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Pepsi [2]
3 years ago
6

A ship leaves a harbor and sails at 23.5 ∘ to the north of due west. After traveling 575 km, how far west is the ship from the h

arbor?
Physics
1 answer:
solniwko [45]3 years ago
6 0

Answer:

D = 527.31 Km

Explanation:

given,

angle of ship, θ = 23.5° N of W

distance travel in the direction = 575 Km

Distance of ship in west from harbor = ?

now,

Distance of the ship in the west direction

D = d cos θ

d = 575 Km

θ = 23.5°

inserting all the values

D = 575 x cos 23.5°

D = 575 x 0.91706

D = 527.31 Km

Hence, the distance travel by the ship in west from harbor is equal to D = 527.31 Km  

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3 years ago
At 20 C, a steel rod of length 40.000 cm and a brass rod
balandron [24]

Answer:

a. stress in steel rod is -9.6 X 10⁷pa while stress in brass rod is -7.2 X 10⁷pa

b. new junction from steel rod is 40.0192cm while new junction from brass rod is 30.024cm

Explanation:

<u>Step 1:</u> <u>identify the given parameters and standard parameters</u>

⇒Steel rod: length of rod = 40.000 cm

                    Young modulus(Υ) = 20 X 10¹⁰ pa

                    coefficient of linear expansion(α) = 1.2 X 10⁻⁵ K⁻¹

                     stress in the rod =?

 ⇒Brass rod: length of rod = 30.000 cm

                    Young modulus(Υ) = 9 X 10¹⁰ pa

                    coefficient of linear expansion(α) = 2.0 X 10⁻⁵ K⁻¹

                     stress in the rod =?

--------------------------------------------------------------------------------------------

<u>Step 2:</u> <u>calculate the stress in each rod</u>

⇒Steel rod: stress in the rod = -Y*α*ΔT

                                                = (-20 X 10¹⁰ pa) (1.2 X 10⁻⁵ K⁻¹)(60-20)K

                                                = (-20 X 10¹⁰ pa) (1.2 X 10⁻⁵ K⁻¹)(40)K

                                                =  -9.6 X 10⁷ pa

--------------------------------------------------------------------------------------------

⇒Brass rod: stress in the rod = -Y*α*ΔT

                                                = (-9 X 10¹⁰ pa) (2.0 X 10⁻⁵ K⁻¹)(60-20)K

                                                = (-9 X 10¹⁰ pa) (2.0 X 10⁻⁵ K⁻¹)(40)K

                                                =  -7.2 X 10⁷ pa

--------------------------------------------------------------------------------------------

∴ stress in steel rod is -9.6 X 10⁷pa while stress in brass rod is -7.2 X 10⁷pa

--------------------------------------------------------------------------------------------

<u>Step 3:</u> <u>calculate the new length of each rod</u>

⇒Steel rod: New length = ΔL + L₀

                                   ΔL = α*L₀*ΔT

                                  ΔL = (1.2 X 10⁻⁵ K⁻¹)(40cm)(40)K

                                   ΔL = 1920 X 10⁻⁵ cm = 0.0192cm

                    New length = ΔL + L₀ = 0.0192cm + 40cm

                    New length = 40.0192cm

--------------------------------------------------------------------------------------------

⇒Brass rod: New length = ΔL + L₀

                                    ΔL = α*L₀*ΔT

                                   ΔL = (2.0 X 10⁻⁵ K⁻¹)(30cm)(40)K

                                   ΔL = 2400 X 10⁻⁵ cm = 0.024cm

                    New length = ΔL + L₀ = 0.024cm + 30cm

                    New length = 30.024cm

--------------------------------------------------------------------------------------------

Therefore, new junction from steel rod is 40.0192cm while new junction from brass rod is 30.024cm

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