3. A model rocket is launched straight upward at 58.8 m/s.
1 answer:
Let us assume that rocket only runs in initial energy and not using its own to flying.
Also , let upward direction is +ve and downward direction is -ve .
Initial velocity , u = 58.8 m/s .
Acceleration due to gravity , .
Final velocity , v - = 0 m/s .
We know , by equation of motion .
Hence, this is the required solution .
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Answer:
Velocity of airplane is 500 km/h
Velocity of wind is 40 km/h
Explanation:
= Velocity of airplane in still air
= Velocity of wind
Time taken by plane to travel 1150 km against the wind is 2.5 hours
Time taken by plane to travel 450 km against the wind is 50 minutes = 50/60 hours
Subtracting the two equations we get
Applying the value of velocity of wind to the first equation
∴ Velocity of airplane in still air is 500 km/h and Velocity of wind is 40 km/h
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:
Energy required = 3169.34 Joules.
Explanation:
The quantity of energy (Q) required can be determined by;
Q = mcΔθ
Where: m is the mass, c is the specific heat and Δθ is the change in temperature.
But, m = 96.7 kg, c = 0.874 J/(kg),
=
and
=
.
So that,
Q = mc( -
)
= 96.7 x 0.874 x ( -
)
= 96.7 x 0.874 x 37.5
= 3169.3425
Q = 3169.34
= 3.2 KJ
The amount of energy required is 3169.34 Joules.