Arrange the phases of the Moon in order of increasing rising time, from the phase with the earliest rising time at 12:00 a.m. to
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In a double-slit interference experiment, the distance y of the maximum of order m from the center of the observed interference pattern on the screen is
where D=5.00 m is the distance of the screen from the slits, and
is the distance between the two slits.
The fringes on the screen are 6.5 cm=0.065 m apart from each other, this means that the first maximum (m=1) is located at y=0.065 m from the center of the pattern.
Therefore, from the previous formula we can find the wavelength of the light:
And from the relationship between frequency and wavelength,
, we can find the frequency of the light:
where D=5.00 m is the distance of the screen from the slits, and
The fringes on the screen are 6.5 cm=0.065 m apart from each other, this means that the first maximum (m=1) is located at y=0.065 m from the center of the pattern.
Therefore, from the previous formula we can find the wavelength of the light:
And from the relationship between frequency and wavelength,
Answer:
a)
b)S= 46.4 cm
Explanation:
Given that
Velocity = 16 Km/s
V= 16,000 m/s
E= 27 mV/m
E=0.027 V/m
d= 22.5 cm
d= 0.225 m
a)
lets time taken by electron is t
d = V x t
0.225 = 16,000 t
b)
We know that
F = m a = E q ------------1
Mass of electron ,m
Charge on electron
So now by putting the values in equation 1
Here initial velocity u= 0 m/s
S=0.464 m
S= 46.4 cm
S is the deflection of electron.
To solve this problem we will apply the concepts related to the calculation of the surface, volume and error through the differentiation of the formulas given for the calculation of these values in a circle. Our values given at the beginning are
The radius then would be
And
PART A ) For the Surface Area we have that,
Deriving we have that the change in the Area is equivalent to the maximum error, therefore
Maximum error:
The relative error is that between the value of the Area and the maximum error, therefore:
PART B) For the volume we repeat the same process but now with the formula for the calculation of the volume in a sphere, so
Therefore the Maximum Error would be,
Replacing the value for the radius
And the relative Error
Answer:
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