A control device that uses a small input current to energize or de-energize the load connected to it.

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

Precipitation is greater over what and less over what? A. Land, sea B. Sea, land

Ksju [112]

I would say letter A to this because we need rain water for crops plants. Also we need it so that we can drink it. Its less great over sea because as it merge with the salt in the ocean it becomes less drinkable and less usable.

3 0

4 years ago

Do you add the initial position and final position to get distance?

mamaluj [8]

You can do d=st (s is speed and t is time) or you can use pythagorean theorem if the distance is or the two positions is a right angle.

4 0

3 years ago

put the following in order of a full lunar month: full moon, new moon, 1st quarter, waxing gibbous, waning gibbous, waxing cresc

ASHA 777 [7]

The same cycle of phases repeats over and over and over and over and over again, so I could start the list with whatever phase I want, and build the list until I get to the same phase I started with.

But the cultures that track their months by the phases of the Moon (traditional Muslim, traditional Jewish, traditional Chinese) all start the new month with the New Moon, so I guess I'll start my list there too.

-- New Moon

-- Waxing Crescent

-- First Quarter

-- Waxing Gibbous

-- Full Moon

-- Waning Gibbous

-- Third Quarter

-- Waning Crescent

-- next New Moon

4 0

3 years ago

in the diagram, q1,q2, and q3 are in a straight line. each of these particles has a charge of -2.35x10^-6 C. particles q1 and q2

julia-pushkina [17]

The magnitude of the net force that is acting on particle q₃ is equal to 6.2 Newton.

Given the following data:

Charge = $-2.35 \times 10^{-6}$ C.

Distance = 0.100 m.

Scientific data:

Coulomb's constant = $8.988\times 10^9 \;Nm^2/C^2$

<h3>How to calculate the net force.</h3>

In this scenario, the magnitude of the net force that is acting on particle q₃ is given by:

F₃ = F₁₃ + F₂₃

Mathematically, the electrostatic force between two (2) charges is given by this formula:

$F = k\frac{q_1q_2}{r^2}$

Where:

q represent the charge.

r is the distance between two charges.

k is Coulomb's constant.

Note: d₁₃ = 2d₂₃ = 2(0.100) = 0.200 meter.

For electrostatic force (F₁₃);

$F_{13} = 8.988\times 10^9 \times \frac{(-2.35 \times 10^{-6} \times [-2.35 \times 10^{-6}])}{0.200^2}\\\\F_{13} = \frac{0.0496}{0.04}$

F₁₃ = 1.24 Newton.

For electrostatic force (F₂₃);

$F_{13} = 8.988\times 10^9 \times \frac{(-2.35 \times 10^{-6} \times [-2.35 \times 10^{-6}])}{0.100^2}\\\\F_{13} = \frac{0.0496}{0.01}$

F₂₃ = 4.96 Newton.

Therefore, the magnitude of the net force that is acting on particle q₃ is given by:

F₃ = 1.24 + 4.96

F₃ = 6.2 Newton.

Read more on charges here: brainly.com/question/14372859

5 0

2 years ago

A control device that uses a small input current to energize or de-energize the load connected to it.​

A control device that uses a small input current to energize or de-energize the load connected to it.