In the morning, if you walk through a grassy lawn or field, quite often your feet will get very wet. What is the liquid generall
1 answer:
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Answer:
a) 1.89 m/s b) 172.1 N
Explanation:
a)
- Applying the work-energy theorem, if we can neglect the friction between truck and road, the total change in kinetic energy must be equal to the work done by the external forces.
- This work, is just 4,475 J.
- So we can write the following equation:
- where m= mass of the truck = 2.5*10³ kg.
- So, we can find the speed v, as follows:
b)
- The work done by the man, is just the horizontal force applied, times the displacement produced by the force horizontally:
- We can solve for F, as follows:
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
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
Answer:
Magnitude of force on wheel B is 4 N
Explanation:
Given that
For wheel A
m= 1 kg
d= 1 m,r= 0.5 m
F=1 N
We know that
T= F x r
T=1 x 0.5 N.m
T= 0.5 N.m
T= I α
Where I is the moment of inertia and α is the angular acceleration
T= I α
0.5= 0.25 α
For Wheel B
m= 1 kg
d= 2 m,r=1 m
Given that angular acceleration is same for both the wheel
T= I α
T= 1 x 2
T= 2 N.m
Lets force on wheel is F then
T = F x r
2 = F x 1
So F= 2 N
Magnitude of force on wheel B is 2 N