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Grace [21]
2 years ago
6

Four distinguishable particles move freely in a room divided into octants (there are no actual partitions). Let the basic states

be given by specifying the octant in which each particle is located. How many basic states are there?
Physics
1 answer:
mafiozo [28]2 years ago
8 0

Answer:

Explanation:

Since the door that leads to the room is opened, this gives room for particles to move into the next identical room and divided into octants. Now the amount of space that can be occupied becomes double, the number of basic states has increased by 404916

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A person is nearsighted with a far point of 75.0 cm. a. What focal length contact lens is needed to give him normal vision
BigorU [14]

Complete Question

The  complete question is  shown on the first uploaded image  

Answer:

a

  f=  -75 \ cm =  - 0.75 \ m

b

  P  =  -1.33 \ diopters

Explanation:

From the question we are told that

    The  image distance is  d_i =  -75 cm

The value of the image is negative because it is on the same side with the corrective glasses

    The  object distance is  d_o =  \infty

The  reason object distance  is because the object father than it being picture by the eye

General focal length is mathematically represented as

              \frac{1}{f}  =  \frac{1}{d_i}  -   \frac{1}{d_o}

substituting values

             \frac{1}{f}  =  \frac{1}{-75}  -   \frac{1}{\infty}

=>         f=  -75 \ cm =  - 0.75 \ m

Generally the power of the corrective lens is  mathematically represented as

        P  =  \frac{1}{f}

substituting values

       P  =  \frac{1}{-0.75}

        P  =  -1.33 \ diopters

7 0
2 years ago
A hockey stick strikes a hockey puck of mass 0.17 kg. If the force exterted on the hockey puck is 35.0 N and there is a force of
GREYUIT [131]

Answer:

a=190\ m/s^2

Explanation:

Mass of a hockey puck, m = 0.17 kg

Force exerted by the hockey puck, F' = 35 N

The force of friction, f = 2.7 N

We need to find the acceleration of the hockey puck.

Net force, F=F'-f

F=35-2.7

F=32.3 N

Now, using second law of motion,

F = ma

a is the acceleration of the hockey puck

a=\dfrac{F}{m}\\\\a=\dfrac{32.3}{0.17}\\\\a=190\ m/s^2

So, the acceleration of the hockey puck is 190\ m/s^2.

5 0
2 years ago
Plug variables expressed in SI units in the kinematic equation given in article: a = -v0^2/(2sg). What value of g you get as exp
Elanso [62]

Answer:

1) acceleration is increased by a factor of four 4X

2) the acceleration increases a factor of 2X

3) the correct answer of 400g

Explanation:

This is a kinematics exercise, where you use the velocity equation to obtain the acceleration, with the final velocity equal to zero.

           v² = v₀² + 2 a x

           0 = v₀² + 2 a x

           a = - v₀² / 2 x

           

In the case of wanting to give the acceleration as a function of g, we can find the relationship between the two quantities

         a / g = - v₀² / (2 x g)

Let's answer the different questions about this equation

1. The initial velocity is doubled, how much the acceleration is worth

           

       a/g = - (2v₀) 2 / 2xg

       a = 4 (-v₀² / 2xg) g

acceleration is increased by a factor of four 4X

2. if the stopping distance is reduced by 2, that is, x = x₀ / 2

we substitute

        a/g = (- v₀² / 2g) 2/x

         

        a =2  (-v₀² / 2x₀g)  g

       

therefore the acceleration increases a factor of 2X

3. the initial velocity of the hockey player is v₀ = 20 m / s and the stopping distance is

x = 5cm = 0.05m

we calculate the acceleration

        a / g = - 20² / (2 0.05)

        a / g = - 4000 / g

        a / g = - 4000 / 9.8 = 408

        a = 408 g

the correct answer of 400g, the value matches exactly if g = 10 m / s2 is taken

6 0
2 years ago
A 1 200-kg car traveling initially at vCi 5 25.0 m/s in an easterly direction crashes into the back of a 9 000-kg truck moving i
sukhopar [10]

Answer:

The velocity of the truck after the collision is 20.93 m/s

Explanation:

It is given that,

Mass of car, m₁ = 1200 kg

Initial velocity of the car, v_{Ci}=25\ m/s

Mass of truck, m₂ = 9000 kg

Initial velocity of the truck, v_{Ti}=20\ m/s

After the collision, velocity of the car, v_{Cf}=18\ m/s

Let v is the velocity of the truck immediately after the collision. The momentum of the system remains conversed.

initial\ momentum=final\ momentum

1200\ kg\times 25\ m/s+9000\ kg\times 20\ m/s=1200\ kg\times 18+9000\ kg\times v

210000-21600=9000\ kg\times v

v=20.93\ m/s

So, the velocity of the truck after the collision is 20.93 m/s. Hence, this is the required solution.

8 0
2 years ago
Juan amd kym have four samples of matter. They are observing and describing the properties of these samples. Which property will
TiliK225 [7]

Answer: The property that will best provide evidence that the samples are solid includes:

--> if the substance has a definite shape,

-->if the substance has a definite volume

--> if it's tightly packed.

Explanation:

According to the kinetic theory of matter, every substance consist of very large number of very small particles called molecules. These molecules, which are made up of atoms that are the smallest particles of a substance that can exist in a free state.

Matter can exist in the following states:

--> Solid state

--> liquid state or

--> Gaseous state.

The general property of a substance that is in gaseous state includes:

--> Definite shape: A substance can be grouped as a solid if it's shape is fixed that is, it doesn't depend on the shape of other materials.

--> Definite volume: A substance can be grouped as a solid if it occupies its own shape. This is due to the force of cohesion among its molecules.

--> Tightly packed: A substance can be grouped as solid if the molecular movements of the particles are negligible.

From the samples under observation by Juan and kym, if the sample that possesses the above described qualities, it is a solid rather than liquid or gas.

4 0
2 years ago
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