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zysi [14]
3 years ago
8

Conclusion of galileo feather and coin experiment

Physics
2 answers:
Novosadov [1.4K]3 years ago
7 0

Answer:

This experiment lets you repeat Galileo's experiment in a vacuum. The free fall of a coin and feather are compared, first in a tube full of air and then in a vacuum. With air resistance, the feathers fall more slowly. In a vacuum, the objects fall at the same rate independent of their respective masses.

Brilliant_brown [7]3 years ago
6 0

Answer:

he also used the law of inertia

Explanation:

moving in a straight line without changing speed

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The gas state of water
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Which is an example of radiation?
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If you drop an object, it accelerates downward at 9.8 m/s2 (in the absence of air resistance). If, instead, you throw it downwar
Korvikt [17]

If you drop an object, it accelerates downward at 9.8 m/s2 (in the absence of air resistance). If instead, you throw it downward, its downward acceleration after release is 9.8 m/s2.

Acceleration is the rate at which an object's velocity with respect to time changes. They are vector quantities and accelerations. The direction of the net force acting on an object determines the direction of its acceleration. Uniform acceleration, non-uniform acceleration, and average acceleration are the three different forms of accelerated motions.

A free-falling object experiences a downward acceleration of 9.8 m/s/s (on Earth). This specific designation is given to the numerical value for an object in free fall because it is such an essential value. The longer an object is in free fall, the faster it descends toward the ground due to gravity. In actuality, an object's velocity rises by 9.8 m/s2, so it reaches 9.8 m/s by the time it begins to fall.

To know more about acceleration refer to: brainly.com/question/14468548

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4 0
2 years ago
A particle (m = 4.3 × 10^-28 kg) starting from rest, experiences an acceleration of 2.4 × 10^7 m/s^2 for 5.0 s. What is its de B
Novay_Z [31]

Answer:

Wavelength, \lambda=1.28\times 10^{-14}\ m

Explanation:

Given that,

Mass of the particle, m=4.3\times 10^{-28}\ kg

Acceleration of the particle, a=2.4\times 10^7\ m/s^2

Time, t = 5 s

It starts from rest, u = 0

The De Broglie wavelength is given by :

\lambda=\dfrac{h}{mv}

v = a × t

\lambda=\dfrac{h}{mat}

\lambda=\dfrac{6.67\times 10^{-34}}{4.3\times 10^{-28}\times 2.4\times 10^7\times 5}

\lambda=1.28\times 10^{-14}\ m

Hence, this is the required solution.

4 0
3 years ago
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