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Alik [6]
2 years ago
7

The gold foil experiment led to the conclusion that each atom in the foil was composed mostly of empty space because most alpha

particles directed at the foil
(1) passed through the foil
(2) remained trapped in the foil
(3) were deflected by the nuclei in gold atoms
(4) were deflected by the electrons in gold atoms

Physics
2 answers:
Sholpan [36]2 years ago
6 0

Answer:

Option 1

Explanation:

The correct answer is option 1

The gold foil experiment was conducted by Rutherford. This experiment was conducted to study the Atom.

In the experiment Alpha rays from the emitter are passed through gold foil and there was a receiver that was present there to intercept the alpha rays.

The outcome of the result was that most of the alpha particle pass through foil undeflected and very few rays revert back on the original path from the heavy mass present at the center.

Later this heavy mass was known Nucleus.

Hence, most alpha particles passed through the foil.

katovenus [111]2 years ago
5 0

Answer:

(1) passed through the foil

Explanation:

Ernest Rutherford conducted an experiment using an alpha particle emitter projected towards a gold foil and the gold foil was surrounded by a fluorescent screen which glows upon being struck by an alpha particle.

  • When the experiment was conducted he found that most of the alpha particles went away without any deflection (due to the empty space) glowing the fluorescent screen right at the point of from where they were emitted.
  • While a few were deflected at reflex angle because they were directed towards the center of the nucleus having the net effective charge as positive.
  • And some were acutely deflected due to the field effect of the positive charge of the proton inside the nucleus. All these  conclusions were made based upon the spot of glow on the fluorescent screen.

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In a house the temperature at the surface of a window is 28.9 °C. The temperature outside at the window surface is 7.89 °C. Heat
Alenkasestr [34]

Answer:

-13.18°C

Explanation:

To develop the problem it is necessary to consider the concepts related to the thermal conduction rate.

Its definition is given by the function

\frac{Q}{t} = \frac{kA\Delta T}{d}

Where,

Q = The amount of heat transferred

t = time

k = Thermal conductivity constant

A = Cross-sectional area

\Delta T = The difference in temperature between one side of the material and the other

d= thickness of the material

The problem says that there is a loss of heat twice that of the initial state, that is

Q_2 = 2*Q_1

Replacing,

kA\frac{\Delta T_m}{x} = 2*kA\frac{\Delta T}{x}

\frac{\Delta T}{x}=2*\frac{\Delta T}{x}

\frac{T_i-T_o}{x} = 2\frac{T_1-T_2}{x}

\frac{28.9-T_o}{x} = 2\frac{28.9-7.86}{x}

Solvinf for T_o,

T_o = -13.18

Therefore the temprature at the outside windows furface when the heat lost per second doubles is  -13.18°C

3 0
2 years ago
Two sinusoidal waves are identical except for their phase. When these two waves travel along the same string, for which phase di
Kamila [148]

Answer:

zero or 2π is maximum

Explanation:

Sine waves can be written

      x₁ = A sin (kx -wt + φ₁)

     x₂ = A sin (kx- wt + φ₂)

When the wave travels in the same direction

      Xt = x₁ + x₂

      Xt = A [sin (kx-wt + φ₁) + sin (kx-wt + φ₂)]

We are going to develop trigonometric functions, let's call

     a = kx + wt

     Xt = A [sin (a + φ₁) + sin (a + φ₂)

We develop breasts of double angles

     sin (a + φ₁) = sin a cos φ₁ + sin φ₁ cos a

    sin (a + φ₂) = sin a cos φ₂ + sin φ₂ cos a

Let's make the sum

     sin (a + φ₁) + sin (a + φ₂) = sin a (cos φ₁ + cos φ₂) + cos a (sin φ₁ + sinφ₂)

to have a maximum of the sine function, the cosine of fi must be maximum

     cos φ₁ + cos φ₂ = 1 +1 = 2

the possible values ​​of each phase are

     φ1 = 0, π, 2π  

     φ2 = 0, π, 2π,  

so that the phase difference of being zero or 2π is maximum

6 0
2 years ago
The diameters of bolts produced by a certain machine are normally distributed with a mean of 0.30 inches and a standard deviatio
enot [183]

Answer:

2.25 %

Explanation:

65-95-99.7 is a rule to remember the precentages that lies around the mean.

at the range of mean (\mu) plus or minus one standard deviation (\sigma), P([\mu-\sigma \leq X \leq \mu+\sigma])\approx 68.3%

at the range of mean plus or minus two standard deviation, P([\mu -2\sigma \leq X \leq \mu+2\sigma])\approx 95.5%

at the range of mean plus or minus three standard deviation, P([\mu - 3\sigma\leq X \leq \mu+3\sigma])\approx 99.7%

So, note that  they are asking about the probability that it is greater than 0.32, that is the mean (0.3) plus two times the standard deviation (0.1) (P(X \leq \mu+2\sigma))  

So we know that the 95.5% is between \mu - 2\sigma = 0.3 -2*0.1 = 0.28 and \mu + 2\sigma = 0.3 +2*0.1 = 0.32, hence approximately the 4.5% (100%-95.5%) is greater than 0.32 or less than  0.28. But half (4.5%/2=2.25%) is greater than 0.32 and the other half is less than 0.28.

So P(X \leq \mu+2\sigma) \approx 2.25%

8 0
2 years ago
An organ pipe is tuned to exactly 384 Hz when the temperature in the room is 20°C. Later, when the air has warmed up to 25°C, th
maksim [4K]

Answer: A. Greater than 384 Hz

Explanation:

The velocity of sound is directly related to the temperature rather it is directly proportional meaning if the temperature decreases the velocity decreases and if temperature increases the velocity increases.

Now, we are given that temperature has risen from 20°C to 25°C meaning it has increases. So it implies that velocity must also increase.

Also, the velocity for organ pipe is directly proportional to its frequency. Now if velocity increases frequency must also increase. In this case, the original frequency is 384 Hz. Now increasing the temperature resulted in increase in velocity and thus increase in frequency.

So option a is correct. i.e. now frequency will be greater than 384 Hz.

3 0
2 years ago
A projectile has an initial horizontal velocity of 15 meters per second and an initial vertical velocity of 25 meters per second
Artyom0805 [142]

Answer:

75 m

Explanation:

The horizontal motion of the projectile is a uniform motion with constant speed, since there are no forces acting along the horizontal direction (if we neglect air resistance), so the horizontal acceleration is zero.

The horizontal component of the velocity of the projectile is

v_x = 15 m/s

and it is constant during the motion;

the total time of flight is

t = 5 s

Therefore, we can apply the formula of the uniform motion to find the horizontal displacement of the projectile:

d= v_x t =(15 m/s)(5 s)=75 m

5 0
2 years ago
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