Protons have a __________ charge
1 answer:
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(a) The launching velocity of the beetle is 6.4 m/s
(b) The time taken to achieve the speed for launch is 1.63 ms
(c) The beetle reaches a height of 2.1 m.
(a) The beetle starts from rest and accelerates with an upward acceleration of 400 g and reaches its launching speed in a distance 0.53 cm. Here g is the acceleration due to gravity.
Use the equation of motion,
Here, the initial velocity of the beetle is u, its final velocity is v, the acceleration of the beetle is a, and the beetle accelerates over a distance s.
Substitute 0 m/s for u, 400 g for a, 9.8 m/s² for g and 0.52×10⁻²m for s.
The launching speed of the beetle is <u>6.4 m/s</u>.
(b) To determine the time t taken by the beetle for launching itself upwards is determined by using the equation of motion,
Substitute 0 m/s for u, 400 g for a, 9.8 m/s² for g and 6.385 m/s for v.
The time taken by the beetle to launch itself upwards is <u>1.62 ms</u>.
(c) After the beetle launches itself upwards, it is acted upon by the earth's gravitational force, which pulls it downwards towards the earth with an acceleration equal to the acceleration due to gravity g. Its velocity reduces and when it reaches the maximum height in its path upwards, its final velocity becomes equal to zero.
Use the equation of motion,
Substitute 6.385 m/s for u, -9.8 m/s² for g and 0 m/s for v.
The beetle can jump to a height of <u>2.1 m</u>
Answer:
<em>1.</em> <em>39068.07 N</em>
<em>2. 19534.036 N</em>
Explanation:
depth of water h = 10 m
atmospheric pressure Patm = 101325 Pa
density of water p = 1000 kg/m^3
acceleration due to gravity g = 9.81 m/s^2
pressure due to depth of water = pgh
P = 1000 x 9.81 x 10 = 98100 Pa
total pressure on the bottom of the tank is Patm + p = 101325 + 98100 = 199425 Pa
Left plug has diameter = 50 cm = 0.5 m
radius = 0.5/2 = 0.25 m
height = 1 cm = 0.01 m
<em>height below tank surface = 10 - 0.01 = 9.99</em>
pressure at this depth = 1000 x 9.81 x 9.99 = 98001.9 Pa
<em>total pressure = Patm + P = 101325 + 98001.9 = 199326.9 Pa</em>
surface area of plug = π = 3.142 x
= 0.196
<em>force required to lift left plug = pressure x area</em>
F = 199326.9 x 0.196 = <em>39068.07 N</em>
<em>The right side is a hemisphere with the same diameter, therefore surface area is half of the left plug</em>
A = 0.196/2 = 0.098
force F required to lift right plug = 199326.9 x 0.098 =<em> 19534.036 N</em>