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2 years ago

1. What part of the Earth do “faults” appear?

Physics

1 answer:

alekssr [168]2 years ago

6 0

Answer:

Faults are found in collisions zones, and tectonic plates push up against mountain ranges for example the Himalayas or the Rocky Mountains.

Explanation:

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How do distinctive rock strata support the theory of continental drift?

Scorpion4ik [409]

“It states that parts of the Earth's crust slowly drift atop a liquid core. The fossil record supports and gives credence to the theories of continental drift and plate tectonics. ... Continental Coastlines appearing to fit together,Fossil Distrubution, Distinctive Rock Strata, and Coal Distrubution.”

Hope this helps!

~Mia

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3 years ago

What method is used to find the number of neutrons in an atom?

Ugo [173]

Answer:

Explanation:

5 0

2 years ago

Read 2 more answers

A closed, rigid tank fitted with a paddle wheel contains 2 kg of air, initially at 300 K. During an interval of 5 minutes, the p

anzhelika [568]

Answer:

The final temperature of the air is $T_2= 605 K$

Explanation:

We can start by doing an energy balance for the closed system

$\Delta KE+\Delta PE+ \Delta U = Q - W$

where

$\Delta KE$ = the change in kinetic energy.

$\Delta PE$ = the change in potential energy.

$\Delta U$ = the total internal energy change in a system.

Q = the heat transferred to the system.

W = the work done by the system.

We know that there are no changes in kinetic or potential energy, so $\Delta KE = 0$ and $\Delta PE=0$

and our energy balance equation is $\Delta U = Q - W$

We also know that the paddle-wheel transfers energy to the air at a rate of 1 kW and the system receives energy by heat transfer at a rate of 0.5 kW, for 5 minutes.

We use this information to calculate the total internal energy change $\Delta U=W+Q$ using the energy balance equation.

We convert the interval of time to seconds $t = 5 \:min = 300\:s$

$\Delta \dot{U}=\dot{W}+ \dot{Q}\\=\Delta U=(W+ Q)\cdot t$

$\Delta U=(1 \:kW+0.5\:kW)\cdot 300\:s\\\Delta U=450 \:kJ$

We can use the change in specific internal energy $\Delta U = m(u_2-u_1)$ to find the final temperature of the air.

We are given that $T_1=300 \:K$ and the air can be describe by ideal gas model, so we can use the ideal gas tables for air to determine the initial specific internal energy $u_1$

$u_1=214.07\:\frac{kJ}{kg}$

Next, we will calculate the final specific internal energy $u_2$

$\Delta U = m(u_2-u_1)\\\frac{\Delta U}{m} =u_2-u_1$

$\frac{\Delta U}{m} =u_2-u_1\\u_2=u_1+\frac{\Delta U}{m}$

$u_2=214.07 \:\frac{kJ}{kg} +\frac{450 \:kJ}{2 \:kg}\\u_2= 439.07 \:\frac{kJ}{kg}$

With the value $u_2=439.07 \:\frac{kJ}{kg}$ and the ideal gas tables for air we make a regression between the values $u = 434.78 \:\frac{kJ}{kg},T=600 \:K$ and $u = 442.42 \:\frac{kJ}{kg}, T=610 \:K$ and we find that the final temperature $T_2$ is 605 K.

3 0

2 years ago

Difference between Pascal’s law and law of flotation

irakobra [83]

Hi, I didn’t understand too well your question, but I hope this helps!

Archimedes principle is based on the weight of the object to push the object upward. Law of floation is the priciple which tells us about the density of the object with the liquid in which it is placed.

5 0

2 years ago

A current of 15.0 A flows through an electric heater operating on 120 V. Compute the power consumption of the electric heater.

Rzqust [24]

Answer:

There will be 1800 W power consumption in heater

Explanation:

We have given current flowing in the heater I = 15 A

Voltage on which heater is operating V = 120 volt

We have to find the power consumption in the heater

We know that power consumption is given by P = VI

So power consumption in heater = 120 × 15 = 1800 W

So there will be 1800 W power consumption in heater

5 0

3 years ago