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3 years ago

What did bohr contribute to modern atomic theory

Physics

2 answers:

ddd [48]3 years ago

5 0

Bohr’s model showed that electron orbits had distinct radii.

Kobotan [32]3 years ago

5 0

Answer:

bohr developed a model of the atom that predicts how it will behave

Explanation:

a p e x :)

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Light in vacuum is incident on the surface of a slab of transparent material. In the vacuum the beam makes an angle of 39.9° wit

wariber [46]

Answer:

n=2.053

Explanation:

We will use Snell's Law defined as:

$n_{1}*Sin\theta_{1}=n_{2}*Sin\theta_{2}$

Where n values are indexes of refraction and $\theta$ values are the angles in each medium. For vacuum, the index of refraction in n=1. With this we have enough information to state:

$1*Sin(39.9)=n_{2}*Sin(18.2)$

Solving for $n_{2}$ yields:

$n_{2}=\frac{Sin(39.9)}{Sin(18.2)}=2.053$

Remember to use degrees for trigonometric functions instead of radians!

6 0

3 years ago

Car À moves at a speed of 8m/s for 43 seconds. Car B moves at a speed of 7 m/s for 50 seconds. Which car traveled a longer dista

nadezda [96]

Distance = (speed) x (time)

Car A: Distance = (8 m/s) x (43 s) = 344 meters

Car B: Distance = (7 m/s) x (50 s) = 350 meters

350 meters is a longer distance than 344 meters.

<em>Car-B traveled a longer distance</em> than Car-A did.

5 0

2 years ago

Read 2 more answers

The gravitational force between two objects is 100 N.

defon

200N is the answer (at least thats what I think)

5 0

3 years ago

A 5.20 kg chunk of ice is sliding at 13.5 m/s on the floor of an ice-covered valley when it collides with and sticks to another

Stella [2.4K]

Answer:

The chunk went as high as

2.32m above the valley floor

Explanation:

This type of collision between both ice is an example of inelastic collision, kinetic energy is conserved after the ice stuck together.

Applying the principle of energy conservation for the two ice we have based on the scenery

Momentum before impact = momentum after impact

M1U1+M2U2=(M1+M2)V

Given data

Mass of ice 1 M1= 5.20kg

Mass of ice 2 M2= 5.20kg

velocity of ice 1 before impact U1= 13.5 m/s

velocity of ice 2 before impact U2= 0m/s

Velocity of both ice after impact V=?

Inputting our data into the energy conservation formula to solve for V

5.2*13.5+5.2*0=(10.4)V

70.2+0=10.4V

V=70.2/10.4

V=6.75m/s

Therefore the common velocity of both ice is 6.75m/s

Now after impact the chunk slide up a hill to solve for the height it climbs

Let us use the equation of motion

v²=u²-2gh

The negative sign indicates that the chunk moved against gravity

And assuming g=9.81m/s

Initial velocity of the chunk u=0m/s

Substituting we have

6.75²= 0²-2*9.81*h

45.56=19.62h

h=45.56/19.62

h=2.32m

5 0

3 years ago

MIT’s robot cheetah can jump over obstacles 46. cm high and has speed of 12.0 km/h. a) If the robot launches itself at an angle

labwork [276]

Answer:

(a) $y_{max}=0.423m$

(b) $\alpha =64.3^{o}$

Explanation:

Given data

$v_{i}=12km/h=3.33m/s\\\alpha =60^{o}\\g=9.8m/s^{2}\\Required\\(a)y_{max}\\(b)Angle$

Solution

For Part (a)

As the velocity component in direction of y is given by:

$v_{yi}=v_{i}Sin\alpha \\v_{yi}=3.33Sin60\\v_{yi}=2.88m/s$

The maximum displacement is given by:

$v_{yf}^{2}=v_{yi}^{2}-2gy_{max}\\ y_{max}=\frac{(2.88)^{2}}{2(9.8)}\\ y_{max}=0.423m$

For Part (b)

To reach y=46cm =0.46m apply:

$0=v_{yi}^{2}-2(9.8)(0.46)\\v_{yi}=3m/s\\As\\Sin\alpha =\frac{v_{yi}}{v_{i}}\\\alpha =Sin^{-1}(\frac{v_{yi}}{v_{i}})\\\alpha =Sin^{-1}(\frac{3}{3.33} )\\\alpha =64.3^{o}$

5 0

3 years ago

Read 2 more answers