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Marizza181 [45]
3 years ago
9

There are several technological applications for the transuranium elements (Z > 92). An important one is in smoke detectors,

which can use the decay of a tiny amount of americium-241 to neptunium-237. What subatomic particle is emitted from that decay process
Physics
1 answer:
natima [27]3 years ago
6 0

Answer:

particle emit 4 unit of mass

Explanation:

Given data

Z > 92

americium-241

neptunium-237

to find out

subatomic particle is emitted from that decay process

solution

we have given that tiny amount decay from americium 241 to  neptunium 237

so total difference is here  (241- 237 ) = 4

so its mean this particle emit 4 unit of mass.

and we say that radioactive substance americium 241 emit alpha rays during decay process

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Father drove 176 km in 7 hours. For the first 92 km, he spent
Verizon [17]

Answer:

28km/h

Explanation:

92km = 4 hours

176 - 92 = 84

84km = 3 hours

84/3 = 28

28km/h

Hopefully this helps you :)

pls mark brainlest ;)

3 0
3 years ago
The index of refraction of silicate flint glass for red light is 1.620 and for violet light is 1.660 . A beam of white light in
chubhunter [2.5K]

Answer:

The angular separation equals 0.35^{o}

Explanation:

We have according to Snell's law

n_{1}sin(\theta _{1})=n_{2}sin(\theta _{2})

\therefore \theta _{2}=sin^{-1}(\frac{n_{1}sin(\theta _{1})}{n_{2}})

Using this equation for both the colors separately we have

\theta _{red}=sin^{-1}(\frac{sin(23.90^{o})}{1.62})\\\\\theta _{red}=14.48^{o}

Similarly for violet light we have

\theta _{violet}=sin^{-1}(\frac{sin(23.90^{o})}{1.660})\\\\\theta _{violet}=14.13^{o}

Thus the angular separation becomes

\Delta \theta =14.48-14.13=0.35^{o}

6 0
4 years ago
In a Rutherford scattering experiment a target nucleus has a diameter of 1.34×10-14 m. The incoming α particle has a mass of 6.6
Rasek [7]

Answer:

E = 2.5 x 10⁻¹⁴ J

Explanation:

given,

diameter = 1.33 x 10⁻¹⁴ m

mass = 6.64 x 10⁻²⁷ kg

wavelength is equal to diameter

de broglie wavelength equal to diameter

         \lambda = \dfrac{h}{mv}

         1.33 \times 10^{-14}= \dfrac{6.626 \times 10^{-34}}{6.64 \times 10^{-27}\times v}

         v= \dfrac{6.626 \times 10^{-34}}{6.64 \times 10^{-27}\times 1.33 \times 10^{-14}}

              v = 7.5 x 10⁶ m/s

Kinetic energy is equal to

     E = \dfrac{1}{2}mv^2

     E = \dfrac{1}{2}\times 6.64 \times 10^{-27}\times (7.5\times 10^6)^2

            E = 2.5 x 10⁻¹⁴ J

8 0
3 years ago
Two small plastic spheres are given positive electrical charges. When they are a distance of 14.8cm apart, the repulsive force b
sdas [7]

Answer:

Explanation:

Case I: They have same charge.

Charge on each sphere = q

Distance between them, d = 14.8 cm = 0.148 m

Repulsive force, F = 0.235 N

Use Coulomb's law in electrostatics

F=\frac{Kq_{1}q_{2}}{d^{2}}

By substituting the values

0.235=\frac{9\times10^{9}q^{2}}{0.148^{2}}

q=7.56\times10^{-7}C

Thus, the charge on each sphere is q=7.56\times10^{-7}C.

Case II:

Charge on first sphere = 4q

Charge on second sphere = q

distance between them, d = 0.148 m

Force between them, F = 0.235 N

Use Coulomb's law in electrostatics

F=\frac{Kq_{1}q_{2}}{d^{2}}

By substituting the values

0.235=\frac{9\times 10^{9}\times 4q^{2}}{0.148^{2}}

q=3.78\times10^{-7}C

Thus, the charge on second sphere is q=3.78\times10^{-7}C and the charge on first sphere is 4q = 4\times 3.78\times 10^{-7}=1.51 \times 10^{-6} C.

6 0
4 years ago
When an electric current flows through a long conductor, each free electron moves
ki77a [65]
A. through a relatively short distance.

The speed is actually called the drift speed of the electron.
7 0
4 years ago
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