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xxTIMURxx [149]
2 years ago
7

What is the wavelength associated with 0.113kg ball traveling with velocity of 43 m/s?

Physics
1 answer:
maks197457 [2]2 years ago
7 0

Answer:

Wavelength = 1.36 * 10^{-34} meters

Explanation:

Given the following data;

Mass = 0.113 kg

Velocity = 43 m/s

To find the wavelength, we would use the De Broglie's wave equation.

Mathematically, it is given by the formula;

Wavelength = \frac {h}{mv}

Where;

h represents Planck’s constant.

m represents the mass of the particle.

v represents the velocity of the particle.

We know that Planck’s constant = 6.6262 * 10^{-34} Js

Substituting into the formula, we have;

Wavelength = \frac {6.6262 * 10^{-34}}{0.113*43}

Wavelength = \frac {6.6262 * 10^{-34}}{4.859}

Wavelength = 1.36 * 10^{-34} meters

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When a substance gains electrons, is it positively or negatively charged?
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Answer:

negative charge

Explanation:

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Two identical peaches, Peach 1 and Peach 2, fall from a cliff at time t=0t=0t, equals, 0 from the same height. Peach 1 is droppe
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Answer: The correct answer is graph A.

Explanation:

See Khan Academy.

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3 years ago
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A small metal object is tied to the end of a string and whirled round a circular path of radius 30cm. The object makes 20 oscill
LUCKY_DIMON [66]

Answer:

(i) The angular speed of the small metal object is 25.133 rad/s

(ii) The linear speed of the small metal object is 7.54 m/s.

Explanation:

Given;

radius of the circular path, r = 30 cm = 0.3 m

number of revolutions, n = 20

time of motion, t = 5 s

(i) The angular speed of the small metal object is calculated as;

\omega = \frac{20 \ rev}{5 \ s} \times \frac{2 \pi \ rad}{1 \ rev} = \frac{40\pi \ rad}{5 \ s} = 8\pi \ rad/s = 25.133 \ rad/s

(ii) The linear speed of the small metal object is calculated as;

v = \omega r\\\\v = 25.133 \ rad/s \ \times \ 0.3 \ m\\\\v = 7.54 \ m/s

6 0
3 years ago
A spring with force constant of 59 N/m is compressed by 1.3 cm in a hockey game machine. The compressed spring is used to accele
Furkat [3]

Answer:

The puck moves a vertical height of 2.6 cm before stopping

Explanation:

As the puck is accelerated by the spring, the kinetic energy of the puck equals the elastic potential energy of the spring.

So, 1/2mv² = 1/2kx² where m = mass of puck = 39.2 g = 0.0392 g, v = velocity of puck, k = spring constant = 59 N/m and x = compression of spring = 1.3 cm = 0.013 cm.

Now, since the puck has an initial velocity, v before it slides up the inclined surface, its loss in kinetic energy equals its gain in potential energy before it stops. So

1/2mv² = mgh where h = vertical height puck moves and g = acceleration due to gravity = 9.8 m/s².

Substituting the kinetic energy of the puck for the potential energy of the spring, we have

1/2kx² = mgh

h = kx²/2mg

= 59 N/m × (0.013 m)²/(0.0392 kg × 9.8 m/s²)

= 0.009971 Nm/0.38416 N

= 0.0259 m

= 2.59 cm

≅ 2.6 cm

So the puck moves a vertical height of 2.6 cm before stopping

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3 years ago
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