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3 years ago

What are the three subatomic particles that make up an atom

Physics

1 answer:

Harrizon [31]3 years ago

5 0

Answer:

Proton, neutron, electron

Explanation:

The atom consists of a nucleus, where almost all the mass is concentrated, and electrons orbiting around the nucleus.

The nucleus consists of two types of particles:

- Proton: it has a mass of $1.6726\cdot 10^{-27} kg$ , and a positive electric charge of +e ( $1.6\cdot 10^{-19}C$ )

- Neutron: it has a mass of $1.6749 \cdot 10^{-27}kg$ , and it has no electric charge

The third particle that makes an atom is the electron, that orbit around the nucleus:

- Electron: it has a mass of $9.1094\cdot 10^{-31}kg$ , and it has a negative electric charge of -e ( $-1.6\cdot 10^{-19}C$ )

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An oil film with refractive index 1.48 and thickness 290 nm is floating on water and illuminated with white light at normal inci

VikaD [51]

Answer:

572.3 nm

Explanation:

$n_{oil}$ = refractive index of the oil film = 1.48

$t_{oil}$ = thickness of the oil film = 290 nm

$\lambda$ = wavelength of the dominant color

m = order

Using the equation

$2 n_{oil} t_{oil} = (m + 0.5) \lambda$

For m = 0

$2 (1.48) (290) = (0 + 0.5) \lambda$

$\lambda$ = 1716.8 nm

For m = 1

$2 (1.48) (290) = (1 + 0.5) \lambda$

$\lambda$ = 572.3 nm

For m = 2

$2 (1.48) (290) = (2 + 0.5) \lambda$

$\lambda$ = 343.4 nm

Hence the dominant color wavelength is 572.3 nm

8 0

3 years ago

Can a goalkeeper at his goal kick a soccer ball into the opponent’s goal without the ball touching the ground? The distance will

zmey [24]

The goalkeeper at his goal cannot kick a soccer ball into the opponent’s goal without the ball touching the ground

Explanation:

Consider the vertical motion of ball,

We have equation of motion v = u + at

Initial velocity, u = u sin θ

Final velocity, v = 0 m/s

Acceleration = -g

Substituting

v = u + at

0 = u sin θ - g t

$t=\frac{usin\theta }{g}$

This is the time of flight.

Consider the horizontal motion of ball,

Initial velocity, u = u cos θ

Acceleration, a =0 m/s²

Time, $t=\frac{usin\theta }{g}$

Substituting

s = ut + 0.5 at²

$s=ucos\theta \times \frac{usin\theta }{g}+0.5\times 0\times (\frac{usin\theta }{g})^2\\\\s=\frac{u^2sin\theta cos\theta}{g}\\\\s=\frac{u^2sin2\theta}{2g}$

This is the range.

In this problem

u = 30 m/s

g = 9.81 m/s²

θ = 45° - For maximum range

Substituting

$s=\frac{30^2\times sin(2\times 45)}{2\times 9.81}=45.87m$

Maximum horizontal distance traveled by ball without touching ground is 45.87 m, which is less than 95 m.

So the goalkeeper at his goal cannot kick a soccer ball into the opponent’s goal without the ball touching the ground

6 0

3 years ago

an object traveling 200 feet per second slows to 50 feet per second in 5 seconds. Calculate the acceleration of the object

emmainna [20.7K]

Answer:

The acceleration of the object is $-30\ m/s^2$

Explanation:

Given:

Initial velocity of object $v_i$ = 200 feet/second

Final velocity of object $v_f$ = 50 feet/second

Time of travel = 5 seconds

To calculate acceleration of the object we will find the rate of change of velocity with respect to time.

So, acceleration $a$ is given by:

$a=\frac{v_f-v_i}{t}$

where $v_f$ represents final velocity, $v_i$ represents initial velocity and $t$ is time of travel.

Plugging in values to evaluate acceleration.

$a=\frac{50-200}{5}$

$a=\frac{-150}{5}$

$a=-30\ m/s^2$

The acceleration of the object is $-30\ m/s^2$ (Answer). The negative sign shows the object is slowing down.

4 0

3 years ago

At the same moment, one rock is dropped and one is theown downwand with an iniial velocily of 29 us frm op of a building that is

Helen [10]

Answer:

The thrown rock will strike the ground $2.42s$ earlier than the dropped rock.

Explanation:

Known Data

$y_{i}=300m$

$y_{f}=0m$
$v_{iD}=0m/s$
$v_{iT}=-29m/s$ , it is negative as is directed downward

Time of the dropped Rock

We can use $y_{f}=y_{i}+v_{iy}t-\frac{gt^{2}}{2}$ , to find the total time of fall, so $0=300m-\frac{(9.8m/s^{2})t_{D}^{2}}{2}$ , then clearing for $t_{D}$ .

$t_{D}=\sqrt[2]{\frac{300m}{4.9m/s^{2}}} =\sqrt[2]{61.22s^{2}} =7.82s$

Time of the Thrown Rock

We can use $y_{f}=y_{i}+v_{iy}t-\frac{gt^{2}}{2}$ , to find the total time of fall, so $0=300-29t_{T}-\frac{(9.8)t_{T}^{2}}{2}$ , then, $0=-4.9t_{T}^{2}-29t_{T}+300$ , as it is a second-grade polynomial, we find that its positive root is $t_{T}=5.4s$

Finally, we can find how much earlier does the thrown rock strike the ground, so $t_{E}=t_{D}-t_{T}=7.82s-5.4s=2.42s$

6 0

3 years ago

A bike rider pedals with constant acceleration to reach a velocity of 7.5 (vf)m/s over a time of 4.5 s(t). during the period of

BigorU [14]

The initial velocity of the bike was 1.67 (vf)m/s. This is found by evaluating 7.5/4.5 which yields the velocity per unit of time which is equivalent to initial velocity.

6 0

3 years ago