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3 years ago

Isaac Newton’s investigations of gravity explained which of the following?

2 answers:

8 0

"Gravity acts on all objects in the universe" is the one among the following choices given in the question that was explained by Isaac Newton’s investigations of gravity. The correct option among all the options that are given in the question is the third option or option "C". I hope that the answer has helped you.

Umnica [9.8K]3 years ago

4 0

C. Gravity acts on all objects in the universe.

Explanation:

Isaac Newton with his studies understood that the force of gravity acts on all objects with mass in the universe, and it is the same force that causes an apple to fall from a tree and that keeps the Earth in motion around the Sun.

Newtons also created a formula that allows us to calculate the force of gravity between two objects:

$F=G\frac{m_1 m_2}{r^2}$

where

G is the gravitational constant

m1 and m2 are the masses of the two objects

r is the separation between the two objects

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OLga [1]

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3 years ago

the lines from the boxes show the positions of two of the parts that are present in both cells in the boxes write the names of t

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3 years ago

Two identical loudspeakers 2.00 m apart are emitting sound waves into a room where the speed of sound is 340 m/s. Abby is standi

Ganezh [65]

Answer:

242.85 Hz

Explanation:

For maximum intensity of sound, the path difference,ΔL = (n + 1/2)λ/2 where n = 0,1,2...

Since Abby is standing perpendicular to one speaker, the path length for the sound from the other speaker to him is L₁ = √(2.00² + 5.50²) = √(4.00 + 30.25) = √34.25 = 5.85 m.

The path difference to him is thus ΔL = 5.85 m - 5.50 m = 0.35 m.

Since ΔL = (n + 1/2)λ/2 and for lowest frequency n = 0,

ΔL = (n + 1/2)λ/2 = (0 + 1/2)λ/2 = λ/4

ΔL = λ = v/f and f = v/4ΔL where f = frequency of wave and v = velocity of sound wave = 340 m/s.

f = 340/(4 × 0.35) = 242.85 Hz

5 0

3 years ago

Consider the motion of a 4.00-kg particle that moves with potential energy given by U(x) = + a) Suppose the particle is moving w

gtnhenbr [62]

Correct question:

Consider the motion of a 4.00-kg particle that moves with potential energy given by

$U(x) = \frac{(2.0 Jm)}{x}+ \frac{(4.0 Jm^2)}{x^2}$

a) Suppose the particle is moving with a speed of 3.00 m/s when it is located at x = 1.00 m. What is the speed of the object when it is located at x = 5.00 m?

b) What is the magnitude of the force on the 4.00-kg particle when it is located at x = 5.00 m?

Answer:

a) 3.33 m/s

b) 0.016 N

Explanation:

a) given:

V = 3.00 m/s

x1 = 1.00 m

x = 5.00

$u(x) = \frac{-2}{x} + \frac{4}{x^2}$