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2 years ago

Which of these is emitted during beta decay ?

Physics

2 answers:

satela [25.4K]2 years ago

4 0

Answer:

C. a small charged particle.

Explanation:

typically beta radiation emits an electron which is a small negativity charged particle.

hope it helps. :)

Vesna [10]2 years ago

3 0

Answer:

C) a small charged particle

Explanation:

AP3X

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The distance between the first and fifth minima of a single-slit diffraction pattern is 0.500 mm with the screen 37.0 cm away fr

WARRIOR [948]

In the single-slit experiment, the displacement of the minima of the diffraction pattern on the screen is given by
$y_n= \frac{n \lambda D}{a}$ (1)
where
n is the order of the minimum
y is the displacement of the nth-minimum from the center of the diffraction pattern
$\lambda$ is the light's wavelength
D is the distance of the screen from the slit
a is the width of the slit

In our problem,
$D=37.0 cm=0.37 m$
$\lambda=530 nm=5.3 \cdot 10^{-7} m$
while the distance between the first and the fifth minima is
$y_5-y_1 = 0.500 mm=0.5 \cdot 10^{-3} m$ (2)

If we use the formula to rewrite $y_5, y_1$ , eq.(2) becomes
$\frac{5 \lambda D}{a} - \frac{1 \lambda D}{a} =\frac{4 \lambda D}{a}= 0.5 \cdot 10^{-3} m$
Which we can solve to find a, the width of the slit:
$a= \frac{4 \lambda D}{0.5 \cdot 10^{-3} m}= \frac{4 (5.3 \cdot 10^{-7} m)(0.37 m)}{0.5 \cdot 10^{-3} m}= 1.57 \cdot 10^{-3} m=1.57 mm$

7 0

3 years ago

What are the three subatomic particles of an atom?

Anestetic [448]

Protons, neutrons, and electrons are the three main subatomic particles found in an atom. Protons have apositive (+) charge. An easy way to remember this is to remember that both proton and positive start with theletter "P." Neutrons have no electrical charge.

5 0

2 years ago

Read 2 more answers

Please help meeeee will mark as brainliest

Katyanochek1 [597]

Answer:

a) pressure means force acted on a body per unit area

6 0

2 years ago

Select the correct answer

Nadya [2.5K]

Answer:

$f= 3.0 \times 10 {}^{8} \div 7.0 \times 10 {}^{7} \\ f = 4.28hz$

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2 years ago

A 6.4-N force pulls horizontally on a 1.5-kg block that slides on a smooth horizontal surface. This block is connected by a hori

Elena L [17]

-- Although it's not explicitly stated in the question,we have to assume that
the surface is frictionless. I guess that's what "smooth" means.

-- The total mass of both blocks is (1.5 + 0.93) = 2.43 kg. Since they're
connected to each other (by the string), 2.43 kg is the mass you're pulling.

-- Your force is 6.4 N.
Acceleration = (force)/(mass) = 6.4/2.43 m/s²
 That's about 2.634 m/s² 

(I'm going to keep the fraction form handy, because the acceleration has to be
used for the next part of the question, so we'll need it as accurate as possible.)

-- Both blocks accelerate at the same rate. So the force on the rear block (m₂) is

Force = (mass) x (acceleration) = (0.93) x (6.4/2.43) = 2.45 N.

That's the force that's accelerating the little block, so that must be the tension
in the string.

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2 years ago