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N76 [4]
3 years ago
15

A student creates a model by placing raisins in bread dough and allowing the dough to rise for several hours. The student sketch

es the model before and after the dough rises.
The student claims that the model accurately describes the motion of the galaxies in the universe. What evidence from the model supports the student’s claim?


As the dough rises, the size of the raisins increases, resembling the expansion of the galaxies.
As the dough rises, the size of the raisins increases, resembling the expansion of the galaxies.
As the dough rises, the size of the raisins increases, resembling the expansion of the galaxies.

As the dough rises, the distance between the raisins decreases, indicating that the galaxies will collide.
As the dough rises, the distance between the raisins decreases, indicating that the galaxies will collide.

As the dough rises, the size of the raisins decreases, resembling the emission of light from stars.
As the dough rises, the size of the raisins decreases, resembling the emission of light from stars.

As the dough rises, the distance between the raisins increases, indicating that the galaxies are moving apart.
Physics
1 answer:
lbvjy [14]3 years ago
3 0

As the dough rises, the distance between the raisins increases, indicating that the galaxies are moving apart. D

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What are the units, if any, of the particle in a box wavefunction. What does this mean?
SIZIF [17.4K]

Answer:

  • [\psi]= [Length^{-3/2}]
  • This means that the integral of the square modulus over the space is dimensionless.

Explanation:

We know that the square modulus of the wavefunction integrated over a volume gives us the probability of finding the particle in that volume. So the result of the integral

\int\limits^{x_f}_{x_0} \int\limits^{yf}_{y_0} \int\limits^{z_f}_{z_0} |\psi|^2 \, dz \,  dy \,  dx

must be dimensionless, as represents a probability.

As the differentials has units of length

[dx]=[dy]=[dz]=[Length]

for the integral to be dimensionless, the units of the square modulus of the wavefunction has to be:

[\psi]^2 = [Length^{-3}]

taking the square root this gives us :

[\psi] = [Length^{-3/2}]

5 0
3 years ago
In a local bar, a customer slides an empty beer mug down the counter for a refill. The height of the counter is 1.42 m. The mug
telo118 [61]

Answer:

a) V_{x}=3.72m/s, b) ∠=-54.83°

Explanation:

In order to solve this problem, we must start with a drawing of the situation, this will help us visualize the problem better. (See picture attached).

a)

Now, the idea is that the beer mug has a horizontal speed and no vertical speed at initial conditions. So knowing this, we can start finding the initial velocity of the mug.

In order to do so, we need to find the time it takes for the mug to reach the ground. We can find it by using the following equation:

y=y_{0}+V_{y0}t+\frac{1}{2}a_{y}t^{2}

We can see from the drawing that y and the initial velocity in y are zero, so we can simplify our formula:

0=y_{0}+\frac{1}{2}a_{y}t^{2}

so we can solve for t, so we get:

t=\sqrt{\frac{-(2)y_{0}}{a}}

so now we can substitute the known values, so we get:

t=\sqrt{\frac{-(2)(1.42)}{-9.8}}

which yields:

t=0.538s

So we can use this value to find the velocity in x:

V_{x}=\frac{x}{t}

When substituting we get:

V_{x}=\frac{2m}{0.538s}

which yields:

V_{x}=3.72m/s

b)

In order to solve part b, we need to find the y-component of the velocity, for which we can use the following formula:

\Delta y=\frac{V_{f}^{2}-V_{0}^{2}}{2a}

We know that V_{0} is zero, so we can simplify the expression:

\Delta y=\frac{V_{yf}^{2}}{2a}

So we can solve the equation for V_{yf}^{2} so we get:

V_{yf}=\sqrt{2\Delta y a}

and when substituting the known values we get:

V_{yf}=\sqrt{2(-1.42m)(-9.8m/s^{2})}

which yields:

V_{yf}=-5.28m/s

Once we got the final velocity in y, we can use it together with the velocity in x to find the angle.

So we can use the following formula:

tan \theta =\frac{V_{y}}{V_{x}}

when solving for theta we get:

\theta = tan^{-1}(\frac{V_{y}}{V_{x}})

We can substitute so we get:

\theta = tan^{-1}(\frac{-5.28m/s}{3.72m/s})

which yields:

\theta = -54.83^{o}

7 0
3 years ago
I didn't understand ​
julsineya [31]
Huh? The answer is A Shadow
4 0
2 years ago
What is the full form of work?​
Pepsi [2]

Answer: JOB

Explanation: They full forms of work are JOB which is JUST OBEY BOSS or Joining Other Bussiness .

7 0
2 years ago
A net force of Fnet acts on an object, causing the object to accelerate at a rat of 9m/s/s. what is the acceleration rate in a s
Readme [11.4K]

The new acceleration is c) 18m/s/s

Explanation:

Net force, mass and acceleration of an object are related by Newton's second law of motion:

F=ma

where

F is the net force on the object

m is its mass

a is its acceleration

In the first trial of this problem, a net force of F_{net} is applied to the object, causing an acceleration of

a=9 m/s^2

Calling the mass of the object 'm', this means that

F_{net} = ma = 9m [N] (1)

In the second trial, the force applied is 2F_{net}, so we have

2F_{net} = ma'

where a' is the new acceleration. Substituting (1) into the second equation, we find:

2(9m) = ma' \rightarrow a' = 18 m/s^2

This is because the acceleration is directly proportional to the force applied: therefore, if the force applied doubled, the acceleration doubles as well.

Learn more about Newton's second law:

brainly.com/question/3820012

#LearnwithBrainly

7 0
3 years ago
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