All categories

2 years ago

A subducting oceanic plate

Physics

1 answer:

beks73 [17]2 years ago

4 0

Maybe this would help understand it better.

<span>Tectonic plates can transport both continental crust and oceanic crust, or they may be made of only one kind of crust. Oceanic crust is denser than continental crust. At a subduction zone, the oceanic crust usually sinks into the mantle beneath lighter continental crust</span>

You might be interested in

how much water is needed to produce 1kwh of electricity at a power plant that is 30% efficient if the temperature increase 10 C

Dimas [21]

The amount of water needed is 287 kg

Explanation:

The amount of energy that we need to produce with the power plant is

$E=1 kWh = (1000W)(1h)=(1000W)(3600s)=3.6\cdot 10^6 J$

We also know that the power plant is only 30% efficient, so the energy produced in input must be:

$E_{in}=\frac{E}{0.30}=\frac{3.6\cdot 10^6}{0.3}=1.2\cdot 10^7 J$

The amount of water that is needed to produce this energy can be found using the equation

$E_{in}=mC\Delta T$

where:

m is the amount of water

$C=4186 J/kg^{\circ}C$ is the specific heat capacity of water

$\Delta T=10^{\circ}C$ is the increase in temperature

And solving for m, we find:

$m=\frac{E_{in}}{C\Delta T}=\frac{1.2\cdot 10^7}{(4186)(10)}=287 kg$

Learn more about specific heat capacity:

brainly.com/question/3032746

brainly.com/question/4759369

#LearnwithBrainly

3 0

2 years ago

A man of 60kg moves in a lift of constant velocity 5m/s .What is the reactive force acting on the man by the elevator?

Flauer [41]

Answer:

588 N

Explanation:

Since the 60 kg is moving at a constant velocity there is no acceleration. In order for the system to be balanced, both the normal force and the force of gravity must be equal. In this case the man has a mass of 60 kg. So to find the force you multiply mass by gravitys constant (9.81). And you end up with an answer of 588.6 but I rounded to 588.

4 0

2 years ago

Read 2 more answers

A harmonic force of maximum value 25 N and frequency of 180 cycles>min acts on a machine of 25 kg mass. Design a support syst

kramer

Answer:

registrar y de h2rhrjlekeek tert2h2jwkwlwlwlwlwlwñlwk

3 0

3 years ago

QUESTION 1 If you do 1,000 J of work on a system and remove 500 J of heat from it, what is the change in its internal energy?

Mazyrski [523]

1. U = Q + W
U = -500 + 1000
U = 500 J

2. The first law of thermodynamic is about the law of conservation of energy where energy in should be equal to energy out.

3. It is the windmill that does not transform energy from heat to mechanical instead it is the transforms the opposite.

4. In a heat engine, work is used to transfer thermal energy from a hot reservoir to a cold one.

5. 5.00 × 10^4 J - 2.00 × 10^4 J = 3.00 × 10^4 J

6. To increase the work done, we raise the temperature of the cold reservoir.

7 0

3 years ago

Read 2 more answers

What fraction of an iceberg is submerged? (ρice = 917 kg/m3, ,ρsea = 1030 kg/m3.)

aksik [14]

Answer:

Choice d. Approximately $89\%$ of the volume of this iceberg would be submerged.

Explanation:

Let $V_\text{ice}$ denote the total volume of this iceberg. Let $V_\text{submerged}$ denote the volume of the portion that is under the liquid.

The mass of that iceberg would be $\rho_\text{ice} \cdot V_\text{ice}$ . Let $g$ denote the gravitational field strength ( $g \approx 9.81\; \rm N \cdot kg^{-1}$ near the surface of the earth.) The weight of that iceberg would be: $\rho_\text{ice} \cdot V_\text{ice} \cdot g$ .

If the iceberg is going to be lifted out of the sea, it would take water with volume $V_\text{submerged}$ to fill the space that the iceberg has previously taken. The mass of that much sea water would be $\rho_\text{sea} \cdot V_\text{submerged}$ .

Archimedes' Principle suggests that the weight of that much water will be exactly equal to the buoyancy on the iceberg. By Archimedes' Principle:

$\text{buoyancy} = \rho_\text{sea} \cdot V_\text{submerged} \cdot g$ .

The buoyancy on the iceberg should balance the weight of this iceberg. In other words:

$\underbrace{\rho_\text{ice} \cdot V_\text{ice} \cdot g}_\text{weight of iceberg} = \underbrace{\rho_\text{sea} \cdot V_\text{submerged} \cdot g}_\text{buoyancy on iceberg}$ .

Rearrange this equation to find the ratio between $V_\text{submerged}$ and $V_\text{ice}$ :

$\begin{aligned} &\frac{V_\text{submerged}}{V_\text{ice}} \\&= \frac{\rho_\text{ice} \cdot g}{\rho_\text{sea} \cdot g}\\ &= \frac{\rho_\text{ice}}{\rho_\text{sea}}\ = \frac{917\; \rm kg \cdot m^{-3}}{1030\; \rm kg \cdot m^{-3}} \approx 0.89 \end{aligned}$ .

In other words, $89\%$ of the volume of this iceberg would have been submerged for buoyancy to balance the weight of this iceberg.

8 0

2 years ago