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Julli [10]
3 years ago
7

On February 15, 2013, Asteroid 2012 DA14 passed within 17,200 miles [mi] of the surface of the Earth at a relative speed of 7.8

kilometers per second [km/s ]. This is considerably closer than the orbit of geosynchronous satellites (26,200 miles). This is the closest recorded approach of an object this large. The asteroid 2012 DA14 was estimated to have a diameter of 30 meters [m] and a specific gravity of 3. If 2012 DA14 had hit the Earth, what is the total amount of energy that would have been released (i.e., what was the kinetic energy of the asteroid)
Physics
1 answer:
xz_007 [3.2K]3 years ago
6 0

Answer:

The total amount of energy that would have been released if the asteroid hit earth = The kinetic energy of the asteroid = 1.29 × 10¹⁵ J = 1.29 PetaJoules = 1.29 PJ

1 PJ = 10¹⁵ J

Explanation:

Kinetic energy = mv²/2

velocity of the asteroid is given as 7.8 km/s = 7800 m/s

To obtain the mass, we get it from the specific gravity and diameter information given.

Density = specific gravity × 1000 = 3 × 1000 = 3000 kg/m³

But density = mass/volume

So, mass = density × volume.

Taking the informed assumption that the asteroid is a sphere,

Volume = 4πr³/3

Diameter = 30 m, r = D/2 = 15 m

Volume = 4π(15)³/3 = 14137.2 m³

Mass of the asteroid = density × volume = 3000 × 14137.2 = 42411501 kg = 4.24 × 10⁷ kg

Kinetic energy of the asteroid = mv²/2 = (4.24 × 10⁷)(7800²)/2 = 1.29 × 10¹⁵ J

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Steam flows steadily through an adiabatic turbine. The inlet conditions of the steam are 4 MPa, 500◦C, and 80 m/s, and the exit
Cerrena [4.2K]

Answer:

a) ΔEC=-23.4kW

b)W=12106.2kW

c)A=0.01297m^2

Explanation:

A)

The kinetic energy is defined as:

\frac{m*vel^2}{2} (vel is the velocity, to differentiate with v, specific volume).

The kinetic energy change will be: Δ (\frac{mvel^2}{2})=\frac{m*vel_2^2}{2}-\frac{m*vel_1^2}{2}

Δ (\frac{mvel^2}{2})=\frac{m}{2}*(vel_2^2-vel_1^2)

Where 1 and 2 subscripts mean initial and final state respectively.

Δ(\frac{mvel^2}{2})=\frac{12\frac{kg}{s}}{2}*(50^2-80^2)\frac{m^2}{s^2}=-23400W=-23.4kW

This amount is negative because the steam is losing that energy.

B)

Consider the energy balance, with a neglective height difference: The energy that enters to the turbine (which is in the steam) is the same that goes out (which is in the steam and in the work done).

H_1+\frac{m*vel_1^2}{2}=H_2+\frac{m*vel_2^2}{2}+W\\W=m*(h_1-h_2)+\frac{m}{2} *(vel_1^2-vel_2^2)

We already know the last quantity: \frac{m}{2} *(vel_1^2-vel_2^2)=-Δ (\frac{mvel^2}{2})=23400W

For the steam enthalpies, review the steam tables (I attach the ones that I used); according to that, h_1=h(T=500C,P=4MPa)=3445.3\frac{kJ}{kg}

The exit state is a liquid-vapor mixture, so its enthalpy is:

h_2=h_f+xh_{fg}=289.23+0.92*2366.1=2483.4\frac{kJ}{kg}

Finally, the work can be obtained:

W=12\frac{kg}{s}*(3445.3-2438.4)\frac{kJ}{kg} +23.400kW)=12106.2kW

C) For the area, consider the equation of mass flow:

m=p*vel*A where p is the density, and A the area. The density is the inverse of the specific volume, so m=\frac{vel*A}{v}

The specific volume of the inlet steam can be read also from the steam tables, and its value is: 0.08643\frac{m^3}{kg}, so:

A=\frac{m*v}{vel}=\frac{12\frac{kg}{s}*0.08643\frac{m^3}{kg}}{80\frac{m}{s}}=0.01297m^2

Download pdf
7 0
3 years ago
In some cases fixture wires may be used for
zalisa [80]

You can use fixture wires: For installation in luminaires where they are enclosed and protected and not subject to bending and twisting and also can be used to connect luminaires to their branch circuit conductors.

<h3>What are some uses of fixture wires?</h3>

Fixture wires are flexible conductors that are used for wiring fixtures and control circuits. There are some special uses and requirements for fixture wires and no fixture can be smaller than 18 AWG

In modern fixtures, neutral wire is white and the hot wire is red or black. In some types of fixtures, both wires will be of the same color.

To know more about fixture wires, refer

brainly.com/question/26098282

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3 0
1 year ago
What is know as law of inertia? ​
schepotkina [342]

Answer:

<em><u>mark</u></em><em><u> </u></em><em><u>me</u></em><em><u> </u></em><em><u>brianliest</u></em><em><u> </u></em><em><u>plz</u></em>

Explanation:

  • Law of inertia, also called Newton's first law, postulate in physics that, if a body is at rest or moving at a constant speed in a straight line, it will remain at rest or keep moving in a straight line at constant speed unless it is acted upon by a force.
  • Law of Inertia states that a body in a state of rest or uniform motion remains in the same state until and unless an external force acts on it.
  • A body continues to be in its state of rest or in uniform motion along a straight line unless an external force is applied on it. This law is also called law of inertia.
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A bar magnet is dropped toward a conducting ring lying on the floor. As the magnet falls toward the ring, does it move as a free
REY [17]

Explanation:

According to the Faraday-Lenz law, a conductive ring generates an induced current due to the change in the magnetic flux caused by the motion of the bar magnet. This induced current generates a magnetic field opposite to the magnetic field of the bar, generating an upward force that opposes the weight of the bar magnet, Therefore, it does not move as a freely  falling object.

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What produces a record of seismic waves?
Dmitrij [34]

Seismograph is the correct answer

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