Two students are balancing on a 10m seesaw. The seesaw is designed so that each side of the seesaw is 5m long. The student on th
1 answer:
Answer:
Option A. 4 m
Explanation:
Please see attached photo for diagram.
In the attached photo, X is the distance from the centre to which the student on the right must sit in order to balance the seesaw.
Clockwise moment = X × 45
Anticlock wise moment = 3 × 60
Clockwise moment = Anticlock wise moment
X × 45 = 3 × 60
X × 45 = 180
Divide both side by 45
X = 180 / 45
X = 4 m
Thus, the student must sit at 4 m from the centre.
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Answer:
9 meters
Explanation:
Given:
Mass of Avi is,
Spring constant is,
Compression in the spring is,
Let the maximum height reached be 'h' m.
Now, as the spring is compressed, there is elastic potential energy stored in the spring. This elastic potential energy is transferred to Avi in the form of gravitational potential energy.
So, by law of conservation of energy, decrease in elastic potential energy is equal to increase in gravitational potential energy.
Decrease in elastic potential energy is given as:
Now, increase in gravitational potential energy is given as:
Now, increase in gravitational potential energy is equal decrease in elastic potential energy. Therefore,
Therefore, Avi will reach a maximum height of 9 meters.
Answer:
The induced emf in the short coil during this time is 1.728 x 10⁻⁴ V
Explanation:
The magnetic field at the center of the solenoid is given by;
B = μ(N/L)I
Where;
μ is permeability of free space
N is the number of turn
L is the length of the solenoid
I is the current in the solenoid
The rate of change of the field is given by;
The induced emf in the shorter coil is calculated as;
where;
N is the number of turns in the shorter coil
A is the area of the shorter coil
Area of the shorter coil = πr²
The radius of the coil = 2.5cm / 2 = 1.25 cm = 0.0125 m
Area of the shorter coil = πr² = π(0.0125)² = 0.000491 m²
E = 14 x 0.000491 x 0.02514
E = 1.728 x 10⁻⁴ V
Therefore, the induced emf in the short coil during this time is 1.728 x 10⁻⁴ V