I think the answer is the first condition sum of forces acting on a body is zero ( ∑ F =0 ) and the second condition sum of torque acting on a body is ( ∑ τ = 0 )
Could you put me as brainliest?
Answer:
λ = 396.7 nm
Explanation:
For this exercise we use the diffraction ratio of a grating
d sin θ = m λ
in general the networks works in the first order m = 1
we can use trigonometry, remembering that in diffraction experiments the angles are small
tan θ = y / L
tan θ =
= sin θ
sin θ = y / L
we substitute
= m λ
with the initial data we look for the distance between the lines
d =
d = 1 656 10⁻⁹ 1.00 / 0.600
d = 1.09 10⁻⁶ m
for the unknown lamp we look for the wavelength
λ = d y / L m
λ = 1.09 10⁻⁶ 0.364 / 1.00 1
λ = 3.9676 10⁻⁷ m
λ = 3.967 10⁻⁷ m
we reduce nm
λ = 396.7 nm
Answer:
0.25 m
Explanation:
We can solve the problem by using the lens equation:

where
f is the focal length
p is the distance of the object from the lens
q is the distance of the image from the lens
In this problem, we have
f = +20 cm=+0.20 m (the focal length is positive for a converging lens)
q = +1.0 m (the image distance is positive for a real image)
Solving the equation for p, we find

Answer:
First, let’s correct the question. Acceleration is the rate of change in velocity. Its unit therefore is ft/sec/sec. If S is the distance traveled for a given duration, S = Vot + (1/2)at^2 where Vo is the initial velocity, a is the acceleration and t is the time. For Vo = 0, a = 6m/sec/sec and t = 3 sec. The distance traveled is S = 0 + (1/2) x 6 x 3^2 = 27 meters
Answer: Angle 59 degree
Explanation: Given that the
n1 = 1.0
n2 = 1.5
Øi = 35 degree
From Snell law, which says that
n1/n2 = sinØ1/ sinØ2
Substitute all the parameters into the formula
1/1.5 = sin 35/sinØ2
Cross multiply
Sin Ø2 = 1.5 sin35
SinØ2 = 1.5 × 0.573 = 0.860
Ø2 = sin^-1(0.860)
Ø2 = 59.36 degree
Ø2 = 59 degree ( approximately)
It has angle 59 degree when passing from air to glass