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nasty-shy [4]
3 years ago
11

How do you reconcile the law of falling bodies (that all objects fall to earth at the same acceleration despite their weight) wi

th the Second Law of Motion which states that the acceleration of a body subjected to an applied force is inversely proportional to its mass?
Physics
1 answer:
ASHA 777 [7]3 years ago
5 0

From the gravity acceleration theorem due to a celestial body or planet, we have that the Force is given as

F = \frac {GMm} {r ^ 2}

Where,

F = Strength

G = Universal acceleration constant

M = Mass of the planet

m = body mass

r = Distance between centers of gravity

The acceleration by gravity would be given under the relationship

g = \frac {F} {m}

g = \frac {GM} {r ^ 2}

Here the acceleration is independent of the mass of the body m. This is because the force itself depended on the mass of the object.

On the other hand, the acceleration of Newton's second law states that

a = \frac {F} {m}

Where the acceleration is inversely proportional to the mass but the Force does not depend explicitly on the mass of the object (Like the other case) and therefore the term of the mass must not necessarily be canceled but instead, considered.

You might be interested in
A spy satellite uses a telescope with a 1.7-m-diameter mirror. It orbits the earth at a height of 180 km.
WINSTONCH [101]

Answer: the minimum spacing that must be there between two objects on the earth's surface if they are to be resolved as distinct objects by this telescope 6.45 cm

Explanation:

Given that;

diameter of the mirror d = 1.7 m

height h = 180 km = 180 × 10³ m

wavelength λ = 500 nm = 5 × 10⁻⁹ m

Now Angular separation from the peak of the central maximum is expressed as;

sin∅= 1.22 λ / d

sin∅ = (1.22 × 5 × 10⁻⁹) / 1.7

 sin∅ = 3.588 × 10⁻⁷

we know that;

 sin∅  = object separation / distance from telescope

object separation =   sin∅ × distance from telescope

object separation = 3.588 × 10⁻⁷  × 180 × 10³

object separation =6.45 × 10⁻² m

then we convert to centimeter

object separation = 6.45 cm

Therefore the minimum spacing that must be there between two objects on the earth's surface if they are to be resolved as distinct objects by this telescope 6.45 cm

5 0
3 years ago
An ant is crawling along a yardstick that is pointed with the 0-inch mark to the east and the 36-inch mark to the west. It start
galina1969 [7]

Answer:

it moves 25 inches.

Explanation:

the east west bit isn't important, ignore it. if an ant starts at 6 then moves to 19 then we need to subtract 19 from 6, that's 13. then it moves to 7. the difference between 19 and 7 is 12. add that to 13 and you get 25. it's important to remember that there is no such thing as negative distance. if it moved, then it counts.

3 0
3 years ago
What is the force weight of a jaguar who jumps 3 meters to a tree branch with 2670 J of work?
fgiga [73]

Answer:

we \: know \: energy \:  =  \: force \:  \times distance \\ e = f \times d \\ so \: f \:  =  \frac{e}{d}  \\ so \: th \: force \: here \:  =  (\frac{2670}{3}) newton \\  = 890newton

Hope it helps

7 0
3 years ago
E=?
WITCHER [35]
E=mc2 or MC Squared as in (mass energy equivalence)
4 0
3 years ago
At TTT = 14 ∘C∘C, how long must an open organ pipe be to have a fundamental frequency of 262 HzHz ? The speed of sound in air is
Stolb23 [73]

Answer:

Length of pipe organ(L) = 0647 m (Approx)

Explanation:

Given:

Temperature (T) = 14°C

Fundamental frequency (F) = 262 Hz

Speed of sound (v) = 331 + 0.60(T) m/s

Find:

Length of pipe organ(L)

Computation:

Speed of sound (v) = 331 + 0.60(14) m/s

Speed of sound (v) = 339.4

Length of pipe organ(L) = Speed of sound (v) / 2(Fundamental frequency)

Length of pipe organ(L) = 339.4 / 2 (262)

Length of pipe organ(L) = 0647 m (Approx)

7 0
3 years ago
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