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3 years ago

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Many radioactive atoms that have large masses undergo radioactive decay by releasing a particle that is identical to a helium-4

nucleus. What changes in the original atom are expected as a result of this natural phenomenon

1 answer:

Ksenya-84 [330]3 years ago

8 0

Answer: Atomic number is reduced by two and mass number is reduced by four

Explanation:

Radioactive decay of the nuclei happens when a nuclei has to attain some sort of stability. It undergoes various decay processes and releases various particles during these decay process.

One of decays is alpha decay in which the nucleus emits alpha particles. Alpha decay is radioactive decay in which a unstable atomic nucleus emits an alpha particle (helium nucleus) and transforms into an atom with an atomic number that is reduced by two and mass number that is reduced by four.

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

where Z = atomic number

A = mass number

X = symbol of element

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A worker exerts a pulling force on a box. The worker exerts this force by attaching a rope to the box and pulling on the rope so

shusha [124]

Answer:

<em>The magnitude of the pulling force is 20.66% of the gravitational force acting on the box</em>

Explanation:

<u>Accelerated Motion</u>

The net force exerted on a body is the (vector) sum of all forces applied to the body. The net force can be decomposed in its rectangular components and the dynamics of the body can be studied in each direction x,y separately.

Let's start off by calculating the acceleration the worker gives to the box when pulling it. The distance traveled by the box initially at rest in a time t at an acceleration a is given by

$\displaystyle x=\frac{at^2}{2}$

Solving for a

$\displaystyle a=\frac{2x}{t^2}$

$\displaystyle a=\frac{2\cdot 10}{5.12^2}$

$a=0.763\ m/s^2$

Now we analyze the geometric of the forces applied to the box. Please refer to the free body diagram provided below.

The forces in the y-axis must be in equilibrium since no movement takes place there, thus, being g the acceleration of gravity:

$T_y+N=m.g$

There Ty is the vertical component of the tension of the rope, N is the normal force, and m is the mass of the box

The decomposition of T gives us

$T_y=Tsin\theta$

$T_x=Tcos\theta$

Solving the above equation for N

$Tsin\theta+N=m.g$

$N=m.g-Tsin\theta\text{..........[1]}$

Now for the x-axis, there are two forces acting on the box, the x-component of the tension and the friction force Fr. Those forces are not equilibrated, thus acceleration is produced:

$Tcos\theta-F_r=m.a$

Recalling that

$F_r=\mu N$

$Tcos\theta-\mu N=m.a$

Replacing N from [1]

$Tcos\theta-\mu (m.g-Tsin\theta)=m.a$

Operating

$Tcos\theta-\mu m.g+\mu Tsin\theta=m.a$

Solving for T

$T(cos\theta+\mu sin\theta)=m.a+\mu m.g$

$\displaystyle T=\frac{m.a+\mu m.g}{cos\theta+\mu sin\theta}$

$\displaystyle T=m\frac{a+\mu g}{cos\theta+\mu sin\theta}$

We don't know the value of m, thus we'll plug in the rest of the data

$\displaystyle T=m\frac{0.763+0.10\cdot 9.8}{cos36.8^o+0.10 sin36.8^o}$

$T=2.0252m$

Dividing by the weight of the box m.g

$T/(m.g)=2.0252/9.8=0.2066$

Thus, the magnitude of the pulling force is 20.66% of the gravitational force acting on the box

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3 years ago

A mass of 0.560 kg is attached to a spring and set into oscillation on a horizontal frictionless surface. The simple harmonic mo

Slav-nsk [51]

Answer:

(a) A = 0.700m

(b) k = 80.6N/m

(c) x = -0.699m

(d) x = -0.350m

(e) t = 0.168s

Explanation:

Given the equation of motion for the spring

X = 0.700cos(12.0t), m = 0.56kg

(a) A = amplitude = 0.700m

(b) The angular velocity ω = 12rad/s

ω = √(k/m)

ω² = k/m

k = m×ω² = 0.56×12² = 80.6N/m

Spring constant k = 80.6N/m

(c) T = 2π/ω = 2π/12

T = 0.524s

At t = T/2 = 0.524/2 = 0.262s

So x = 0.700cos(12×0.262) = –0.699m

(d) At t = 2/3×T = 2×0.524/3 = 0. 349s

x = 0.700cos(12×0.349) = –0.350m

(e) to find t at x = -0.300m

–0.300 = 0.700cos(12t)

–0.300/0.700 = cos(12t)

cos(12t) = –0.429

12t = cos-¹(-0.429)

12t = 2.01

t = 2.01/12

t = 0.168s

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The digestive system digests and makes nutrients out of food while the circulatory system distributes and circulates the nutrients i believe?

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If a body of mass 10 kg is pushed with a force of 15N, then the acceleration of the body is

sladkih [1.3K]

Answer:

1.5m/s2

Explanation:

F=ma

a=F/m

a=15/10

a=1.5m/s2

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3 years ago