Q1. Fill in the blanks .(a) Monsoon winds carry .................... (dust/rain)(b) Moving air can provide a ...................
1 answer:
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Refer to the diagram shown below.
Vector a is 2.8 cm long and 60° above the x-axis in the first quadrant. Therefore it s represented by
Vector b is 1.9 cm long and 60° below the x-axis in the fourth quadrant. Therefore it is represented by
The vectors may be written as
a = 2.8*(cos60°, sin60°) = (1.445, 2.5028)
b = 1.9*(cos60°, -sin60°) = (0.950, -1.6454)
Part (a)
a + b = (1.445+0.950, 2.5028-1.6454)
= (2.395, 0.8574)
The magnitude of this vector is
√[2.395² + 0.8574² ] = 2.5438 cm
Its direction is
tan⁻¹ (0.8574/2.395) = 19.7° above the x-axis, in the first quadrant.
The resultant vector is shown in the figure below.
Part (b)
Similarly obtain
a - b = (0.495, 4.1485)
The magnitude is 4.1777.
The direction is 83° above the x-axis, in the 1st quadrant.
The resultant vector is shown in the figure below.
Part (c)
Similarly, obtain
b - a = (-0.495, -4.1483)
The magnitude is 4.1777.
The direction is 83° below the negative x-axis, in the 3rd quadrant.
The resultant vector is shown in the figure below.
Vector a is 2.8 cm long and 60° above the x-axis in the first quadrant. Therefore it s represented by
Vector b is 1.9 cm long and 60° below the x-axis in the fourth quadrant. Therefore it is represented by
The vectors may be written as
a = 2.8*(cos60°, sin60°) = (1.445, 2.5028)
b = 1.9*(cos60°, -sin60°) = (0.950, -1.6454)
Part (a)
a + b = (1.445+0.950, 2.5028-1.6454)
= (2.395, 0.8574)
The magnitude of this vector is
√[2.395² + 0.8574² ] = 2.5438 cm
Its direction is
tan⁻¹ (0.8574/2.395) = 19.7° above the x-axis, in the first quadrant.
The resultant vector is shown in the figure below.
Part (b)
Similarly obtain
a - b = (0.495, 4.1485)
The magnitude is 4.1777.
The direction is 83° above the x-axis, in the 1st quadrant.
The resultant vector is shown in the figure below.
Part (c)
Similarly, obtain
b - a = (-0.495, -4.1483)
The magnitude is 4.1777.
The direction is 83° below the negative x-axis, in the 3rd quadrant.
The resultant vector is shown in the figure below.
The appropriate value in blank given is Δf = 5.5 x Hertz.
We have vibrational - rotational spectrum Hydrochloric Acid.
We have to investigate the estimated separation between absorption peaks and fill the blank.
<h3>What is vibrational - rotational spectrum ?</h3>Rotational–vibrational spectroscopy is a branch of molecular spectroscopy. It deals with the infrared and Raman spectra of molecules in the gaseous phase.
According to the question -
The estimated separation between absorption peaks in the vibrational-rotational spectrum of HCl is denoted by Δf and is equal to -
Δf = 5.5 x Hertz
Hence, the appropriate value in blank given is Δf = 5.5 x Hertz.
To learn more about vibrational-rotational spectrum, visit the link below-
#SPJ4
Answer:
a) 0.28 m or 28 cm is the minimum height above ground the fish reaches.
b) at the height of 0.484 m height , the pufferfish will eventually come to rest.
c) There exists two types of energy remain at the equilibrium point in the system. These are :
Gravitational potential energy = 23.72J
Spring potential energy = 0.384 J
Explanation:
Given that :
Mass of the pufferfish m =5kg
initial height of the fish h =5.5m
length of the spring l =0.5m
Spring constant K =3000N/m
a)
Assuming no energy loss to friction, what is the minimum height above the ground that the pufferfish reaches?
Lets assume that the minimum height the fish reaches is = x meters
Now by using the conservation of energy; we realize that :
Initial total energy = final total energy
Gravitational potential energy =
Gravitational potential energy' + Spring potential energy (kinetic energy is zero in both cases)
Replacing our given values into the above equation; we have :
269.5 = 47.5 x + 1500(0.5 -x )²
269.5 = 47.5 x + 1500(0.25 - x²)
269.5 = 47.5 x + 375 - 1500 x²
269.5 - 375 = 47.5 x - 1500 x²
-105.5 = 47.5 x - 1500 x²
-105.5 + 1500 x² - 47.5 x = 0
1500 x² - 47.5 x - 105.5 = 0
By using quadratic equation and taking the positive value;
x = 0.28 m or 28 cm is the minimum height above ground the fish reaches.
b)
At the equilibrium position the weight of fish will be equal to the force applied by the spring thus
mg = kx
substituting our given values ; we have:
(5)(9.8) = 3000x
x = 61.22
x = 0.016m : so this is the compression in the spring
Now; to determine the height the pufferfish gets to before it eventually come to rest; we have
(0.5-0.016) m = 0.484m
therefore, at the height of 0.484 m height , the pufferfish will eventually come to rest.
c)
There exists two types of energy remain at the equilibrium point in the system. These are :
Gravitational potential energy = mgh' = (5)(9.8)(0.484)
= 23.72J
and spring potential energy