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3 years ago

Which nucleus completes the following equation?

Physics

2 answers:

IgorLugansk [536]3 years ago

4 0

Answer: $^{50}_{25}Mn\rightarrow ^{0}_1e+ ^{50}_{24}Cr$

The chromium nucleus will complete the reaction.

Explanation:

The given reaction is a type of radioactive nuclei decay:

Positron emission: It is a type of decay process, in which a proton gets converted to neutron and an electron neutrino. This is also known as $\beta ^+$ -decay. In this the mass number remains same.

$_A^Z\textrm{X}\rightarrow _{Z-1}^A\textrm{Y}+_{+1}^0e$

Now according to the given reaction:

$^{50}_{25}Mn\rightarrow ^{0}_1e+ ^{50}_{24}Cr$

The chromium nucleus will complete the reaction.

Olegator [25]3 years ago

4 0

Answer:

50 25 Mn-0 1 e+ 50 24 Cr

Explanation:

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Two boats start together and race across a 48-km-wide lake and back. Boat A goes across at 48 km/h and returns at 48 km/h. Boat

nika2105 [10]

Answer:

Part 1)

Boat A will win the race

Part 2)

Boat A will win the race by 48 km as the 2nd boat will reach the other end while boat A will just touches the finish line

Part 3)

average velocity must be zero

Explanation:

As we know that the distance moved by the boat is given as

$d = 48 km$

now the time taken by the boat to move to and fro is given as

$t = \frac{d}{v}$

$t = \frac{48 + 48}{48}$

$t = 2 hrs$

Time taken by Boat B to cover the distance

$t = \frac{48}{24} + \frac{48}{72}$

$t = 2.66 h$

Part 1)

Boat A will win the race

Part 2)

Boat A will win the race by 48 km as the 2nd boat will reach the other end while boat A will just touches the finish line

Part 3)

Since the displacement of Boat A is zero

so average velocity must be zero

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3 years ago

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ivanzaharov [21]

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3 years ago

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The potential difference across a variable resistor is 11V and the current flowing through it is 0.4A.

klemol [59]

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Find the angle for the third-order maximum for 591 nm wavelength light falling on a diffraction grating having 1460 lines per ce

Marat540 [252]

Answer:

15.32°

Explanation:

We have given the wavelength $\lambda =591nm=591\times 10^{-9}m$

Diffraction grating is 1460 lines per cm

So $d=\frac{10^{-2}}{1460}=6.71\times 10^{-6}m$ (as 1 m=100 cm )

For maximum diffraction

$dsin\Theta =m\lambda$ here m is order of diffraction

So $6.71\times 10^{-6}sin\Theta =3\times 591\times 10^{-9}$

$sin\Theta =0.264$

$\Theta =15.32^{\circ}$

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2 years ago

A river flows with a speed of 0.600 m/s. A student first swims upriver 0.500 km, then turns around and returns to his starting p

DerKrebs [107]

Answer:

a) 1111.0 seconds

b) 833.3 s

c) Because of proportions

Explanation:

a) Total time of round trip is the sum of time upriver and time downriver

$t_{total}=t_{up}+t_{down}$

Time upriver is calculated with the net speed of student and 0.500 km:

$t_{up}=\frac{d_{istance}}{|v_{swimmer}|} ;\\v_{swimmer}=v_{relative to river}+v_{river}=-1.2+0.6=-0.6 m/s\\t_{up}=\frac{500 m}{0.6 m/s}=833.3 s$

(Becareful with units 0.5 km= 500m) Similarly of downriver:

$t_{down}=\frac{d_{istance}}{|v_{swimmer}|} ;\\v_{swimmer}=1.2+0.6=1.8 m/s\\t_{down}=\frac{500 m}{1.8 m/s}=277.7 s$

So the sum is:

$t_{total}=1111.0s$

b) Still water does not affect student speed, so total time would be simply:

$t_{total}=\frac{1000 m}{1.2 m/s}=833.3 s$

c) For the upriver trip, student moved half the distance in half speed of the calculation in b), so it kept the same ratio and therefore, same time. So the aditional time is actually the downriver.

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2 years ago