d = 0.221 mm is the spacing (in mm) between the two slits..
here,
wavelength = 589 nm
D = 150 cm = 1.5 m
y = 4 mm
then ,
y = mλD/d
for m = `1
0.004 = 589*10^-9 * 1.5/d
d = 2.21*10^-4 m
d = 0.221 mm
slit spacing is 0.221 mm
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In a typical hydrogen bond, as atoms move further up the energy ladder, The distance between the energy levels decreases. For Scientists Working at Such Small Scales, these “charging times” are likely candidates for the cause of the distance effect.
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Answer:
Parts of a transverse wave include:
Crest - the highest point on the wave
Trough - the lowest point on the wave
Wavelength - the distance from one crest to the next crest, or one trough to the next trough
Amplitude - the displacement of the wave from the midpoint to the highest point (crest) or the lowest point (trough)
Explanation:
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Complete question:
The recoil of a shotgun can be significant. Suppose a 3.6-kg shotgun is held tightly by an arm and shoulder with a combined mass of 15.0 kg. When the gun fires a projectile with a mass of 0.040 kg and a speed of 380 m/s, what is the recoil velocity of the shotgun and arm–shoulder combination?
Answer:
The recoil velocity of the shotgun and arm–shoulder combination is 1.013 m/s
Explanation:
Given;
combined mass of the shotgun and arm–shoulder, m₁ = 15 kg
mass of the projectile, m₂ = 0.04 kg
speed of the projectile, u₂ = 380 m/s
let the recoil velocity of the shotgun and arm–shoulder combination = u₁
Apply the principle of conservation of linear momentum;
m₁u₁ + m₂u₂ = 0
m₁u₁ = - m₂u₂

Therefore, the recoil velocity of the shotgun and arm–shoulder combination is 1.013 m/s
Answer: I think the answer <em><u>MIGHT</u></em> be: Boron, germanium, and tellurium....
Explanation:I know this cuz I had this question on my unit test and I had to look it up online....Hope this helps
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Answer:
y = 0.99 m
Explanation:
This is a projectile launching exercise, let's start by finding the components of the initial velocity, using trigonometry
cos θ = v₀ₓ / v₀
sin θ = v_{oy} / v₀
v₀ₓ = vo cos θ
v_{oy} = I go sin θ
v₀ₓ = 15 cos 30 = 12.99 m / s
v_{oy} = 15 sin 30 = 7.5 m / s
Let's find the time it takes to travel x = 18 m
x = v₀ₓ t
t = x / v₀ₓ
t = 18 / 12.99
t = 1,385 s
at this point it is at a height of
y = v_{oy} - ½ g t²
y = 7.5 1.385 - ½ 9.8 1.385²
y = 0.99 m
therefore the camera must place the foot 99 cm from the ground