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OverLord2011 [107]
4 years ago
7

Choose the correct statement regarding energy production in the interior of a star and energy production in a power plant. A sta

r uses fusion as an energy source by building larger atoms from smaller atoms. A nuclear power plant uses fusion as an energy source by building larger atoms from smaller atoms. A nuclear power plant uses fusion as an energy source by breaking apart atoms. A star uses fusion as an energy source by breaking apart atoms.
Physics
2 answers:
sergij07 [2.7K]4 years ago
8 0

Answer:

A star uses fusion as an energy source by building larger atoms from smaller atoms.

Explanation:

Nuclear fission and fusion are two processes at which an atomic nucleus is changed to produce energy. Fission is the process splitting heavy atomic into lighter atomic nuclei.

So, fusion is the combination of smaller atoms to form larger atoms and star uses this as source of energy.

Fusion is the process at which light atomic nuclei are merged or fused together to form heavier nuclei.

The energy source for all stars is nuclear fusion. In a nuclear fusion reaction, the nuclei of two atoms combine to create a new atom. Most commonly, in the core of a star, two hydrogen atoms fuse to become a helium atom.

katrin2010 [14]4 years ago
7 0

Answer:

Option A) A star uses fusion as an energy source by building larger atoms from smaller ones

Explanation:

The energy source of stars is by nuclear fusion. Nuclear fusion is the process whereby two smaller nuclei combine to form a larger nucleus. The sun is an example of a star which is made up of mostly hydrogen and helium atoms. Two hydrogen atoms combine in the core of the star to form a helium atom.

Due to the fusion of hydrogen atoms that takes place in the sun to produce a helium atom, a great amount of heat is provided for the sun to utilize.

Nuclear power plants use heat produced during nuclear fission to heat water. In nuclear fission, atoms are split apart to form smaller atoms, thereby releasing energy.

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5 0
4 years ago
Read 2 more answers
Imagine that the satellite described in the problem introduction is used to transmit television signals. You have a satellite TV
Korolek [52]

Complete Question

A satellite in geostationary orbit is used to transmit data via electromagnetic radiation. The satellite is at a height of 35,000 km above the surface of the earth, and we assume it has an isotropic power output of 1 kW (although, in practice, satellite antennas transmit signals that are less powerful but more directional).

Imagine that the satellite described in the problem introduction is used to transmit television signals. You have a satellite TV reciever consisting of a circular dish of radius R which focuses the electromagnetic energy incident from the satellite onto a receiver which has a surface area of 5 cm2.

How large does the radius R of the dish have to be to achieve an electric field vector amplitude of 0.1 mV/m at the receiver?

For simplicity, assume that your house is located directly beneath the satellite (i.e. the situation you calculated in the first part), that the dish reflects all of the incident signal onto the receiver, and that there are no losses associated with the reception process. The dish has a curvature, but the radius R refers to the projection of the dish into the plane perpendicular to the direction of the incoming signal.

Give your answer in centimeters, to two significant figures.

Answer:

 The radius  of  the dish is R = 18cm

Explanation:

  From the question we are told that

     The radius of the orbit is  = R = 35,000km = 35,000 *10^3 m

    The power output of the power is  P = 1 kW = 1000W

   The electric vector amplitude is given as E = 0.1 mV/m = 0.1 *10^{-3}V/m

    The area of thereciever  is   A_R = 5cm^2

Generally the intensity of the dish is mathematically represented as

         I = \frac{P}{A}

Where A is the area orbit which is a sphere so this is obtained as

          A = 4 \pi r^2

              = (4 * 3.142 * (35,000 *10^3)^2)

              =1.5395*10^{16} m^2

  Then substituting into the equation for intensity

          I_s  =  \frac{1000}{1.5395*10^{16}}

            = 6.5*10^ {-14}W/m2

 Now the intensity received by the dish can be mathematically evaluated as

              I_d = \frac{1}{2}  * c \epsilon_o E_D ^2

  Where c is thesped of light with a constant value  c = 3.0*10^8 m/s

              \epsilon_o is the permitivity of free space  with a value  8.85*10^{-12} N/m

              E_D is the electric filed on the dish

So  since we are to assume to loss then the intensity of the satellite is equal to the intensity incident on the receiver dish

