Answer:
0.004 m
Explanation:
For light passing through a single slit, the position of the nth-minimum in the diffraction pattern is given by
where
is the wavelength
D is the distance of the screen from the slit
d is the width of the slit
Therefore, the width of the central maximum is equal to twice the value of y for n=1 (first minimum):
where we have
is the wavelength
D = 2.0 m is the distance of the screen
is the width of the slit
Substituting, we find
The planet closest to the sun that has a dense iron core and no moons is : B. mercury
From the options above, only mercury and venus are the one that doesn't have moons. Between the two, mercury is the one that has a dense iron core
hope this helps
Answer:
a) w = 9.599 10⁴ rad / s
, b) v = 3.35 10¹⁶ m / s
, c) a = 3.22 10²¹ m / s²
Explanation:
For this exercise we must use the relation of angular kinematics
a) angular velocity, the distance remembered in orbit between time (period)
w = 2π r / T
w = 2 π 3.59 10¹¹ / 2.35 10⁷
w = 9.599 10⁴ rad / s
b) linear and angular velocity are related by the equation
v = w r
v = 9,599 10⁴ 3.49 10¹¹
v = 3.35 10¹⁶ m / s
c) the centripetal acceleration is
a = v² / r = w² r
a = (9,599 10⁴)² 3.49 10¹¹
a = 3.22 10²¹ m / s²
Answer:
d = 2,042 10-3 m
Explanation:
The laser diffracts in the circular slit, so the process equation is
d sin θ= m λ
The first diffraction minimum occurs for m = 1
We can use trigonometry in the mirror
tan θ = Y / L
Where L is the distance from the Moon to Earth
Since the angle is extremely small
tan θ = sin θ / cos θ
Cos θ = 1
tant θ = sin θ = y / L
We replace
d y / L = λ
d = λ L / y
Let's calculate
d = 532 10⁻⁹ 3.84 10⁶/1 10³
d = 2,042 10-3 m
