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nekit [7.7K]
3 years ago
11

Acceleration of a free-falling object in a frictionless environment increases as a function of time.

Physics
1 answer:
Digiron [165]3 years ago
6 0

A free-falling object is an object moving under the effect of gravitational forces alone

The correct option to select for the True or False question is False

The reason the above selected option is correct is as follows:

According to Newton's second law of motion, we have;

Force = Mass × Acceleration

The force of gravity is F_{g} =G \cdot \dfrac{M \cdot m}{r^{2}}

Where;

G \cdot \dfrac{M }{r^{2}} = Acceleration \ due \ to \ gravity , \ g \approx 9.81 m/s^2

m  = The mass of the object

∴ The force acting on an object in free fall, F_g = m × g

Therefore the acceleration of an object in free fall is the constant acceleration due to gravity, and it therefore, does not change with time

The correct option for the question, acceleration of a free-falling object in a frictionless environment increases as a function of time is <u>False</u>

<u></u>

Learn more about object in free fall here:

brainly.com/question/13712424

brainly.com/question/11698474

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Infrared radiation has wavelengths ranging from about 800 nm to 1 mm. what is the frequency of radiation of wavelength 960 nm? a
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The correct answer to your question and how to solve it is
 
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If a cup of coffee has temperature 95∘C95∘C in a room where the temperature is 20∘C,20∘C, then, according to Newton's Law of Coo
lina2011 [118]

Answer:

T = 76.39°C

Explanation:

given,

coffee cup temperature = 95°C

Room temperature= 20°C

expression

T( t ) = 20 + 75 e^{\dfrac{-t}{50}}

temperature at t = 0

T( 0 ) = 20 + 75 e^{\dfrac{-0}{50}}

T(0) = 95°C

temperature after half hour of cooling

T( t ) = 20 + 75 e^{\dfrac{-t}{50}}

t = 30 minutes

T( 30 ) = 20 + 75 e^{\dfrac{-30}{50}}

T( 30 ) = 20 + 75 \times 0.5488

T(30) = 61.16° C

average of first half hour will be equal to

T = \dfrac{1}{30-0}\int_0^30(20 + 75 e^{\dfrac{-t}{50}})\ dt

T = \dfrac{1}{30}[(20t - \dfrac{75 e^{\dfrac{-t}{50}}}{\dfrac{1}{50}})]_0^30

T = \dfrac{1}{30}[(20t - 3750e^{\dfrac{-t}{50}}]_0^30

T = \dfrac{1}{30}[(20\times 30 - 3750 e^{\dfrac{-30}{50}} + 3750]

T = \dfrac{1}{30}[600 - 2058.04 + 3750]

T = 76.39°C

4 0
3 years ago
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