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defon
2 years ago
5

Do the proton and the neutron have exactly the same mass? How do the masses of the proton and the neutron compare to the mass of

the electron? Which particles make the greatest contribution to the mass of an atom? Which particles make the greatest contribution to the chemical properties of an atom?
Physics
1 answer:
True [87]2 years ago
6 0

Answer:

Explanation:

mas of proton = 1.673 x 10^-27 kg

mass of neutron = 1.675 x 10^-27 kg

mass of electron = 9.1 x 10^-31 kg

(a) the mass of proton and the mass of neutron is not exactly same.

(b) mass of proton = 1838.46 x mass of electron

mass of neutron = 1840.65 x mass of electron

(c) As proton and neutrons are more massive than the electrons, so the protons and the neutrons has the greatest contribution of mass of an atom.

(d) The greatest contribution to the chemical properties of an atom is due to the electrons.

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A speeding motorist traveling down a straight highway at 100 km/h passes a parked police car. It takes the police constable 1.0
Lubov Fominskaja [6]

Answer:

t = 7.5 s

Explanation:

The distance traveled by the car at the time of meeting of the two cars must be the same. First, we calculate the distance traveled by the police car. For that we use 2nd equation of motion. Here, we take the time when police car starts to be reference. So,

s₁ = Vi t + (0.5)gt²

where,

s₁ = distance traveled by police car

Vi = Initial Velocity = 0 m/s

t = time taken

Therefore,

s₁ = (0 m/s)(t) + (0.5)(9.8 m/s²)t²

s₁ = 4.9 t²

Now, we calculate the distance traveled by the car. For constant speed and time to be 1 second more than the police car time, due to car starting time, we get:

s₂ = Vt = V(t + 1)

where,

s₂ = distance traveled by car

V = Velocity of car = (100 km/h)(1000 m/1 km)(1 h/ 3600 s) = 27.78 m/s

Therefore,

s₂ = 27.78 t + 27.78

Now, we know that at the time of meeting:

s₁ = s₂

4.9 t² = 27.78 t + 27.78

4.9 t² - 270.78 t - 27.78 = 0

solving the equation and choosing the positive root:

t = 6.5 s

since, we want to know the time from the moment car crossed police car. Therefore, we add 1 second of starting time in this.

t = 6.5 s + 1 s

<u>t = 7.5 s</u>

6 0
3 years ago
two tugboats pull a disabled supertanker. each tug exerts a constant force of 2.20×106 n , one at an angle 19.0 ∘ west of north,
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The amount of work the two tugboats completed on the supertanker W = 3.12 × 10^9 joules .

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7 0
1 year ago
You are trying to overhear a juicy conversation, but from your distance of 24.0m , it sounds like only an average whisper of 40.
Neporo4naja [7]

Answer:

The distance is r_2  =  0.24 \  m

Explanation:

From the question we are told that

       The  distance from the conversation is r_1    =  24.0 \ m

       The  intensity of  the sound at your position is  \beta _1 =  40 dB

        The  intensity at the sound at the new position is  \beta_2 =  80.0dB

Generally the intensity in  decibel is  is mathematically represented as

      \beta  =  10dB log_{10}[\frac{d}{d_o} ]

The intensity is  also mathematically represented as

      d =  \frac{P}{A}

So

    \beta  =  10dB *  log_{10}[\frac{P}{A* d_o} ]

=>   \frac{\beta}{10}  =  log_{10} [\frac{P}{A (l_o)} ]

From the logarithm definition

=>    \frac{P}{A  *  d_o}  =  10^{\frac{\beta}{10} }

=>      P =  A (d_o ) [10^{\frac{\beta }{ 10} } ]

Here P is the power of the sound wave

 and  A is the cross-sectional area of the sound wave  which is generally in spherical form

Now the power of the sound wave at the first position is mathematically represented as

               P_1 =  A_1 (d_o ) [10^{\frac{\beta_1 }{ 10} } ]

Now the power of the sound wave at the second  position is mathematically represented as

               P_2 =  A_2 (d_o ) [10^{\frac{\beta_2 }{ 10} } ]

Generally  power of the wave is constant at both positions  so  

    A_1 (d_o ) [10^{\frac{\beta_1 }{ 10} } ]  = A_2 (d_o ) [10^{\frac{\beta_2 }{ 10} } ]

      4 \pi r_1 ^2   [10^{\frac{\beta_1 }{ 10} } ]  = 4 \pi r_2 ^2   [10^{\frac{\beta_2 }{ 10} } ]

        r_2 =  \sqrt{r_1 ^2 [\frac{10^{\frac{\beta_1}{10} }}{ 10^{\frac{\beta_2}{10} }} ]}

       substituting value

        r_2 =   \sqrt{ 24^2 [\frac{10^{\frac{ 40}{10} }}{10^{\frac{80}{10} }} ]}

        r_2  =  0.24 \  m

     

7 0
3 years ago
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