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SIZIF [17.4K]
2 years ago
10

How does the today's model of the atom DIFFER from the Rutherford’s model? 25 Points

Physics
2 answers:
Yuliya22 [10]2 years ago
8 0

Answer:

The Rutherford Model shows an atom with electrons orbiting a fixed, positively charged nucleus in set, predictable paths. The Bohr model shows electrons travel in defined circular orbits around the nucleus.

Explanation:

Goryan [66]2 years ago
5 0
Maybe it differs from a atoms?
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A proton, mass 1.67 × 10−27 kg and charge +1.6 × 10−19 c, moves in a circular orbit perpendicular to a uniform magnetic field of
stellarik [79]

Time taken by proton to complete one complete circular orbit= 7.28 x 10⁻⁸ s

Explanation:

For proton, the centripetal force required for circular motion is provided by the magnetic force,

so Fm= Fc

q v B = m v²/r

m= mass of charged particle

v= velocity

B =magnetic field

q= charge

r= radius of circular path

v= q B r/m

now v= r ω

ω= angular velocity

ω r = q B r /m

ω=q B /m

now ω= 2π/T where T =time period

so 2π/T=q B/m

T= 2 πm/q B

T= 2π (1.67 x 10⁻²⁷)/ [( 1.6 x 10⁻¹⁹)* (0.9)]

T= 7.28 x 10⁻⁸ s

6 0
3 years ago
Why does a rolling sphere slows down?​
Iteru [2.4K]
There’s frictional force acting on the sphere, which causes it to gradually slow down, and eventually come to a stop.
4 0
3 years ago
Now she is out on a hike and comes to the left bank of a river. There is no bridge and the right bank is 10.0 mm away horizontal
Harrizon [31]

Answer:

a. 8.96 m/s b. 1.81 m

Explanation:

Here is the complete question.

a) A long jumper leaves the ground at 45° above the horizontal and lands 8.2 m away.

What is her "takeoff" speed  v 0 ?

b) Now she is out on a hike and comes to the left bank of a river. There is no bridge and the right bank is 10.0 m away horizontally and 2.5 m, vertically below.  

If she long jumps from the edge of the left bank at 45° with the speed calculated in part a), how long, or short, of the opposite bank will she land?

a. Since she lands 8.2 m away and leaves at an angle of 45 above the horizontal, this is a case of projectile motion. We calculate the takeoff speed v₀ from R = v₀²sin2θ/g. where R = range = 8.2 m.

So, v₀ = √gR/sin2θ = √9.8 × 8.2/sin(2×45) = √80.36/sin90 = √80.36 = 8.96 m/s.

b. We use R = v₀²sin2θ/g to calculate how long or short of the opposite bank she will land. With v₀ = 8.96 m/s and θ = 45

R = 8.96²sin(2 × 45)/9.8 = 80.2816/9.8 = 8.192 m.

So she land 8.192 m away from her bank. The distance away from the opposite bank she lands is 10 - 8.192 m = 1.808 m ≅ 1.81 m

8 0
3 years ago
Why a body in uniform velocity is zero acceleration ?​
seropon [69]
Uniform velocity means no Net force and therefore no acceleration. Acceleration only happens when the velocity changes.
4 0
3 years ago
Consider the sections of two circuits illustrated above. Select True or False for all statements. After connecting c and d to a
Basile [38]

Answer:

Follows are the solution to the given question:

Explanation:

In this question the missing file of the circuit is not be which is defined in attached file please find it.

In Option 1, this statement is true because the current is on R_3 \  and \ R_4, that is the same.

In option 2, this statement is false because Rcd=R_3+R_4 therefore it implies that Rcd is always larger then R_3.

In option 3,  this statement is true because the voltage of R_1 \ and \ R_2 is always equal.

In option 4, this statement is true because Rab= \frac{1}{(\frac{1}{R1}+\frac{1}{R2})}=\frac{R1R2}{(R1+R2)}. \frac{R2}{(R1+R2)}is always smaller then 1 therefore,R_1 \times (\frac{R2}{(R1+R2)}) is always equal to R1.

5 0
2 years ago
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