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3 years ago

13

Models are particularly useful in relativity and quantum mechanics, where conditions are outside those normally encountered by h

umans. What is a model?

1 answer:

mylen [45]3 years ago

4 0

Answer:

A model is defined as a structure used to represent an object, usually of a different scale.

Explanation:

In quantum mechanics and particle physics, many of the particles are subatomic, meaning that they are smaller than atoms. This is where a model would be useful. A model could help people to visualise what the particle looks like, and in general would make it easier to understand the behaviour of such a particle.

You might be interested in

A 77.0 kg woman slides down a

statuscvo [17]

Assuming the woman starts at rest, she descends the slide with acceleration <em>a</em> such that

(20.3 m/s)² = 2 <em>a</em> (42.6 m) → <em>a</em> ≈ 4.84 m/s²

which points parallel to the slide.

The only forces acting on her, parallel to the slide, are

• the parallel component of her weight, <em>w</em> (//)

• friction, <em>f</em>, opposing her descent and pointing up the slide

Take the downward sliding direction to be positive. By Newton's second law, the net force in the parallel direction acting on the woman is

∑ <em>F</em> (//) = <em>w </em>(//) - <em>f</em> = <em>ma</em>

where <em>m</em> = 77.0 kg is the woman's mass.

Solve for <em>f</em> :

<em>mg</em> sin(42.3°) - <em>f</em> = <em>ma</em>

<em>f</em> = <em>m</em> (<em>g</em> sin(42.3°) - <em>a</em>)

<em>f</em> = (77.0 kg) ((9.80 m/s²) sin(42.3°) - 4.84 m/s²) ≈ 135 N

Compute the work <em>W</em> done by friction: multiply the magnitude of the friction by the length of the slide.

<em>W</em> = (135 N) (42.6 m) ≈ 5770 N•m = 5770 J

3 0

3 years ago

Which would ultimately cause the armature in an electric motor to spin?

malfutka [58]

It would be the electromagnet because that is how it spins

4 0

4 years ago

Read 2 more answers

Imagine you derive the following expression by analyzing the physics of a particular system: v2=v20+2ax. The problem requires so

lisabon 2012 [21]

As per kinematics equation we are given that

$v^2 = v_o^2 + 2ax$

now we are given that

a = 2.55 m/s^2

$v_0 = 21.8 m/s$

$v = 0$

now we need to find x

from above equation we have

$0^2 = 21.8^2 + 2(2.55)x$

$0 = 475.24 + 5.1 x$

$x = 93.2 m$

so it will cover a distance of 93.2 m

7 0

3 years ago

Read 2 more answers

19) A child on a sled starts from rest at the top of a 15.0° slope. If the trip to the bottom takes 15.2 s,

sveticcg [70]

Answer: 288.8 m

Explanation:

We have the following data:

$t=15.2 s$ is the time it takes to the child to reach the bottom of the slope

$V_{o}=0$ is the initial velocity (the child started from rest)

$\theta=15\°$ is the angle of the slope

$d$ is the length of the slope

Now, the Force exerted on the sled along the ramp is:

$F=ma$ (1)

Where $m$ is the mass of the sled and $a$ its acceleration

In addition, if we draw a free body diagram of this sled, the force along the ramp will be:

$F=mg sin \theta$ (2)

Where $g=9.8 m/s^{2}$ is the acceleration due gravity

Then:

$ma=mg sin \theta$ (3)

Finding $a$ :

$a=g sin \theta$ (4)

$a=9.8 m/s^{2} sin(15\°)$ (5)

$a=2.5 m/s^{2}$ (6)

Now, we will use the following kinematic equations to find $d$ :

$V=V_{o}+at$ (7)

$V^{2}=V_{o}^{2}+2ad$ (8)

Where $V$ is the final velocity

Finding $V$ from (7):

$V=at=(2.5 m/s^{2})(15.2 s)$ (9)

$V=38 m/s$ (10)

Substituting (10) in (8):

$(38 m/s)^{2}=2(2.5 m/s^{2})d$ (11)

Finding $d$ :

$d=288.8 m$

6 0

3 years ago

Mercury has a mass of 3.3 e 23 kg and a radius of 2.44 e 6 m. Find Mercury's

valentina_108 [34]

Answer:

11.) g = 3.695 m/s^2

12.) g = 8.879 m/s^2

13.) E = 8127 N/C

Explanation:

11.) Given that the

Mercury mass M = 3.3 × 10^23kg

Radius r = 2.44 ×10^6 m

Gravitational constant G = 6.67408 × 10^-11 m3kg-1 s^-2

Gravitational field strength g can be calculated by using the formula below

g = GM/r^2

Substitutes all the parameters into the formula

g = (6.67408 × 10^-11 × 3.3 × 10^23)/(2.44×10^6)^2

g = 2.2×10^13/5.954×10^12

g = 3.695 m/s^2

12.) Given that the

Venus mass M = 4.87×10^24kg

Radius r = 6.05 × 10^6 m

Using the same formula for gravitational field strength g

g = GM/R2

Substitute all the parameters into the formula

g = (6.67408 × 10^-11 × 4.87×10^24)/(6.05×10^6)^2

g = 3.25×10^14/3.66×10^13

g = 8.879 m/s^2

13.) Given that the

Charge = 2.26 nC = 2.26×10^-9

Distance d = 0.05m

Electric field strength E can be calculated by using the formula below

E = Kq/d^2

Where

K = electrostatic constant 8.99 × 10^9 Nm2/C2

Substitutes all the parameters into the formula

E = (8.99 × 10^9 × 2.26×10^-9)/0.05^2

E = 20.3174/2.5×10^-3

E = 8126.96 N/C

7 0

4 years ago