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3 years ago

What is the minimum uncertainty in the energy state of an atom if an electron remains in this state for 10 s

Physics

1 answer:

Dafna11 [192]3 years ago

6 0

Answer:

The minimum uncertainty in the energy state of an atom is $1.0557\times10^{-35}\ J$

Explanation:

Given that,

Time t = 10 s

We need to calculate the minimum uncertainty in the energy state

Using Hisen burg of uncertainty principle equation

$\Delta E\Delta t=\dfrac{h}{2\pi}$

$\Delta E=\dfrac{h}{2\pi}\times\dfrac{1}{\Delta t}$

Where, E = energy

t = time

Put the value into the formula

$\Delta E=\dfrac{6.63\times10^{-34}}{2\times3.14}\times\dfrac{1}{10}$

$\Delta E=1.0557\times10^{-35}\ J$

Hence, The minimum uncertainty in the energy state of an atom is $1.0557\times10^{-35}\ J$

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Sliva [168]

Answer:

Explanation:

Given

Height of ceiling is $h=3.6\ m$

Initial speed of Putty $u=9.5\ m/s$

Speed of Putty just before it strike the ceiling is given by

$v^2-u^2=2as$

where v=final velocity

u=initial velocity

a=acceleration

s=displacement

$v^2-9.5^2=2\times (-9.8)\times 3.6$

$v^2=19.69$

$v=4.43\ m/s$

time taken by putty to reach the ceiling

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$4.43=9.5-9.8\times t$

$t=\frac{5.07}{9.8}$

$t=0.517\ s$

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3 years ago

An airplane accelerates from rest to its required takeoff speed with an acceleration 5.0 m/s^2 after travelling on the runway a

Alex Ar [27]

Answer:

Explanation:

.75 km = 750 m

acceleration a = 5 m/s²

distance s = 750 m

initial velocity u = 0

Final velocity v = ?

v² = u² + 2as

v² = 0 + 2 x 5 x 750

v = 86.60 m /s.

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3 years ago

2. It is now 10:29 a.m., but when the bell rings at 10:30 a.m. Suzette will be late for French class for the third time this wee

Alex17521 [72]

Answer:

62 seconds
no

Explanation:

The total travel time Suzette experiences is the sum of the times in each hallway. Using

time = distance/speed

we can add the times.

(35.0 m)/(3.50 m/s) +(48.0 m)/(1.20 m/s) +(60 m)/(5.0 m/s)

= 10 s + 40 s + 12 s

= 62 s

It takes Suzette 62 seconds to get to class. She does not beat the bell.

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3 years ago

An oscillating block - spring system has a mechanical energy of 1.10 j, and amplitude of 11.0 cm, and a maximum speed of 1.7 m/s

ioda

The total mechanical energy of the block-spring system is given by the sum of the potential energy and the kinetic energy of the block:
$E=U+K= \frac{1}{2}kx^2 + \frac{1}{2}mv^2$
where
k is the spring constant
x is the elongation/compression of the spring
m is the mass of the block
v is the speed of the block

At the point of maximum displacement of the spring, the velocity of the block is zero: v=0, so the kinetic energy is zero and the mechanical energy is just potential energy of the spring:
$E= \frac{1}{2}kA^2$ (1)
where we used x=A, the amplitude (which is the maximum displacement of the spring).
Since we know
A = 11.0 cm= 0.11 m
E = 1.10 J
We can re-arrange (1) to find the spring constant:
$k= \frac{2E}{A^2} = \frac{2 \cdot 1.10 J}{(0.11 m)^2}=181.8 N/m$

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3 years ago

A dog is chasing a cat towards a tree. The cat has a 10-yard lead and runs at 6 yards per second. The dog runs at a speed of 8 y

oksano4ka [1.4K]

Answer : Cat will reach tree first.

Explanation :

It is given that,

Total distance between dog and tree is 30 yards. The cat has 10 yards lead. So, the distance between cat and tree is 20 yards.

Speed of cat is 6 Yards/ sec.

Speed of dog is 8 Yards/sec.

speed of cat is, $S_c=\dfrac{distance\ covered\ by\ cat}{time\ taken}$