A water wave has a speed of 22.0 meters/second if the wave frequency is 0.0680 hertz what is the wavelength

Describe the movement on the distance / time graph.

love history [14]

The distance increases per minute, levels out, then continued increasing.

8 0

3 years ago

2) A Ship has an area of cross - section at the water line of 2000m²

STatiana [176]

Answer:

a. 2 m

b. 0.15 m

Explanation:

(a) By what deck does the ship sink in fresh.

water, when it loads a cargo of 4000 tonnes

We know that the upward force , U on the ship equal the weight of fresh water displaced, W = ρVg where ρ = density of fresh water = 1000 kg/m³, V = volume of water displaced and g = acceleration due to gravity.

So, U = W = ρVg

This upward force must equal the weight of the ship, W' = mg where m = mass of ship = 4000 tonnes = 4000 × 1000 kg = 4 × 10⁶ kg

So. U = W'

ρVg = mg

V = m/ρ = 4 × 10⁶ kg/10³ kg/m³ = 4 × 10³ m³

Now the volume of water displaced equals the volume occupied by the ship floating ship of cross-sectional area, A = 2000 m³ and depth in water h.

So, V = Ah

Thus h = V/A = 4 × 10³ m³/2000 m³ = 2 m

So, the ship sinks to a depth of 2 m in fresh water.

(b) if the ship + Cargo has a displacement tonnage

of 12300 tonnes; by what amount will the ship

rise in the water when it sails from fresh water

into Seawater (density of Sea water - 1025kgm⁻³

We know that the upward force , U' on the ship in fresh water equals the weight of fresh water displaced, W" = ρV'g where ρ = density of fresh water = 1000 kg/m³, V' = volume of water displaced and g = acceleration due to gravity.

So, U = W" = ρV'g

This upward force must equal the weight of the ship + cargo, W₁ = m₁g where m₁ = mass of ship + cargo = 12300 tonnes = 12300 × 1000 kg = 1.23 × 10⁷ kg

So. U' = W₁

ρV'g = m₁g

V' = m₁/ρ = 1.23 × 10⁷ kg/10³ kg/m³ = 1.23 × 10⁴ m³

Now the volume of water displaced equals the volume occupied by the ship floating ship of cross-sectional area, A = 2000 m³ and depth in fresh water h'.

So, V' = Ah'

Thus h = V'/A = 1.23 × 10⁴ m³/2000 m³ = 6.15 m

So, the ship sinks to a depth of 0.6 m in fresh water.

Also,

We know that the upward force , U" on the ship in sea water equals the weight of sea water displaced, W₂ = ρ'V"g where ρ' = density of sea water = 1025 kg/m³, V"= volume of water displaced and g = acceleration due to gravity.

So, U" = W₂ = ρ'V"g

This upward force must equal the weight of the ship + cargo, W₁ = m₁g where m₁ = mass of ship + cargo = 12300 tonnes = 12300 × 1000 kg = 1.23 × 10⁷ kg

So. U" = W₁

ρ'V"g = m₁g

V" = m₁/ρ' = 1.23 × 10⁷ kg/1.025 × 10³ kg/m³ = 1.2 × 10⁴ m³

Now the volume of water displaced equals the volume occupied by the ship floating ship of cross-sectional area, A = 2000 m³ and depth in sea water h".

So, V" = Ah"

Thus h" = V'/A = 1.2 × 10⁴ m³/2000 m³ = 6 m

So, the ship sinks to a depth of 6 m in fresh water.

So, the rise in height from fresh water to sea water is Δh = h' - h" = 6.15 m - 6 m = 0.15 m

6 0

3 years ago

A 2.5kg object oscillates at the end of a vertically hanging light spring once every 0.65s .

sesenic [268]

Answer:

Explanation:

By the general expresion for this problem, we have:

y(t) = A*cos(w*t+∅)

since: w = 2π/T = 2π/0,65

For the initial conditions:

y(0) = (0.17cm)*cos(w*0+∅) = + 0.17 m ---> cos(∅) = 1 ---> ∅ = 0°

Then:

A) y(t) = (0.17 m)*cos((2π/0,65)*t)

B part

This means, find the first solution for:

y(t) = (0.17 m)*cos((2π/0,65)*t) = 0 > (equilibrium position)

then: cos((2π/0,65)*t) = 0 ---> (2π/0,65)*t = π/2 ---> t = 0.1625 sec

C part

By definition: (Velocity) v = dy/dt

Then, deriving: v = dy/dt = - (0.17 m)*(2π/0,65)*sin((2π/0,65)*t)

The maximum velocity ocurrs when sin((2π/0,65)*t) = ±1, then (in absolute value): Vmax = 1,64 m/s

D part

By definition: (Aceleration) A = dv/dt

Then, deriving: v = dv/dt = - (0.17 m)*(2π/0,65)²*cos((2π/0,65)*t)

The maximum aceleration ocurrs when cos((2π/0,65)*t) = ±1, then (in absolute value): Amax = 15,88 m/s²

E part

For the acceleration, ocurres by all the solution when:

cos((2π/0,65)*t) = cos(phase) = ±1, this means: phase = {π, 2π, 3π, ...}, all π multiples.

Then, for the position:

y(t) = (0.17 m)*±1= {+0.17 m ; -0.17 m}

For the velocity, ocurres by all the solution when:

sin((2π/0,65)*t) = sin(phase) = ±1, this means: phase = {π/2, π, 3π/2, ...}, all π/2 multiples.

Then, for the position:

y(t) = (0.17 m)*cos(phase)= 0

6 0

3 years ago

If the velocity of gas molecules is doubled the its kinetic energy will be

Stolb23 [73]

Answer: Quadrupled

Explanation:

Because when it’s doubled they are going to make another pair so 2+2=4

3 0

3 years ago

State Graham's law and state in which sector it falls

boyakko [2]

Answer:

states that the rate of diffusion of gases is inversely proportional to their square roots

5 0

2 years ago

Read 2 more answers