An experiment is conducted such that an applied force is exerted on a 5kg object as it travels across a horizontal surface in wh
2 answers:
If an experiment is conducted such that an applied force is exerted on an object, a student could use the graph to determine the net work done on the object.
The graph of the net force exerted on the object as a function of the object’s distance traveled is attached below.
- A student could use the graph to determine the net work done on the object by Calculating the area bound by the line of best fit and the horizontal axis from 0m to 5m
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Answer:
The average emf that will be induced in the wire loop during the extraction process is 37.9 V
Explanation:
The average emf induced can be calculated from the formula
Where is the number of turns
is the change in magnetic flux
is the time interval
The change in magnetic flux is given by
Where is the final magnetic flux
and is the initial magnetic flux
Magnetic flux is given by the formula
Where is the magnetic field
is the area
and is the angle between the magnetic field and the area.
Initially, the magnetic field and the area are pointed in the same direction, that is,
From the question,
B = 1.5 T
and radius = 15.0 cm = 0.15 m
Since it is a circular loop of wire, the area is given by
∴
∴
( NOTE: )
Wb
For
The field pointed upwards, that is . Since
Then
Hence,
From the question
Here, = 1
Hence,
becomes
Hence, the average emf that will be induced in the wire loop during the extraction process is 37.9 V.
(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>