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zepelin [54]
4 years ago
8

The luxury liner Queen Elizabeth 2 has a diesel-electric powerplant with a maximum power of 90 MW at a cruising speed of 31.5 kn

ots. What forward force is exerted on the ship at this highest attainable speed? (1 knot = 0.514 m/s. MW = Megawatts)
Physics
1 answer:
AlexFokin [52]4 years ago
5 0

Answer:

5558643.69 N

Explanation:

F = Force

v = Velocity = 31.5 knots

Converting to m/s

1\ knot=0.514\ m/s

31.5\ knot=31.5\times 0.514\ m/s=16.191\ m/s

Power is given by

P=Fv\\\Rightarrow F=\frac{P}{v}\\\Rightarrow F=\frac{90\times 10^6}{16.191}\\\Rightarrow F=5558643.69\ N

The forward force is exerted on the ship at this highest attainable speed is 5558643.69 N

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Seberkas cahaya monokromatik dijatuhkan pada kisi difraksi dengan 5000 goresan/cm menghasilkan spectrum garis terang orde kedua
Vlada [557]

Answer:

500 nm

Explanation:

In this problem, we have a diffraction pattern created by light passing through a diffraction grating.

The formula to find a maximum in the pattern produced by a diffraction grating is the following:

d sin \theta = m\lambda

where:

d is the distance between the lines in the grating

\theta is the angle at which the maximum is located

m is the order of the maximum

\lambda is the wavelength of the light used

In this problem we have:

\theta=30^{\circ} is the angle at which is located the 2nd-order bright line, which is the 2nd maximum

n = 5000 lines/cm is the number of lines per centimetre, so the distance between two lines is

d=\frac{1}{d}=\frac{1}{5000}=2\cdot 10^{-4} cm = 2\cdot 10^{-6} m

Re-arranging the equation for \lambda, we find the wavelength of the light used:

\lambda=\frac{d sin \theta}{m}=\frac{(2\cdot 10^{-6})(sin 30^{\circ})}{2}=5\cdot 10^{-7} m = 500 nm

4 0
3 years ago
А)
loris [4]

Image of the plate is missing, so i have attached it.

Answer:

A) t = 7.854 mm

B) σ = 26.67 MPa

Explanation:

A) From shearing of rivet, formula for pressure is;

P = τ•A_rivets

Where;

τ is allowable stress

A_rivets is Area of rivet

We are given:

τ = 60 MPa

Diameter; d = 20 mm

A_rivets = πd²/4 = π × 20²/4 = 100π

Thus;

P = 60 × 100π

P = 6000π N

From bearing of plate material, we can calculate pressure as;

P = σ_b•A_b

We are given;

σ_b = 120 MPa

A_b is area of plate = 20t

Where t is the thickness

Thus;

6000π = 120 × 20t

t = 6000π/(120 × 20)

t = 7.854 mm

B) Largest average tensile stress is given by the formula;

σ = P/A

Where A = 110t - 20t

A = 90t

A = 90 × 7.854

Thus;

σ = 6000π/(90 × 7.854)

σ = 26.67 MPa

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