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zysi [14]
3 years ago
10

What is the mass number of the particle emitted from the nucleus during beta minus decay? What kind of charge does the particle

emitted from the nucleus during beta minus decay have?
Physics
1 answer:
Lisa [10]3 years ago
5 0

Beta decay is exclusively the emission of electrons ... or its
oppositely charged relative, the positron ... from an atomic
nucleus.

Beta-minus is the emission of an electron, with a negative charge.

The change in the mass of the nucleus, if any, is minuscule. 
The mass number of a nucleus is the sum of its protons and
neutrons, and the mass of an emitted electron is something
like 0.0005 the mass of one of those. 

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vekshin1
<span>The Badminton World Federation</span>
4 0
4 years ago
Read 2 more answers
I’M DESPERATE PLZ HELP!! 40 POINTS!!
Ierofanga [76]

Answer:

a. The acceleration of the hockey puck is -0.125 m/s².

b. The kinetic frictional force needed is 0.0625 N

c. The coefficient of friction between the ice and puck, is approximately 0.012755

d. The acceleration is -0.125 m/s²

The frictional force is 0.125 N

The coefficient of friction is approximately 0.012755

Explanation:

a. The given parameters are;

The mass of the hockey puck, m = 0.5 kg

The starting velocity of the hockey puck, u = 5 m/s

The distance the puck slides and slows for, s = 100 meters

The acceleration of the hockey as it slides and slows and stops, a = Constant acceleration

The velocity of the hockey puck after motion, v = 0 m/s

The acceleration of the hockey puck is obtained from the kinematic equation of motion as follows;

v² = u² + 2·a·s

Therefore, by substituting the known values, we have;

0² = 5² + 2 × a × 100

-(5²) = 2 × a × 100

-25 = 200·a

a = -25/200 = -0.125

The constant acceleration of the hockey puck, a = -0.125 m/s².

b. The kinetic frictional force, F_k, required is given by the formula, F = m × a,

From which we have;

F_k = 0.5 × 0.125 = 0.0625 N

The kinetic frictional force required, F_k = 0.0625 N

c. The coefficient of friction between the ice and puck, \mu_k, is given from the equation for the kinetic friction force as follows;

F_k = \mu_k \times Normal \ force \ of \ hockey \ puck = \mu_k \times 0.5 \times 9.8

\mu_k = \dfrac{0.0625}{0.5 \times 9.8} \approx 0.012755

The coefficient of friction between the ice and puck, \mu_k ≈ 0.012755

d. When the mass of the hockey puck is 1 kg, we have;

Given that the coefficient of friction is constant, we have;

The frictional force F_k = 0.012755 \times 1 \times 9.8 = 0.125  \ N

The acceleration, a = F_k/m = 0.125/1 = 0.125 m/s², therefore, the magnitude of the acceleration remains the same and given that the hockey puck slows, the acceleration is -0.125 m/s² as in part a

The frictional force as calculated here,  F_k  = 0.125  \ N

The coefficient of friction \mu_k ≈ 0.012755 is constant

5 0
3 years ago
What is a plane mirror? state the characteristics of the image formed by a plane mirror ​
Nadusha1986 [10]
A plane mirror always forms a virtual image. the image and the object are the same distance from a flat mirror, the image size is the same as the object, and the image is upright!
3 0
3 years ago
The displacement of particle moving along on x axis is given by x = 18t + 5.0t^2, where x is in meters and t is in seconds Calcu
Masja [62]

Answer:

38 m/s

43 m/s

Explanation:

x = 18t + 5.0t²

The instantaneous velocity is the first derivative:

v = 18 + 10.t

At t = 2.0:

v = 18 + 10.(2.0)

v = 38 m/s

The average velocity is the change in position over change in time.

v = Δx / Δt

v = [ (18t₂ + 5.0t₂²) − (18t₁ + 5.0t₁²) ] / (t₂ − t₁)

Between t = 2.0 and t = 3.0:

v = [ (18(3.0) + 5.0(3.0)²) − (18(2.0) + 5.0(2.0)²) ] / (3.0 − 2.0)

v = [ (54 + 45) − (36 + 20.) ] / 1.0

v = 99 − 56

v = 43 m/s

6 0
3 years ago
How many protons are in one kg of lead? b) How many electrons are in one kg of lead? c) What is the total negative charge of the
mina [271]

Answer:

Number of protons = 2.3836\times 10^{26}

Number of electrons = 2.3836\times 10^{26}

Charge on these electrons = -3.8185\times 10^{7}\ C

Explanation:

1 atom of lead contains 82 protons and 82 electrons

1 mole = 6.023\times 10^{23} atoms

Thus,

1 mole of lead contains 82\times 6.023\times 10^{23} protons

Also, 1 mole of lead weighs 207.2 g

So,

207.2 g of lead contains 82\times 6.023\times 10^{23} protons

1 kg = 1000 g

So,

1000 g or 1 kg of lead contains \frac {82\times 6.023\times 10^{23}}{207.2}\times 1000 protons

<u>Number of protons = 2.3836\times 10^{26}</u>

Also, for neutral atom, number of protons = number of electrons.

Thus,

<u>Number of electrons = 2.3836\times 10^{26}</u>

Also,

Charge of 1 electron = - 1.602\times 10^{-19}\ C

Charge on these electrons = -2.3836\times 10^{26}\times 1.602\times 10^{-19}\ C

<u>Charge on these electrons = -3.8185\times 10^{7}\ C</u>

3 0
3 years ago
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