Name and describe the apparatus used by Cavendish to discover the universal gravitation constant
1 answer:
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Answer:
(a) 1500 m
(b) 2827.43m
Explanation:
Given that the time for one cycle of the swing is 1 s
The radius of the swing, R= 3.0m
The angle covered, \theta, by each swing is a quarter of the circle. i.e.
Speed of running of women =5 m/s.
Time of running = 5 minutes= 5 x 60 secondes= 300 s.
(a) As distance= (speed) x (time)
So, the required distance= 5 x 300 m= 1500 m.
(b) As there are two swings in one cycle, so, the distance covered in one swing is the length of the circular shown in the figure.
As arc length, , where \theta is the angle, in radian, subtended by the arc at the center, and R is the radius of curvature of the arc.
So, the distance covered by the child in 1 swing .
In 1 cycle, there are 2 swings, so distance covered in 1 cycle .
Now, in 1 second there is 1 cycle, so in 5 minutes there will be 300 cycles.
So, the total distance covered by the child
.
Answer:
(a) The volume of water is 100 cm³
(b) The volume of the rock is 20 cm³
(c) The density of the rock is 30 g/cm³
Explanation:
The given parameters of the perspex box are;
The area of the base of the box, A = 10 cm²
The initial level of water in the box, h₁ = 10 cm
The mass of the rock placed in the box, m = 600 g
The final level of water in the box, h₂ = 12 cm
(a) The volume of water in the box, 'V', is given as follows;
V = A × h₁
∴ The volume of water in the box, V = 10 cm² × 10 cm = 100 cm³
The volume of water in the box, V = 100 cm³
(b) When the rock is placed in the box the total volume, , is given by the sum of the rock,
, and the water, V, is given as follows;
=
+ V
= A × h₂
∴ = 10 cm² × 12 cm = 120 cm³
The total volume, = 120 cm³
The volume of the rock, =
- V
∴ = 120 cm³ - 100 cm³ = 20 cm³
The volume of the rock, = 20 cm³
(c) The density of the rock, ρ = (Mass of the rock, m)/(The volume of the rock)
∴ The density of the rock, ρ = 600 g/(20 cm³) = 30 g/cm³
Answer:
Explanation:
From the question we are told that:
Frequency of 3rd harmonics
Frequency of 5th harmonics
Generally the equation for Wavelength at 3rd Harmonics is mathematically given by
Therefore
Generally the equation for Wavelength at 1st Harmonics is mathematically given by
Therefore
Generally the equation for the frequency of the first harmonic is mathematically given by