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Y_Kistochka [10]
3 years ago
9

How did the atomic model changed from daltons model in 1803 to schrodingers model of 1926? In 3 to 4 sentences, describe the ch

anges and explain whether the discoveries were independent or cited previous discoveries.
Physics
1 answer:
sasho [114]3 years ago
5 0

Explanation:

Niels Bohr improved Rutherford's model. Using mathematical ideas, he showed that electrons occupy shells or energy levels around the nucleus. The Dalton model has changed over time because of the discovery of subatomic particles .

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The rates of obesity have more than doubled for children between the ages of 6 and 11 since the middle of the twentieth century.
Illusion [34]

Answer:

True

Explanation:

Obesity is a growing concern in children as rates skyrocket to an all time high

8 0
3 years ago
A ball is launched at an angle of 10 degrees from a 20 meter tall building with a speed of 4 m/s. How long is the ball in the ai
shutvik [7]

Answer:

d) None of the above

Explanation:

v_{o} = inituial velocity of launch = 4 m/s

θ = angle of launch = 10 deg

Consider the motion along the vertical direction

v_{oy} = initial velocity along vertical direction = 4 Sin10 = 0.695

m/s

a_{y} = acceleration along the vertical direction = - 9.8 m/s²

y = vertical displacement = - 20 m  

t = time of travel

using the equation

y=v_{oy} t+(0.5)a_{y} t^{2}

- 20 = (0.695) t + (0.5) (- 9.8) t²

t = 2.1 sec

consider the motion along the horizontal direction

x = horizontal displacement

v_{ox} = initial velocity along horizontal direction = 4 Cos10 = 3.94 m/s

a_{x} = acceleration along the horizontal direction = 0 m/s²

t = time of travel =  2.1 s

Using the kinematics equation

x =v_{ox} t+(0.5)a_{x} t^{2}

x = (3.94) (2.1) + (0.5) (0) (2.1)²

x = 8.3 m

Consider the motion along the vertical direction

v_{oy} = initial velocity along vertical direction = 4 Sin10 = 0.695

m/s

a_{y} = acceleration along the vertical direction = - 9.8 m/s²

y_{o} =initial vertical position at the time of launch = 20 m  

y = vertical position at the maximum height = 20 m

v_{fy} = final velocity along vertical direction at highest point = 0 m/s

using the equation

{v_{fy}}^{2}= {v_{oy}}^{2} + 2 a_{y}(y - y_{o})

0^{2}= 0.695^{2} + 2 (- 9.8)(y - 20)

y = 20.02 m

h = height above the starting height

h = y - y_{o}

h = 20.02 - 20

h = 0.02 m

7 0
3 years ago
suppose a ball is thrown vertically upward. Eight seconds later it returns to its point of release. What is the initial velocity
valentinak56 [21]
The ball took half of the total time ... 4 seconds ... to reach its highest
point, where it began to fall back down to the point of release.

At its highest point, its velocity changed from upward to downward. 
At that instant, its velocity was zero.

The acceleration of gravity is 9.8 m/s².  That means that an object that's
acted on only by gravity gains 9.8 m/s of downward speed every second. 

-- If the object is falling downward, it moves 9.8 m/s faster every second.

-- If the object is tossed upward, it moves 9.8 m/s slower every second.

The ball took 4 seconds to lose all of its upward speed.  So it must have
been thrown upward at  (4 x 9.8 m/s)  =  39.2 m/s .

(That's about  87.7 mph straight up.  Somebody had an amazing pitching arm.)
6 0
3 years ago
A race car travels 40 m/s around a banked (45° with the horizontal) circular (radius = 0.20 km) track. What is the magnitude of
OLEGan [10]

Answer:

c)F_{net} = 0.640 kN

Explanation:

As we know that resultant force is the net force that is acting on the system

As per Newton's II law we know that net force is product of mass and acceleration

so we will have

F_{net} = ma

here we know

m = 80 kg

for circular motion acceleration is given as

a_c = \frac{v^2}{R}

a_c = \frac{40^2}{200} = 8 m/s^2

now we have

F_{net} = 80 \times 8

F_{net} = 640 N

F_{net} = 0.640 kN

7 0
3 years ago
Describe how the student should measure the time taken for the toy parachute to
PolarNik [594]

The time must be measured with respect to gravity. As it falls, it has free fall that is the force acting on it will be the gravity.With the distance in account, d = 1/2 gt²

t = √(2d/g)

4 0
3 years ago
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