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Viktor [21]
3 years ago
8

If it takes you 10 seconds to increase from 0 km/h to 50 km/h, what is your acceleration?

Physics
2 answers:
ZanzabumX [31]3 years ago
4 0
Acceleration is change in velocity over change in time. Your Δv is +13.9, since you increased speed by 50 km/h which is 13.9 m/s, and your Δt is 10s. 13.9/10 = 1.39 m/s^2, the standard units for acceleration. Make sense?
Licemer1 [7]3 years ago
3 0
In order to figure this out you use the equation acceleration = final velocity - initial velocity / time. or a = vf-vi/t. So a = 50 km/h - 0 km/h / 10 seconds. Since 50 - 0 is 50, 50 divided by 10 is 5. So you acceleration is 5 km/h plus the direction you are traveling. So your answer should look like a=5km/h + direction
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Due to the wave nature of light, light shined on a single slit will produce a diffraction pattern? Green light (520 nm) is shine
TiliK225 [7]

Answer:

Yes, it will produce a diffraction pattern.

a. 3.9 mm b. 1.95 mm

Explanation:

The light shined from a single slit will produce a diffraction pattern because,  the wavefront act as wavelets which generates its own wave according to Huygens principle. This therefore causes the diffraction pattern.

Given

wavelength of green light, λ = 520 nm = 520 × 10⁻⁹ m = 5.20 × 10⁻⁷ m

width of slit, d = 0.440 mm = 0.44 × 10⁻³ m = 4.4 × 10⁻⁴ m

Distance of slit from central maximum , D = 1.65 m

Distance of first minimum from central maximum, y = ?

a. The relationship between the slit width and wavelength is given by [tex} dsinθ = mλ [/tex]where d = slit width, θ = angular distance from central maximum, λ = wavelength of light and m = ±1, ±2, ±3...

The relationship between y and D is given by tanθ = y/D

Since θ is small, sinθ ≈ θ ≈ tanθ

so, dθ = mλ ⇒ θ = mλ/d = y/D

Therefore, y = mλD/d

Now, for the first minimum above the slit, m = +1 and for the first minimum below the slit, m = -1. So, y₁ =  λD/d and y₋₁ =  -λD/d. So, the width of the central maximum Δy is the difference between the first minima below and above the central maximum. So, Δy = y₁ - y₋₁ = λD/d -(-λD/d) = 2λD/d

Substituting the values from above, Δy= 2 × 5.20 × 10⁻⁷ × 1.65/4.4 × 10⁻⁴ =  3900 × 10⁻⁶ m = 3.9 × 10⁻³ m = 3.9 mm

b. The first order fringe is the fringe located between the first minimum and the second minimum. From dsinθ = mλ and tanθ = y/D when θ is small, sinθ ≈ θ ≈ tanθ. So, y = mλD/d. Let m= 1 and m=2 be the first and second minima respectively. So,y₁ =  λD/d and y₂ =  2λD/d. The difference Δy₁ = y₂ - y₁ is the width of the first order fringe. Therefore, Δy₁ = 2λD/d - λD/d= λD/d. Substituting the values from above, we have

λD/d= 5.20 × 10⁻⁷ × 1.65/4.4 × 10⁻⁴= 1.95 × 10⁻³ m = 1.95 mm

7 0
3 years ago
What is the speed of a bobsled whose distance-time graph indicates that it traveled 113m in 29s?
Elza [17]
In this item, we are asked to determine the speed of the bobsled given the distance traveled and the time it takes to cover the certain distance. This can mathematically be expressed as,
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<em>ANSWER: 3.90 m/s</em>
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