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2 years ago

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In 1932, James Chadwick bombarded beryllium-9 with alpha particles. One product was a neutron. This led to the direct experiment

al verification of the existence of neutrons. The other product of Chadwick's nuclear reaction was carbon-12. boron-12. helium-4. nitrogen-14.

1 answer:

Fed [463]2 years ago

7 0

Answer: The other product of Chadwick's nuclear reaction was carbon-12.

Explanation:

The isotopic representation of an atom is: $_Z^A\textrm{X}$

where,

Z = Atomic number of the atom

A = Mass number of the atom

X = Symbol of the atom

In a nuclear reaction, the total mass and total atomic number remains the same.

For the given reaction:

$^{9}_{4}\textrm{Be}+^4_2\textrm{He}\rightarrow ^{A}_{Z}\textrm{Kr}+^1_0\textrm{n}$

To calculate A:

Total mass on reactant side = total mass on product side

9 + 4= A + 1

A = 12

To calculate Z:

Total atomic number on reactant side = total atomic number on product side

4 + 2 = Z + 0

Z = 6

The isotopic symbol is $_{6}^{12}\textrm{C}$ . Thus The other product of Chadwick's nuclear reaction was carbon-12.

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Two stones resembling diamonds are suspected of being fakes. To determine if the stones might be real, the mass and volume of ea

dimaraw [331]

Answer:

stone A is diamond.

Explanation:

given,

Volume of the two stone = 0.15 cm³

Mass of stone A = 0.52 g

Mass of stone B = 0.42 g

Density of the diamond = 3.5 g/cm³

So, to find which stone is gold we have to calculate the density of both the stone.

We know,

$density$ $\density = \dfrac{mass}{volume}$

density of stone A

$\rho_A = \dfrac{0.52}{0.15}$

$\rho_A = 3.467\ g/cm^3$

density of stone B.

$\rho_B = \dfrac{0.42}{0.15}$

$\rho_B = 2.8\ g/cm^3$

Hence, the density of the stone A is the equal to Diamond then stone A is diamond.

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2 years ago

PLEEEEEAAASSSEEE HELP ME

Xelga [282]

Answer:

1) A. 0.44 m/s East + 0.33 m/s North

2) A. 0 m/s²

3) A. a scalar calculated as distance divided by time.

4) B. 31 km per hour

Explanation:

1) Velocity is DISPLACEMENT over time.

at 1 m/s, total time of walking is 9000 seconds

displacement is 3000 m north and 5000 - 1000 = 4000 m east

4000 m/ 9000 s = 0.44 m/s E

3000 m/ 9000s = 0.33 m/s N

2) constant speed means no acceleration

3) A. a scalar calculated as distance divided by time.

4) displacement 50 km N and 80 km W

v = √(50² + 80²) / (1 + 2) = 31.446603... km/hr

3 0

2 years ago

Which is a better conductor, a flagpole or a flag? Why?

Novosadov [1.4K]

Answer:

flagpole

Explanation:

if it is about electricity then its flagpole

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2 years ago

Read 2 more answers

A p-type Si sample is used in the Haynes-Shockley experiment. The length of the sample is 2 cm, and two probes are separated by

Airida [17]

Answer:

Mobility of the minority carriers, $\mu_{n} =1184.21 cm^{2} /V-sec$

Diffusion coefficient for minority carriers, $D_{n} = 29.20 cm^2 /s$

Verified from Einstein relation as $\frac{D_{n} }{\mu_{n} } = 25 mV$

Explanation:

Length of sample, $l_{s} = 2 cm$

Separation between the two probes, L = 1.8 cm

Drift time, $t_{d} = 0.608 ms$

Applied voltage, V = 5 V

Mobility of the minority carriers ( electrons), $\mu_{n} = \frac{V_{d} }{E}$

Where the drift velocity, $V_{d} = \frac{L}{t_{d} }$

$V_{d} = \frac{1.8}{0.608 * 10^{-3} } \\V_{d} = 2960.53 cm/s$

and the Electric field strength, $E = \frac{V}{l_{s} }$

E = 5/2

E = 2.5 V/cm

Mobility of the minority carriers:

$\mu_{n} = 2960.53/2.5\\\mu_{n} =1184.21 cm^{2} /V-sec$

The electron diffusion coefficient, $D_{n} = \frac{(\triangle x)^{2} }{16 t_{d} }$

$\triangle x = (\triangle t )V_{d}$ , where Δt = separation of pulse seen in an oscilloscope in time( it should be in micro second range)

$\triangle x = \frac{(\triangle t) L}{t_{d} } \\\triangle x = \frac{180*10^{-6} * 1.8}{0.608*10^{-3} }\\\triangle x =0.533 cm$

$D_{n} = \frac{0.533^{2} }{16 * 0.608 * 10^{-3} }\\D_{n} = 29.20 cm^2 /s$

For the Einstein equation to be satisfied, $\frac{D_{n} }{\mu_{n} } = \frac{KT}{q} = 0.025 V$

$\frac{D_{n} }{\mu_{n} } = \frac{29.20}{1184.21} \\\frac{D_{n} }{\mu_{n} } = 0.025 = 25 mV$

Verified.

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2 years ago

rodikova [14]

Answer:

See the explanation below.

Explanation:

12) When an object is falling, how does the objects velocity change? what formula do you use?

The speed of a falling object is increased by a value of 9.81 meters per second per second. That is if we throw any body regardless of mass from a considerable height, its speed in the first second will be 9.81[ m/ s] , in the next second will be equal to 19.62 [m/s] in the next will be equal to 29.43 [m/ s].

The formula is:

$v=v_{0}+g*t$

where:

vo = initial velocity = 0

g = gravity = 9.81[m/s^2]

t = time [s]

13)

what is a falling speed at 6s, 9s, 112s?

v = 0 + (9.81*6) = 58.86[m/s]

v = 0 + (9.81*9) = 88.29 [m/s]

v = 0 + (9*112) = 1098.72 [m/s]

14)

If an object is falling at 65 [m/s]. How long has it been falling ? what is the formula that you use?

The formula is the same:

$v=v_{o}+g*t$

65 = 0 + 9.81*t

t = 65/9.81

t = 6.62[s]

15)

What formula is used to determine the distance an object is falling ?

$y = y_{o}+v_{o}*t + 0.5*9.81*t^{2}$

where:

y = distance [m]

yo = initial distance, in most of the cases and depending of the reference point it will be eqaul to zero

vo = initial velocity, if it is free fall, then = 0

t = time [s]

g = gravity = 9.81[m/s^2]

This equation will be reduce to:

$y = 0.5*g*t^{2}$

16)

using the times given in problem 13. Determine the distance fallen for each.

y = 0.5*9,81*(6)^2 = 176.58 [m]

y = 0.5*9,81*(9)^2 = 397.3 [m]

y = 0.5*9,81*(112)^2 = 61528.3 [m]

17)

If an object has fallen a distance of 87.3 [m]. How long was it falling?

87.3 = 0.5*9.81*t^2

$t=\sqrt{\frac{87.3}{0.5*9.81} }\\ t=4.21[s]$

4 0

2 years ago