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3 years ago

Buhrs atomic model differed from ruthofords because it explained that

2 answers:

6 0

Buhrs atomic model differed from ruthofords because it explained that electrons exist in specified energy levels surrounding the nucleus. This means that, Ruthoford believed that electrons can't do very much. However, Buhrs' model showed that electrons are much more powerful than anyone else believes they can be.

inessss [21]3 years ago

5 0

Answer:

Explanation:

Rutherford's model shows most of an atom mass is in the middle with a positive charge while the negatively charged electrons are outside in a cloud. Bohr improved on his model and explained that the electrons move in certain orbits around the central mass (called the nucleus.)

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Two identical mandolin strings under 200 N of tension are sounding tones with frequencies of 590 Hz. The peg of one string slips

Slav-nsk [51]

To solve this problem it is necessary to apply the concepts related to frequency and vibration of strings. Mathematically the frequency can be expressed as

$f = \frac{v}{\lambda}$

Then the relation between two different frequencies with same wavelength would be

$\frac{f'}{f} = \frac{v'/\wavelength}{v/\wavelength}$

$\frac{f'}{f} = \frac{v'}{v}$

The beat frequency heard when the two strings are sounded simultaneously is

$f_{beat} = f-f'$

$f_{beat} = f(1-\frac{f'}{f})$

$f_{beat} = f(1-\frac{v'}{v})$

We have the velocity of the transverse waves in stretched string as

$v = \sqrt{\frac{T}{\mu}}$

$v = \sqrt{\frac{200N}{\mu}}$

And,

$v' = \sqrt{\frac{196N}{\mu}}$

Therefore the relation between the two is,

$\frac{v'}{v} = \sqrt{\frac{192}{200}}$

$\frac{v'}{v} = \sqrt{0.96}$

Finally substituting this value at the frequency beat equation we have

$f_{beat} = 590(1-\sqrt{0-96})$

$f_{beat} = 11.92Hz$

Therefore the beats per second are 11.92Hz

4 0

2 years ago

What percentage of the original kinetic energy is convertible to internal energy?

schepotkina [342]

There would be very less percentage loss of the kinetic energy during the conversion to internal energy, assuming that there is less air in the surroundings. Also, the friction will contribute to the conversion where if it is, the percentage loses is negligible.

7 0

3 years ago

A car drives 215 km east and then 45 km north. What is the magnitude of the car’s displacement? Round your answer to the nearest

34kurt

Displacement = (straight-line distance between the start point and end point) .

Since the road east is perpendicular to the road north,
the car drove two legs of a right triangle, and the magnitude
of its final displacement is the hypotenuse of the triangle.

Length of the hypotenuse = √ (215² + 45²)

= √ (46,225 + 2,025)

= √ 48,250

= 219.7 miles .

8 0

3 years ago

Read 2 more answers

Suppose that you have been chosen for a space mission to a distant planet. Due to the length of time you'll be away from Earth y

Contact [7]

Answer:

I should be active for 15 hours to meet the physical activity requirement.

Explanation:

Since time dilates in moving objects, we use the formula t = t₀/√(1 - β²) where t = time in space vehicle, t₀ = time on earth = 9 hours and β = v/c where v = speed of space vehicle = 0.8c.

So, t = t₀/√(1 - β²)

t = 9/√(1 - (v/c)²)

= 9/√(1 - (0.8c/c)²)

= 9/√(1 - (0.8)²)

= 9/√(1 - (0.64)

= 9/√0.36

= 9/0.6

= 15 hr

So, according to a timer on the space vehicle, I should be active for 15 hours to meet the physical activity requirement.

8 0

3 years ago

What is the momentum of a 45-kg quarterback moving eastward at 15 m/s?

puteri [66]

Answer:

Given

mass (m) =45kg

velocity (v) =15m/s

momentum (p) =?

Form

p=mv

=45x 15

p=675kg.m/s

the momentum =675kg.m/s

4 0

3 years ago