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3 years ago

Which missing item would complete this beta decay reaction?

1 answer:

8 0

For us to understand the missing item that would complete beta decay reaction, we need to achieve in depth understanding of chemical formulas and nuclear symbols. Next is to have great comprehension of the following points:
<span>1.) Neutron in nucleus breaks down and changes into a proton.
2) Then it emits an electron, as well as an anti-neutrino which go into space.
3) Lastly, atomic number continuously goes UP while mass number remains unchanged.</span>

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How can you find average speed

BabaBlast [244]

Average speed =

(total distance covered)

divided by

(total time spent covering the distance)

4 0

3 years ago

With a diameter that's 11 times larger than Earth's, _______ is the largest planet.

Mkey [24]

<span>With a diameter that's 11 times larger than Earth's, JUPITER is the largest planet.</span>

8 0

3 years ago

A uniform rod of mass M and length L can pivot freely at one end. Initially, the rod is oriented vertically above the pivot, in

Leya [2.2K]

Answer:

The speed of its center of mass = $\sqrt{\frac{3}{2}gL}$

Explanation:

Consider the potential energy at the level of center of mass of rod below the pivot=0

Mass of uniform rod=M

Length of rod=L

The rotational inertia about the end of a uniform rod= $\frac{1}{3}ML^2$

Kinetic energy at the level of center of mass of rod below the pivot= $\frac{1}{2}I\omega^2$

Kinetic energy at the level of center of mass of rod above the pivot=0

Potential energy at the level of center of mass of rod above the pivot=mgh

We have to find the center of mass ( in terms of g and L).

According to conservation of law of energy

Initial P.E+Initial K.E=Final P.E+Final K.E

$mgh+0=0+\frac{1}{2} I\omega^2$

Where $K.E=\frac{1}{2} I\omega^2$

I=Moment of inertia

$\omega$ =Angular velocity

Substitute the values then we get

$MgL=\frac{1}{2}\times \frac{1}{3}ML^2\omega^2$

$\omega^2=\frac{6g}{L}$

Now, we know that $\omega=\frac{v}{r}$ , $r=\frac{L}{2}$

Substitute the values then we get

$\frac{v^2}{(\frac{L}{2})^2}=\frac{6g}{L}$

$\frac{v^24}{L^2}=\frac{6g}{L}$

$v^2=\frac{6g\times L^2}{4L}$

$v^2=\frac{3gL}{2}$

$v=\sqrt{\frac{3}{2}gL}$

Hence, the speed of its center of mass = $\sqrt{\frac{3}{2}gL}$

4 0

4 years ago

A large chunk of ice breaks off an Alaskan glacier and lands in the ocean.

siniylev [52]

Answer:

b hope this helps

Explanation:

8 0

3 years ago

An energy plant produces an output potential of 1500 kV and serves a city 143 km away. A high-voltage transmission line carries

jekas [21]

Answer:

2123.55 $/hr

Explanation:

Given parameters are:

$V_{plant} = 1500$ KV

L = 143 km

I = 500 A

$\rho = 2.4$ $\Omega / km$

So, we will find the voltage potential provided for the city as:

$V_{wire} =IR = I\rho L = 1500*2.4*143 = 514.8$ kV

$V_{city} = V_{plant}- V_{wire} = 1500-514.8 = 985.2$ kV

Then, we will find dissipated power because of the resistive loss on the transmission line as:

$P = I^2R = I^2\rho L=500^2*2.4*143 = 8.58*10^7$ W

Since the charge of plant is not given for electric energy, let's assume it randomly as $x = \frac{\dollar 0.081}{kW.hr}$

Then, we will find the price of energy transmitted to the city as:

$Cost = P * x = 8.58*10^7 * 0.081 * 0.001 = 6949.8$ $/hr

To calculate money per hour saved by increasing the electric potential of the power plant:

Finally,

$I_{new} = P/V_{new} = I/1.2\\P_{new} = I_{new}^2R_{wire}\\Cost = P_{new}/1.44=6949.8/1.44 = 4826.25$ $/hr

The amount of money saved per hour = $6949.8 - 6949.8/1.44 = 2123.55$ $/hr

Note: For different value of the price of energy, it just can be substituted in the equations above, and proper result can be found accordingly.

3 0

3 years ago