In the diagram, what is happening to the temperature at Point B? Question 6 options: A. The temperature is rising as the molecul
es break apart from each other B. The temperature is not rising because the heat is being used to break the connections between the molecules C. The temperature is dropping as the molecules break apart from each other D. The temperature is rising as the substance melts E. The temperature is not rising because the molecules are slowing down
2 answers:
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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:
13.309 m/s²
Explanation:
Length from shoulder to hand, l = 30 cm = 0.3 m
initial velocity, u = 1 m/s
final velocity, v = 2.5 m/s
time, t = 3 s
Let the tangential acceleration is a.
by using first equation of motion
v = u + at
2.5 = 1 + 3 a
a = 0.5 m/s²
Let the centripetal acceleration is a'.
a' = v'²/l
a' = 2 x 2 / 0.3
a' = 13.3 m/s²
The tangential acceleration and the centripetal acceleration are both perpendicular to each other. So, the net acceleration is given by
A = 13.309 m/s²