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Sav [38]
3 years ago
6

A molecule of roughly spherical shape has a mass of 6.10 × 10-25 kg and a diameter of 0.70 nm. The uncertainty in the measured p

osition of the molecule is equal to the molecular diameter. What is the minimum uncertainty in the speed of this molecule? (h = 6.626 × 10-34 J · s)
Physics
1 answer:
Ilia_Sergeevich [38]3 years ago
4 0

Answer:

\Delta v = 0.123 m/s

Explanation:

As per the law of uncertainty we know that

\Delta p \times \Delta x = \frac{h}{4\pi}

now we know that

\Delta x = 0.70 nm

m = 6.10 \times 10^{-25}

also we have

h = 6.626 \times 10^{-34} J.s

now we will have

m\Delta v \times \Delta x = \frac{h}{4\pi}

(6.10 \times 10^{-25})\Delta v \times (0.70 \times 10^{-9}) = \frac{6.626 \times 10^{-34}}{4\pi}

\Delta v= 0.123 m/s

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Andreas93 [3]

Answer:

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Explanation:

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w.d = 2000Nm

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3 years ago
Two disks of identical mass but different radii (r and 2r) are spinning on frictionless bearings at the same angular speed ?0, b
cestrela7 [59]

Answer:

Explanation:

Moment of inertia of a disc = 1/2 M R²

Since mass is same for both and radius are r and 2r, their moment of inertia can be in the ratio of 1: 4 . Let them be I and 4I . Angular speed are ω₀ and   - ω₀ .

We shall apply law of conservation of angular momentum .

initial total angular momentum

I x ω₀ - 4I x ω₀ = - 3Iω₀

Let final common angular momentum be ω

total final angular momentum = ( I + 4I ) ω

Applying law of conservation of angular momentum

( I + 4I ) ω =  - 3Iω₀

ω = - 3 / 5 ω₀ .

b )

Initial total rotational K E

= 1/2 I ω₀² + 1/2 4I ω₀²

= 1/2 x5I ω₀²

Final total rotational K E

= 1/2 ( I + 4I ) ( - 3 / 5 ω₀ )²

= 1/2 x 9 / 5 I ω₀²

= 9 / 10I ω₀²

change in rotational kinetic energy = 9 / 10I ω₀² - 1/2 x5I ω₀²

(9/10 - 5/2) xI ω₀²

=( .9 - 2.5 )I ω₀²

= - 1.6 I ω₀² Ans

8 0
3 years ago
Driving Zone 7 corresponds with
stepan [7]
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zone 2 = your left lane
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3 years ago
A car is traveling at 100 km/h when the driver sees an accident 80 m ahead and slams on the brakes. what minimum constant decele
Effectus [21]
Assuming the driver starts slamming the brakes immediately, the car moves by uniformly decelerated motion, so we can use the following relationship
2aS = v_f^2 - v_i^2 (1)
where 
a is the deleceration
S is the distance covered after a time t
v_f is the velocity at time t
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The accident is 80 m ahead of the car, so the minimum deceleration required to avoid the accident is the value of a such that S=80 m and v_f=0 (the car should stop exactly at S=80 m to avoid the accident). Using these data, we can solve  the equation (1) to find a:
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4 0
3 years ago
Woman pulls a 6.87 kg suitcase,
GenaCL600 [577]

Answer:

2.24 m/s

Explanation:

resolving force of 29.2 N in x component

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as force of friction is 12.7 N hence net force which produces acceleration is

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by Newton 's law a=f/m

a= 2.9/6.87=0.422 m/s^2

now equation of motion is

v^2= U^2+2as

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v^2=5.00

v=2.24 m/s

4 0
3 years ago
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