m/s² for g, and 6.385 m/s for v provides:
The beetle takes 1.62 ms to launch itself.
(c) Upon launching, the beetle is subject to gravitational force, drawing it downward with an acceleration matching gravity g. Its ascent speed decreases until it reaches zero at the peak height.
Use the equation of motion:
Inputting 6.385 m/s for u, -9.8 m/s² for g, and 0 m/s for v yields:
The beetle reaches a height of 2.1 m.
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solution:
the spring force applied by a spring with spring constant k can be expressed as
where k acts as the spring constant
and x indicates the spring's deformation
to determine the work completed by the spring
the amount of work done by the spring when moving from x=0 to x=L
substituting the limits x=0 and x=L
we derive the work done in terms of k and L
Answer:
3.4 x 10⁴ m/s
Explanation:
Analyze the circular path of the electron
B = magnetic field = 80 x 10⁻⁶ T
m = mass of an electron = 9.1 x 10⁻³¹ kg
v = speed in the radial direction
r = radius of the circular trajectory = 2 mm = 0.002 m
q = charge of an electron = 1.6 x 10⁻¹⁹ C
For the electron’s circular movement
qBr = mv
(1.6 x 10⁻¹⁹) (80 x 10⁻⁶) (0.002) = (9.1 x 10⁻³¹) v
v = 2.8 x 10⁴ m/s
Now, consider the electron's movement in a straight line:
v' = speed in linear motion
x = distance traveled horizontally = 9 mm = 0.009 m
t = duration = =
= 4.5 x 10⁻⁷ sec
Using the formula
x = v' t
0.009 = v' (4.5 x 10⁻⁷)
v' = 20000 m/s
v' = 2 x 10⁴ m/s
The resultant speed is given by
V = sqrt(v² + v'²)
V = sqrt((2.8 x 10⁴)² + (2 x 10⁴)²)
v = 3.4 x 10⁴ m/s