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4 years ago

An alpha particle is identical to a(n) _____.

Physics

2 answers:

Ber [7]4 years ago

7 0

Answer:

Helium nucleus

Explanation:

Alpha particles have two protons and two neutrons, wich is the configuration of an Helium Nucleus.

The atomic number of an element it is the number of protons that the element has in its nucleus, this number is uniqe to every element. Helium is the second element in the periodic table so Its atomic number is 2, This means it has two protons and the same amount of neutrons, wich makes alpha particles identical to a helium nucleus, not to a Helium atom, because a Helium atom has electrons and alpha particles don't have electrons.

Alexxx [7]4 years ago

3 0

Helium atom, in other words, it consistis of a particle having four protons and two neutrons.

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A 5 kg block is released from rest at the top of a quarter- circle type curved frictionless surface. The radius of the curvature

malfutka [58]

Answer:

a. 186.2 J b. 8.63 m/s c. 190 m d. 43.2 s e. 186.2 J

Explanation:

a. From conservation of energy, the potential energy loss of block = kinetic energy gain of the block.

So, U + K = U' + K' where U = initial potential energy of block = mgh, K = initial kinetic energy of block = 0, U' = final potential energy of block at bottom of curve = 0 and K' = final kinetic energy of block at bottom of curve.

So, mgh + 0 = 0 + K'

K' = mgh where m = mass of block = 5 kg, g = acceleration due to gravity = 9.8 m/s², h = initial height above the ground of block = radius of curve = 3.8 m

So, K' = 5 kg × 9.8 m/s² × 3.8 m = 186.2 J

b. Since the kinetic energy of the block K = 1/2mv² where m = mass of block = 5 kg, v = velocity of block at bottom of curve

So, v = √(2K/m)

= √(2 × 186.2 J/5 kg)

= √(372.4 J/5 kg)

= √(74.48 J/kg)

= 8.63 m/s

c. To find the stopping distance, from work-kinetic energy principles,

work done by friction = kinetic energy change of block.

So ΔK = -fd where ΔK = K" - K' where K" = final kinetic energy = 0 J (since the block stops)and K' = initial kinetic energy = 186.2 J, f = frictional force = μmg where μ = coefficient of kinetic friction = 0.02, m = mass of block = 5 kg, g = acceleration due to gravity = 9.8 m/s² and d = stopping distance

ΔK = -fd

K" - K' = - μmgd

d = -(K" - K')/μmg

Substituting the values of the variables, we have

d = -(0 J - 186.2 J)/(0.02 × 5 kg × 9.8 m/s²)

d = -(- 186.2 J)/(0.98 kg m/s²)

d = 190 m

d. Using v² = u² + 2ad where u =initial speed of block = 8.63 m/s, v = final speed of block = 0 m/s (since it stops), a = acceleration of block and d = stopping distance = 190 m

So, a = (v² - u²)/2d

substituting the values of the variables, we have

a = (0² - (8.63 m/s)²)/(2 × 190 m)

a = -74.4769 m²/s²/380 m

a = -0.2 m/s²

Using v = u + at, we find the time t that elapsed while the block is moving on the horizontal track.

t = (v - u)/a

t =(0 m/s - 8.63 m/s)/-0.2 m/s²

t = - 8.63 m/s/-0.2 m/s²

t = 43.2 s

e. The work done by friction W = fd where

= μmgd where f = frictional force = μmg where μ = coefficient of kinetic friction = 0.02, m = mass of block = 5 kg, g = acceleration due to gravity = 9.8 m/s² and d = stopping distance = 190 m

W = 0.02 × 5 kg × 9.8 m/s² × 190 m

W = 186.2 J

5 0

3 years ago

A 5kg bucket of water is raised from a well with a rope. If the upward acceleration of the bucket is 3m/s/s, find the force exer

mojhsa [17]

Apply Newton's second law to the bucket's vertical motion:

F = ma

F = net force, m = mass of the bucket, a = acceleration of the bucket

Let us choose upward force to be positive and downward force to be negative. The net force F is the difference of the tension in the rope lifting the bucket and the weight of the bucket, i.e.:

F = T - W

F = net force, T = tension, W = weight

The weight of the bucket is given by:

W = mg

W = weight, m = mass, g = gravitational acceleration

Make some substitutions:

F = T - mg

T - mg = ma

Isolate T:

T = ma + mg

T = m(a+g)

Given values:

m = 5kg, a = 3m/s², g = 9.81m/s²

Plug in and solve for T:

T = 5(3+9.81)

T = 64.05N

3 0

3 years ago

What happens to water when it changes to ice?

mamaluj [8]

When water changes to ice, its density decreases and its volume increases.

7 0

3 years ago

When you observe the world and see something that you don't understand, you make a hypothesis that you try to prove. This either

Whitepunk [10]

Just like the water cycle, rocks undergo changes of form in a rock cycle. A metamorphic rock can become an igneous rock, or a sedimentary rock can become a metamorphic one. Unlike the water cycle, you can’t see the process happening on a day-to-day basis. Rocks change very slowly under normal conditions, but sometimes catastrophic events like a volcanic eruption or a flood can speed up the process. So what are the three types of rocks, and how do they change into each other? Keep reading to find out!

5 0

4 years ago

9. A statue is to be scaled down isomorphically (It will have its size changed without changing its shape). It starts with an in

Fynjy0 [20]

Answer:

m = 1.45 kg

Explanation:

For this exercise we look for size reduction in height

Reduction = y / y₀

Reduction = 2.15 / 6.75

Reduction = 0.3185

As the statue should not be deformed, all reduction has the same factor.

Let's use the concept of density

ρ = m / V

Initial statue

ρ = m₀ / V₀

It is reduced

V = x y z

V = 0.3185 x₀ 0.3185 y₀ 0.3185 z₀

V = 0.3185³ V₀

Density is

ρ = m / V

ρ = m / 0.3185³ V₀

As the density remains constant we can match them

m₀ / Vo = m / 0.3185³ V₀

m = 0.3185³ m₀

Let's calculate

m = 0.3185³ 45

m = 0.03231 45

m = 1.45 kg

5 0

3 years ago