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aliina [53]
2 years ago
10

The structure of the atomOn the basis of Rutherford's experimental observations, which of the following statements predicts the

structure of the atom?1. In an atom, negatively charged electrons are small particles held within a positively charged sphere.2. In an atom, negatively charged electrons are dispersed in the space surrounding the positively charged nucleus of an atom.3. In an atom, all of the positive and negative charges are randomly distributed.4. In an atom, most of the mass and the positive charge are located in a small core within the atom called the nucleus.2. In an atom, negatively charged electrons are dispersed in the space surrounding the positively charged nucleus of an atom.4. In an atom, most of the mass and the positive charge are located in a small core within the atom called the nucleus.
Physics
1 answer:
irga5000 [103]2 years ago
7 0

In rutherford experiment he bombarded highly energetic alpha particles towards the gold foil

He observed that most of the alpha particles pass through the foil without any deviation while very few of the alpha particles are there which reflect back to its own path

while very less in number was there which were deflected from there path

so he concluded that the positive charge of atom is concentrated inside the nucleus and it is small space in tom and electrons are surrounded them

so correct answer would be

4. In an atom, most of the mass and the positive charge are located in a small core within the atom called the nucleus.

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if a runners power is 400 watts as she runs, how much chemical energy does she convert into other forms in 10 minutes
AnnZ [28]

Answer:

Energy converted = 240000\,Joules = 240\, kJoules

Explanation:

Recall that Power is the rate at which energy is transferred therefore defined by the mathematical formula: Power\,=\,\frac{Energy\,transferred}{time}

Since the information on the power of the runner is given, as well as the time the energy conversion takes place, we can then use this equation to find how much energy is been converted. Notice that we just need to change the given time *10 minutes) into the appropriate units  (seconds)to get the answer in SI units of energy (Joules). The conversion of 10 minutes into seconds is done by multiplying : 10 minutes * 60 seconds/minute = 600 seconds.

We use this then to find the energy converted by the runner:

Power\,=\,\frac{Energy\,transferred}{time}\\400 \,W = \frac{E}{600\,sec} \\400 \,W * 600\,sec=E\\E=240000\,Joules = 240\, kJoules

3 0
3 years ago
1 poi
tatuchka [14]

Answer:

true

Explanation:

for example assume you are setting in a moving bus and when someone see you from the ground you are in motion but for some who is with you in the bus you are not in motion.

6 0
2 years ago
Two generators use the same magnetic field and operate at the same frequency. Each has a single-turn circular coil. One generato
Volgvan

Answer:

The coil radius of other generator is 5.15 cm

Explanation:

Consider the equation for induced emf in a generator coil:

EMF = NBAω Sin(ωt)

where,

N = No. of turns in coil

B = magnetic field

A = Cross-sectional area of coil = π r²

ω = angular velocity

t = time

It is given that for both the coils magnetic field, no. of turn and frequency is same. Since, the frequency is same, therefore, the angular velocity, will also be same. As, ω = 2πft.

Therefore, EMF for both coils or generators will be:

EMF₁ = NBπr₁²ω Sin(ωt)

EMF₂ = NBπr₂²ω Sin(ωt)

dividing both the equations:

EMF₁/EMF₂ = (r₁/r₂)²

r₂ = r₁ √(EMF₂/EMF₁)

where,

EMF₁ = 1.8 V

EMF₂ = 3.9 V

r₁ = 3.5 cm

r₂ = ?

Therefore,

r₂ = (3.5 cm)√(3.9 V/1.8 V)

<u>r₂ = 5.15 cm</u>

3 0
2 years ago
What is the energy of a photon that has the same wavelength as an electron having a kinetic energy of 15 ev?
serg [7]

Answer: 6.268(10)^{-16}J

Explanation:

The kinetic energy of an electron K_{e} is given by the following equation:

K_{e}=\frac{(p_{e})^{2} }{2m_{e}}   (1)

Where:

K_{e}=15eV=2.403^{-18}J=2.403^{-18}\frac{kgm^{2}}{s^{2}}

p_{e} is the momentum of the electron

m_{e}=9.11(10)^{-31}kg  is the mass of the electron

From (1) we can find p_{e}:

p_{e}=\sqrt{2K_{e}m_{e}}    (2)

p_{e}=\sqrt{2(2.403^{-18}J)(9.11(10)^{-31}kg)}  

p_{e}=2.091(10)^{-24}\frac{kgm}{s}   (3)

Now, in order to find the wavelength of the electron \lambda_{e}   with this given kinetic energy (hence momentum), we will use the De Broglie wavelength equation:

\lambda_{e}=\frac{h}{p_{e}}    (4)

Where:

h=6.626(10)^{-34}J.s=6.626(10)^{-34}\frac{m^{2}kg}{s} is the Planck constant

So, we will use the value of p_{e} found in (3) for equation (4):

\lambda_{e}=\frac{6.626(10)^{-34}J.s}{2.091(10)^{-24}\frac{kgm}{s}}    

\lambda_{e}=3.168(10)^{-10}m    (5)

We are told the wavelength of the photon  \lambda_{p} is the same as the wavelength of the electron:

\lambda_{e}=\lambda_{p}=3.168(10)^{-10}m    (6)

Therefore we will use this wavelength to find the energy of the photon E_{p} using the following equation:

E_{p}=\frac{hc}{lambda_{p}}    (7)

Where c=3(10)^{8}m/s  is the spped of light in vacuum

E_{p}=\frac{(6.626(10)^{-34}J.s)(3(10)^{8}m/s)}{3.168(10)^{-10}m}  

Finally:

E_{p}=6.268(10)^{-16}J    

4 0
3 years ago
Qqqqqqqqqqqqqqqqqqqqqqqqqqqq
tester [92]
Great question the answer is -25x.
3 0
2 years ago
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