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Nana76 [90]
3 years ago
6

What type of radiation does nuclear fission produce?

Physics
2 answers:
saw5 [17]3 years ago
8 0

Answer:

beta

Explanation:

netineya [11]3 years ago
4 0

Anwser: Beta.

Explanation: Fission products emit beta radiation, while actinides

primarily emit alpha radiation many of each also emit gamma radiation.

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In the 2008 Olympics, Jamaican sprinter Usain Bolt shocked the world as he ran the 100-meter dash in 9.69 seconds. Determine Usa
-Dominant- [34]

Answer:

His average speed was 10.3199 m/s.

Explanation:

4 0
3 years ago
Queremos un cilindro de simple efecto que utilice en su funcionamiento un volumen de aire a presión atmosférica de 13,122 litros
zhuklara [117]

Answer:

1) El diámetro es de aproximadamente 913,987 cm.

2) La fuerza del cilindro es 5576850 kgf

Explanation:

1) Los parámetros dados son;

El volumen del aire = 13,122 litros = 13122000 cm³

La presión de trabajo = 8.5 kgf / cm²

La longitud del cilindro = 20 cm.

Por lo tanto, tenemos;

El área de la base del cilindro = π · r² = 13122000 cm³ / (20 cm) = 656100 cm²

r = √ (656100 / π) ≈ 456,994 cm

El diámetro = 2 × r ≈ 2 × 456.994 ≈ 913.987 cm

El diámetro ≈ 913,987 cm

2) La fuerza del cilindro = El área de la base del cilindro × La presión de trabajo

∴ La fuerza del cilindro = 656100 cm² × 8.5 kgf / cm² = 5576850 kgf

La fuerza del cilindro = 5576850 kgf

3 0
3 years ago
When aluminum foil is formed into a loose ball, it can float on water. But when the ball of foil is pounded flat with a hammer,
docker41 [41]
Air caught in the ball of foil makes the ball less dense than water
8 0
3 years ago
A 4.9-MeV (kinetic energy) proton enters a 0.28-T field, in a plane perpendicular to the field. Part APart complete What is the
BartSMP [9]

Answer:

r=1.14m

Explanation:

\theta is the angle between the velocity and the magnetic field. So, the magnetic force on the proton is:

F_m=qvBsen\theta\\F_m=qvBsen(90^\circ)\\F_m=qvB

A charged particle describes a semicircle in a uniform magnetic field. Therefore, applying Newton's second law to uniform circular motion:

F_m=F_c\\qvB=F_c(1)

F_c is the centripetal force and is defined as:

F_c=m\frac{v^2}{r}

Here v is the proton's speed and r is the radius of the circular motion. Replacing this in (1) and solving for r:

qvB=\frac{mv^2}{r}\\r=\frac{mv^2}{qvB}\\r=\frac{mv}{qB}

Recall that 1 J is equal to 6.242*10^{12}MeV, so:

4.9MeV*\frac{1J}{6.242*10^{12}MeV}=7.85*10^{-13}J

We can calculate v from the kinetic energy of the proton:

K=\frac{mv^2}{2}\\\\v=\sqrt{\frac{2K}{m}}\\v=\sqrt{\frac{2(7.85*10^{-13}J)}{1.67*10^{-27}kg}}\\v=3.06*10^{7}\frac{m}{s}

Finally, we calculate the radius of the proton path:

r=\frac{mv}{qB}\\r=\frac{1.67*10^{-27}kg(3.06*10^{7}\frac{m}{s})}{1.6*10^{-19}C(0.28T)}\\r=1.14m

8 0
3 years ago
When compared to
ElenaW [278]
I think it is B, because the sun’s size is pretty average
3 0
2 years ago
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