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Vika [28.1K]
4 years ago
6

A large atom decays and emits a particle. After the reaction is complete, the atom’s mass has changed substantially. What kind o

f particle has most likely been emitted?
alpha particle
beta particle
positron
photon
Physics
2 answers:
Montano1993 [528]4 years ago
5 0

Answer:

alpha particle

Explanation:

In a process where alpha particle is emitted, a heavier nuclei decays into lighter nuclei. The alpha particle released has a charge of +2 units.

_Z^A\textrm{X}\rightarrow _{Z-2}^{A-4}+_2^4\alpha

The atomic mass of the resulting lighter nuclei reduces by 4.

The process in which beta particle is emitted, a neutron gets converted into a proton and an electron releasing a beta-particle. The beta particle released carries a charge of -1 units.

_Z^A\textrm{X}\rightarrow _{Z+1}^A\textrm{Y}+_{-1}^0\beta

The mass of the resultant nuclei remains same.

The process in which a proton gets converted to neutron and an electron neutrino, a positron is released. This particle carries a charge of +1 units.

_Z^A\textrm{X}\rightarrow _{Z-1}^A\textrm{Y}+_{+1}^0e

The mass of the resultant nuclei remains unchanged.

In gamma ray emission,

_Z^A\textrm{X}^*\rightarrow _Z^A\textrm{X}+_0^0\gamma

A photon is emitted. An unstable nuclei gets stabilized by release of photon. There is no change in the nuclei.

Thus, in a reaction of decay of a large atom, release of alpha particle would result in substantial change in atomic's mass.

LiRa [457]4 years ago
3 0
The answer your looking for is, A. alpha particle.
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erastova [34]

Answer:

The angular acceleration of the wheel is -6.54 rad/s²

Explanation:

We'll use the equations of motion for this.

w = 2πf

f = 75 rpm = 1.25 rps = 1.25 rev/s

w₀ = initial angular velocity = 2π × 1.25 = 7.85 rad/s

w = final angular velocity = 0 rad/s

t = 1.2 s

α = ?

w = w₀ + αt

0 = 7.85 + 1.2α

α = 7.85/1.2 = - 6.54 rad/s²

6 0
4 years ago
If the air temperature is 12 °c and the vapor pressure is the same as the saturation vapor pressure, the relative humidity is:_
aliina [53]

The relative humidity is 100%.

To find the answer, we have to know about the relative humidity.

<h3>What is the relative humidity?</h3>
  • The amount of water vapor in the air is referred to as humidity.
  • The gaseous and invisible condition of water is called water vapor.
  • Humidity can be expressed in a variety of ways, including absolute humidity, relative humidity, mixing ratio, and dew point temperature.
  • Relative humidity can be expressed as,

                         Rel.Humidity=\frac{P_{vapor}}{P_{sat}}*100

  • Here, it is given that, the vapor pressure is the same as the saturation vapor pressure, thus,

                   Rel.Humidity=\frac{P_{sat}}{P_{sat}}*100=100

Thus, we can conclude that, the relative humidity is 100%.

Learn more about the  relative humidity here:

brainly.com/question/21494654

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3 0
2 years ago
Clearly, Atwood’s machine has a lot of systematic error that would not be present if we were to simplify the experiment. What is
Sphinxa [80]

Answer:

Better Equilibrium Maintenance for better accuracy...

Explanation:

In the Galileo's experiment, there is no utilization of two equal masses at a time. However, as we can see in a Atwood Machine, there are two equal masses involved that make the whole system to be in a state of equilibrium and ultimately the better measurements of acceleration due to gravity.

4 0
3 years ago
An 80-kg quarterback jumps straight up in the air right before throwing a 0.43-kg football horizontally at 15 m/s . How fast wil
Oksanka [162]

Answer:

vBxf = 0.08625m/s

Explanation:

This is a problem about the momentum conservation law. The total momentum before equals the total momentum after.

p_f=p_i

pf: final momentum

pi: initial momentum

The analysis of the momentum conservation is about a horizontal momentum (x axis). When the quarterback jumps straight up, his horizontal momentum is zero. Then, after the quarterback throw the ball the sum of the momentum of both quarterback and ball must be zero.

Then, you have:

m_Qv_{Qxi}+m_{Bxi}v_{Bxi}=m_Qv_{Qxf}+m_{Bxf}v_{Bxf}    (1)

mQ: the mass of the quarterback = 80kg

mB: the mass of the football = 0.43kg

(vQx)i: the horizontal velocity of quarterback before throwing the ball = 0m/s

(vBx)i: the horizontal velocity of football before being thrown = 0m/s

(vQx)f: the horizontal velocity of quarterback after throwing the ball = ?

(vBx)f: the horizontal velocity of football after being thrown = 15 m/s

You replace the values of the variables in the equation (1), and you solve for (vBx)f:

0\ kgm/s=-(80kg)(v_{Bxf})+(0.46kg)(15m/s)\\\\v_{Bxf}=\frac{(0.46kg)(15m/s)}{80kg}=0.08625\frac{m}{s}

Where you have taken the speed of the quarterback as negative because is in the negative direction of the x axis.

Hence, the speed of the quarterback after he throws the ball is 0.08625m/s

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Musya8 [376]

Answer:

Actin.

Arp2/3.

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Coronin.

Dystrphin.

Elastin.

F-spondin.

Fibronectin.

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Explanation:

5 0
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