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blondinia [14]
2 years ago
10

An atom of neon has a radius rNe = 38. pm and an average speed in the gas phase at 25°C of 350.⁢/ms. Suppose the speed of a neon

atom at 25°C has been measured to within 0.010%. Calculate the smallest possible length of box inside of which the atom could be known to be located with certainty. Write your answer as a multiple of rNe and round it to 2 significant figures. For example, if the smallest box the atom could be in turns out to be 42.0 times the radius of an atom of neon, you would enter "42.rNe" as your answer.
Physics
1 answer:
V125BC [204]2 years ago
6 0

Answer:

                1.2* 10³ rNe.

Explanation:

Given speed of neon=350 m/s

Un-certainity in speed= (0.01/100) *350 =0.035 m/s

As per heisenberg uncertainity principle

              Δx*mΔv ≥\frac{h}{4\pi }     ..................(1)

  mass of neon atom    =\frac{20*10^{-3} }{6.22*10^{-23} } =3.35*10^{-26} kg

substituating the values in eq. (1)

        Δx =4.49*10^{-8} m

In terms of rNe i.e 38 pm= 38*10^{-12}

               Δx=\frac{4.49*10^{-8} }{38*10^{-12} }

                   =0.118*10^{4}* (rNe)

                   =1.18*10³ rN

                   = 1.2* 10³ rNe.

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m_a_m_a [10]

Answer:

halved

Explanation:

The velocity of the a wave is obtained by multiplying the frequency and wavelength.

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

Where

v = Velocity

f = Frequency

\lambda = Wavelength

The velocity here is constant. So, if the frequency is doubled the wavelength is halved.

6 0
2 years ago
A spring balance consists of a pan that hangs from a spring. A damping force Fd = −bv is applied to the balance so that when an
Citrus2011 [14]

Answer:

b ≈ 64 Kg/s

Explanation:

Given

Fd = −bv

m = 2.5 kg

y = 6.0 cm = 0.06 m

g = 9.81 m/s²

The object in the pan comes to rest in the minimum time without overshoot. this means that damping is critical (b² = 4*k*m).

m is given and we find k from the equilibrium extension of 6.0 cm (0.06 m):

∑Fy = 0 (↑)

k*y - W = 0    ⇒   k*y - m*g = 0   ⇒   k = m*g / y

⇒   k = (2.5 kg)*(9.81 m/s²) / (0.06 m)

⇒   k = 408.75 N/m

Hence, if

b² = 4*k*m    ⇒     b = √(4*k*m) = 2*√(k*m)

⇒     b = 2*√(k*m) = 2*√(408.75 N/m*2.5 kg)

⇒     b = 63.9335 Kg/s ≈ 64 Kg/s

5 0
2 years ago
A certain factory whistle can be heard up to a distance of 2.5 km. Assuming that the acoustic output of the whistle is uniform i
enyata [817]

Answer:

Emitted power will be equal to 7.85\times 10^{-5}watt

Explanation:

It is given factory whistle can be heard up to a distance of R=2.5 km = 2500 m

Threshold of human hearing I=10^{-12}W/m^2

We have to find the emitted power

Emitted power is equal to P=I\times A

P=I\times 4\pi R^2

P=10^{-12}\times 4\times 3.14\times  2500^2=7.85\times 10^{-5}watt

So emitted power will be equal to 7.85\times 10^{-5}watt

4 0
2 years ago
A hiker caught in a rainstorm absorbs 1.00 L of water in her clothing. If it is windy so that the water evaporates quickly at 20
masya89 [10]

Answer:

(A) Q = 2.26×10⁶J

(B) ΔT = 9°C

(C)

Explanation:

We have been given the mass of the hiker, the volume of water from which we can calculate the mass knowing that the density if water is 1000kg/m³.

Evaporation is a phase change and occurs at a constant temperature. We would use the latent heat of vaporization to calculate the amount of heat evaporated.

We would then equate this to the heat change it brings about in the hiker's body and then calculate the temperature drop.

See the attachment below for full solution.

6 0
2 years ago
As shown in the figure below, a bullet is fired at and passes through a piece of target paper suspended by a massless string. Th
NikAS [45]

Answer:

M = 0.730*m

V = 0.663*v

Explanation:

Data Given:

v_{bullet, initial} = v\\v_{bullet, final} = 0.516*v\\v_{paper, initial} = 0\\v_{paper, final} = V\\mass_{bullet} = m\\mass_{paper} = M\\Loss Ek = 0.413 Ek

Conservation of Momentum:

P_{initial} = P_{final}\\m*v_{i} = m*0.516v_{i} + M*V\\0.484m*v_{i} = M*V .... Eq1

Energy Balance:

\frac{1}{2}*m*v^2_{i} = \frac{1}{2}*m*(0.516v_{i})^2 + \frac{1}{2}*M*V^2 + 0.413*\frac{1}{2}*m*v^2_{i}\\\\0.320744*m*v^2_{i} = M*V^2\\\\M = \frac{0.320744*m*v^2_{i} }{V^2}  ....... Eq 2

Substitute Eq 2 into Eq 1

0.484*m*v_{i} = \frac{0.320744*m*v^2_{i} }{V^2} *V  \\0.484 = 0.320744*\frac{v_{i} }{V} \\\\V = 0.663*v_{i}

Using Eq 1

0.484m*v_{i} = M* 0.663v_{i}\\\\M = 0.730*m

7 0
2 years ago
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