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Gnoma [55]
2 years ago
14

Calculate the de Broglie wavelength of a 0.56 kg ball moving with a constant velocity of 26 m/s (about 60 mi/h)

Physics
1 answer:
PilotLPTM [1.2K]2 years ago
6 0

The de Broglie wavelength of a 0.56 kg ball moving with a constant velocity of 26 m/s is 4.55×10⁻³⁵ m.

<h3>De Broglie wavelength:</h3>

The wavelength that is incorporated with the moving object and it has the relation with the momentum of that object and mass of that object. It is inversely proportional to the momentum of that moving object.

λ=h/p

Where, λ is the de Broglie wavelength, h is the Plank constant, p is the momentum of the moving object.

Whereas, p=mv, m is the mass of the object and v is the velocity of the moving object.

Therefore, λ=h/(mv)

λ=(6.63×10⁻³⁴)/(0.56×26)

λ=4.55×10⁻³⁵ m.

The de Broglie wavelength associated with the object weight 0.56 kg moving with the velocity of 26 m/s is λ=4.55×10⁻³⁵ m.

Learn more about de Broglie wavelength on

brainly.com/question/15330461

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The majority of water on Earth exists as salt water, and only a very small percentage exists as fresh water. More than half of t
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2 years ago
After landing on an unfamiliar planet, a space explorer constructs a simple pendulum of length 55.0 cm. She finds the pendulum m
forsale [732]

Answer:

Acceleration due to gravity will be g=5.718m/sec^2

Explanation:

We have given length of pendulum l = 55 cm = 0.55 m

It is given that pendulum completed 100 swings in 145 sec

So time taken by pendulum for 1 swing =\frac{145}{100}=1.45sec

We have to find the acceleration due to gravity at that point

We know that time period of pendulum;um is given by

T=2\pi \sqrt{\frac{l}{g}}

So 1.45=2\times 3.14\times \sqrt{\frac{0.55}{g}}

\sqrt{\frac{0.55}{g}}=0.230

Squaring both side

{\frac{0.3025}{g}}=0.0529

g=5.718m/sec^2

So acceleration due to gravity will be g=5.718m/sec^2

3 0
4 years ago
The moon orbits Earth approximately once a month. <br><br> True<br><br> False
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Answer:true

Explanation:

It takes the moon approximately a month to complete an orbit round the earth

5 0
4 years ago
A satellite orbits the Earth in the same direction it rotates in a circular orbit above the equator a distance of 250 km from th
UkoKoshka [18]

Answer:

Clock on the satellite is slower than the one present on the earth = 29.376 s

Given:

Distance of satellite from the surface, d = 250 km

Explanation:

Here, the satellite orbits the earth in circular motion, thus the necessary centripetal force is provided by the gravitation force and is given by:

\frac{mv^{2}}{R} = \frac{GMm}{R^{2}}

where

v =  velocity of the satellite

R = radius of the earth = 6350 km = 6350000 m

G = gravitational constant = 6.674\times 10^{- 11} m^{3}/ks-s^{2}

M = mass of earth = 5.972\times 10^{24} kg

Therefore, the above eqn can be written as:

v = \sqrt{\frac{GM}{R}}

Now, for relativistic effects:

\frac{v}{c} = \sqrt{\frac{GM}{Rc^{2}}} = 26.41\times 10^{- 6}

Now,

r = R + 250

\frac{v_{surface}}{c} = {\frac{1}{c}\frac{2\pi R}{24} = 1.54\times 10^{-6}

Ratio of rate of satellite clock to surface clock:

\frac{\sqrt{1 - \frac{v^{2}}{c^{2}}}}{\sqrt{1 - \frac{v_{surface}^{2}}{c^{2}}}} = 3.43\times 10^{-10}

Clock on the satellite is slower than the one present on the earth:

3.43\times 10^{-10}\times 24\times 3600 = 29.376 s

4 0
3 years ago
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