The distance between slit and the screen is 1.214m.
To find the answer, we have to know about the width of the central maximum.
<h3>How to find the distance between slit and the screen?</h3>
- It is given that, wavelength 560 nm passes through a slit of width 0. 170 mm, and the width of the central maximum on a screen is 8. 00 mm.
- We have the expression for slit width w as,
where, d is the distance between slit and the screen, and a is the slit width.
- Thus, distance between slit and the screen is,
Thus, we can conclude that, the distance between slit and the screen is 1.214m.
Learn more about the width of the central maximum here:
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Answer:
Explanation:
Given that,
Mass of star M(star) = 1.99×10^30kg
Gravitational constant G
G = 6.67×10^−11 N⋅m²/kg²
Diameter d = 25km
d = 25,000m
R = d/2 = 25,000/2
R = 12,500m
Weight w = 690N
Then, the person mass which is constant can be determined using
W =mg
m = W/g
m = 690/9.81
m = 70.34kg
The acceleration due to gravity on the surface of the neutron star is can be determined using
g(star) = GM(star)/R²
g(star) = 6.67×10^-11 × 1.99×10^30 / 12500²
g (star) = 8.49 × 10¹¹ m/s²
Then, the person weight on neutron star is
W = mg
Mass is constant, m = 70.34kg
W = 70.34 × 8.49 × 10¹¹
W = 5.98 × 10¹³ N
The weight of the person on neutron star is 5.98 × 10¹³ N
Answer:
5.7 x 10^12 C
Explanation:
Let the charge on earth and moon is q.
mass of earth, Me = 5.972 x 10^24 kg
mass of moon, Mm = 7.35 x 10^22 kg
Let d be the distance between earth and moon.
the gravitational force between them is
The electrostatic force between them is
According to the question
1 % of Fg = Fe
q = 5.7 x 10^12 C
Thus, the charge on earth and the moon is 5.7 x 10^12 C.
I would have to say B failed because I think I read something about it being only 2law not 3
Are breathing hard. This is because cardiovascular exercise makes the heart beat faster which in turn creates a need for more air.
