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lidiya [134]
3 years ago
13

Which of the following supports the claim that the atom is like a solid positive cookie with negative electrons embedded within

it? (This model is known as the Plum Pudding Model of the atom, and is illustrated to the right).
a. Law of Constant Composition, Law of Multiple Proportions, Law of Conservation of Mass Alpha particles are scattered at a variety of angles (over 90 degrees) when bombarded at gold foil.
b. When light from hydrogen emissions passes through a diffracting grating, there are distinct bands of color.
c. The Cathode Ray Tube experiment, in which the ray was attracted to the south pole of the magnet.
Physics
1 answer:
allochka39001 [22]3 years ago
8 0

Answer:

a. Law of Constant Composition, Law of Multiple Proportions, Law of Conservation of Mass Alpha particles are scattered at a variety of angles (over 90 degrees) when bombarded at gold foil.

Explanation:

Which of the following supports the claim that the atom is like a solid positive cookie with negative electrons embedded within it? (This model is known as the Plum Pudding Model of the atom, and is illustrated to the right).

a. Law of Constant Composition, Law of Multiple Proportions, Law of Conservation of Mass Alpha particles are scattered at a variety of angles (over 90 degrees) when bombarded at gold foil.

b. When light from hydrogen emissions passes through a diffracting grating, there are distinct bands of color.

c. The Cathode Ray Tube experiment, in which the ray was attracted to the south pole of the magnet.

the plumbudding model of the atom was postulated by  JJ Thompson and plum pudding model. . ... Thomson had discovered that atoms are composite objects, made of pieces with positive and negative charge, and that the negatively charged electrons within the atom were very small compared to the entire atom.

so a. correctly typifies the thompson model of the atom

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E) Thermal energy is released during
vichka [17]

Answer:

e) True, f) False

Explanation:

e) Let consider a close system, that is, a system with no mass interactions with surroundings. Then, we get the following expression by the First Law of Thermodynamics:

Q_{net,in} - W_{net, out} = \Delta U (1)

Where:

Q_{net, in} - Net input heat, measured in joules.

W_{net, out} - Net output work, measured in joules.

\Delta U - Change in thermal energy, measured in joules.

Please notice that work comprises all kind of work (i.e. mechanical, electric, magnetic), whereas heat comprises all heat interactions including chemical and radioactive phenomena.

If thermal energy is released, then \Delta U < 0, which is caused by three scenarios:

(i) Q_{net,in} < 0, W_{net, out} < 0, |Q_{net,in}|>|W_{net,out}|

(ii) Q_{net, in} > 0, W_{net,out} > 0, |Q_{net,in}|

(iii) Q_{net,in}< 0, W_{net, out}>0

In the case Q_{net,in} > 0, W_{net, out}, the thermal energy of the system is increased. Therefore, thermal energy is released during some energy conversions. Answer: True

f) A liquid solidifies when temperature goes below point of fusion, meaning a realease of heat with no work interactions. That is:

Q_{net, in} = \Delta U, Q_{net, in} < 0 (2)

If Q_{net, in} < 0, then  \Delta U < 0. Then, if a liquid absorbs heat energy, then thermal energy is increase and the liquid does not solidifies. Answer: False.

7 0
3 years ago
A wheel accelerates from rest to 34.7 rad/s at a rate of 47.0 rad/s^2. Through what angle (in radians) did the wheel turn while
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12.8 rad

Explanation:

The angular displacement \theta through which the wheel turned can be determined from the equation below:

\omega^2 = \omega_0^2 + 2\alpha\theta (1)

where

\omega_0 = 0

\omega = 34.7\:\text{rad/s}

\alpha = 47.0\:\text{rad/s}^2

Using these values, we can solve for \theta from Eqn(1) as follows:

2\alpha\theta = \omega^2 - \omega_0^2

or

\theta = \dfrac{\omega^2 - \omega_0^2}{2\alpha}

\:\:\:\:= \dfrac{(34.7\:\text{rad/s})^2 - 0}{2(47.0\:\text{rad/s}^2)}

\:\:\:\:= 12.8\:\text{rad}

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2 years ago
Which best explains why a black T-shirt gets warmer in the Sun than a white T-shirt?
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Answer:

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3 years ago
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(a) What is the minimum width of a single slit (in multiples of λ ) that will produce a first minimum for a wavelength λ ? (b) W
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Answer:

The minimum value of width for first minima is λ

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The  minimum value of width for 1000 minima is 1000λ

Explanation:

Given that,

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D\sin\theta=n\lambda

D=\dfrac{n\lambda}{\sin\theta}

Where, D = width of slit

\lambda = wavelength

Put the value into the formula

D=\dfrac{n\lambda}{\sin\theta}

Here, \sin\theta should be maximum.

So. maximum value of \sin\theta is 1

Put the value into the formula

D=\dfrac{1\times\lambda}{1}

D=\lambda

(b). If the minimum number  is 50

Then, the width is

D=\dfrac{50\times\lambda}{1}

D=50\lambda

(c). If the minimum number  is 1000

Then, the width is

D=\dfrac{1000\times\lambda}{1}

D=1000\lambda

Hence, The minimum value of width for first minima is λ

The  minimum value of width for 50 minima is 50λ

The  minimum value of width for 1000 minima is 1000λ

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