Answer:
m = 0.59 kg.
Explanation:
First, we need to find the relation between the frequency and mass on a spring.
The Hooke's law states that
And Newton's Second Law also states that
Combining two equations yields
The term that determines the proportionality between acceleration and position is defined as angular frequency, ω.
And given that ω = 2πf
the relation between frequency and mass becomes
.
Let's apply this to the variables in the question.
Answer:
F - M a force exerted by scales on student
M a = M (9.8 + 4.9) m/s^2 upwards chosen as positive
a = 1.5 g net acceleration of student due to force of scales
W =M g weight of student (actual weight)
Wapp = M 1.5 * g apparent weight (on scales) of student
Answer:
5.59 m/s2
Explanation:
F = 1900 N
m = 340 kg
F = ma
Therefore, a = 1900/340 = 5.59
If the width of the slit is increased by a factor of two , then the central maxima will get decreased by half.
since , width of the central maxima is inversely proportional to the width of the slit . Which implies if width of the slit get increased then width of the central maxima will get decreased and vice versa.
central maxima = (2 * wavelength * D) / width of the slit
= (2 * lambda * D) / d
if initially width of the slit = d
then, after increasing it by a factor of two it become = 2d
central maxima = (2 * lambda * D) / 2d
= (lambda * D) / d
If the width of the slit is increased by a factor of two , then the central maxima will get decreased by half.
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