      Now making the eletric field intensity the subject of the formula

                  E_D = \sqrt{\frac{2 * I_d}{c * \epsilon_o} }

substituting values

                 E_D = \sqrt{\frac{2 * 6.5*10^{-14}}{3.0*10^{8} * 8.85*10^{-12}} }

                       = 7*10^{-6} V/m

The incident power on the dish is what is been reflected to the receiver

                P_D = P_R

Where P_D is the power incident on the dish which is mathematically represented as

              P_D = I_d A_d

                   = \frac{1}{2}  c \epsilon_o E_D^2  (\pi R^2)

And  P_R is the power incident on the dish which is mathematically represented as

                 P_R = I_R A_R

                       = \frac{1}{2} c \epsilon_o E_R^2 A_R

Now equating the two

                \frac{1}{2}  c \epsilon_o E_D^2  (\pi R^2) =  \frac{1}{2} c \epsilon_o E_R^2 A_R

   Making R the subject we have

                   R = \sqrt{\frac{E_R^2 A_R}{\pi E_D^2} }

Substituting values

                   R = \sqrt{\frac{(0.1 *10^{-3})^2 * 5}{\pi (7*10^{-6})^ 2} }

                     R = 18cm

8 0
3 years ago
The mean distance of an asteroid from the Sun is 2.98 times that of Earth from the Sun. From Kepler's law of periods, calculate
lutik1710 [3]

Answer:

The asteroid requires 5.14 years to make one revolution around the Sun.

Explanation:

Kepler's third law establishes that the square of the period of a planet will be proportional to the cube of the semi-major axis of its orbit:

T^{2} = a^{3} (1)

Where T is the period of revolution and a is the semi-major axis.

In the other hand, the distance between the Earth and the Sun has a value of 1.50x10^{8} Km. That value can be known as well as an astronomical unit (1AU).

But 1 year is equivalent to 1 AU according with Kepler's third law, since 1 year is the orbital period of the Earth.

For the special case of the asteroid the distance will be:

a = 2.98(1.50x10^{8}Km)

a = 4.47x10^{8}Km

That distance will be expressed in terms of astronomical units:

4.47x10^{8}Km.\frac{1AU}{1.50x10^{8}Km} ⇒ 2.98AU

Finally, from equation 1 the period T can be isolated:

T = \sqrt{a^{3}}

T = \sqrt{(2.98)^{3}}  

T = \sqrt{26.463592}

T = 5.14AU

Then, the period can be expressed in years:

5.14AU.\frac{1yr}{1AU} ⇒ 5.14 yr

T = 5.14 yr

Hence, the asteroid requires 5.14 years to make one revolution around the Sun.

8 0
3 years ago
A 60 [Hz] high-voltagepower line carries a current of 1000 [A]. The power line is at a height of 50 [m] above the earth. What is
omeli [17]

Answer:

The magnitude of the magneticfield B[T] at a point on the surface of the earth directly below the power line is B=1.496x10^(-6) T.

Explanation:

The distance from the wire to a point in the surface is the heigth of the wire.

The formula for the magnetic field on any point at distance R from a wire conducting alternating current is:

B=\frac{\mu_0I}{2\pi R} =\frac{(4\pi\cdot 10^{-7}T\cdot m/A)(1000\,A)}{2\pi \cdot 50\,m} =1.496\cdot10^{-6}\,T

4 0
3 years ago
What is the force on an object with a mass of 25 kg and an acceleration of 5 m/s/s? 1 point
Oduvanchick [21]

Answer:125N

Explanation:

Mass =25kg

Acceleration =5m/s/s

Force=mass x acceleration

Force=25 x 5

Force=125N

7 0
3 years ago
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