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Mila [183]
2 years ago
11

According to newton's law of gravitation, what affects the force of attraction between two objects? distance between them angle

between them their shape their color
Physics
1 answer:
OlgaM077 [116]2 years ago
3 0

The force of attraction between two objects Mass and distance.

<h3>What is newton's law of gravitation?</h3>

Every particle in the cosmos attracts every other particle with a force that is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centres, according to Newton's law of universal gravitation.

Inductive reasoning, as described by Isaac Newton, was used to deduce this general physical law from actual facts. It was created by Newton and is a component of classical mechanics. Philosophiae Naturalis Principia Mathematica, also known as "the Principia," was originally published on July 5, 1687. In April 1686, when Newton gave Book 1 of the unpublished book to the Royal Society, Robert Hooke said that Newton had learned the inverse square law from him.

According to the law, every point mass attracts every other point mass when a force applies along the line that intersects the two points, in today's parlance. The force is inversely equal to the square of the separation between the masses and directly proportional to their product.

F = G\frac{m_{1} m_{2}}{r^{2}  }

to learn more about newton's law of gravitation go to - brainly.com/question/9373839

#SPJ4

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A speaker at an open-air concert emits 600 W of sound power, radiated equally in all directions.
Marina86 [1]

Answer

Given,

Sound Power, P = 600 W

a) Intensity of sound,

    r = 5 m

   we know,

     I = \dfrac{P}{A}

     I = \dfrac{P}{4\pi r^2}

     I = \dfrac{600}{4\pi\times 5^2}

            I = 1.91 W/m²

b) Sound intensity

  \beta = 10 log(\dfrac{I}{I_0})

       I₀ = 10⁻¹²

  \beta = 10 log(\dfrac{1.91}{10^{-12}})

  \beta = 123\ dB

c) Sound experienced by the Phil

    β₁ = 123 - 23 = 100 \ dB

distance from the sound

again using sound intensity formula

  \beta = 10 log(\dfrac{I}{I_0})

  100= 10 log(\dfrac{I}{10^{-12}})

  I = 10^{-12}\times 10^{10}

where, I = P/A

  \dfrac{P}{4\pi r^2} = 10^{-2}

  r^2 = \dfrac{600}{0.04\times \pi}

        r = 69.09 m

Distance where Phil will experience same sound intensity without earplug is equal to r = 69.09 m

7 0
4 years ago
Can anyone help me solve number 5
Shtirlitz [24]
What subject is this?
6 0
3 years ago
Light can travel through outer space from the Sun to the Earth. This shows that
PIT_PIT [208]

Answer:

c. that light can travel in a vacuum

6 0
4 years ago
A grocery shopper tosses a(n) 9.4 kg bag of rice into a stationary 19.9 kg grocery cart. The bag hits the cart with a horizontal
IRINA_888 [86]

The final speed of the cart and the bag is 2.2 m/s

Explanation:

We can solve this problem by using the law of conservation of momentum: in fact, in absence of external forces, the total momentum of the bag and the cart must be conserved before and after the collision.

Mathematically:

p_i = p_f\\m_1 u_1 + m_2 u_2 = (m_1+m_2)v  

where:  

m_1 = 9.4 kg is the mass of the bag of rice

u_1 = 7.0 m/s is the initial velocity of the bag

m_2 = 19.9 kg is the mass of the cart

u_2 =0 m/s is the initial velocity of the cart (at rest)

v is the final combined velocity of the bag and the cart

Re-arranging the equation and solving for v, we find:  

v=\frac{m_1 u_1 + m_2 u_2}{m_1+m_2}=\frac{(9.4)(7.0)+0}{9.4+19.9}=2.2 m/s

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5 0
3 years ago
Two large parallel conducting plates carrying opposite charges of equal magnitude are separated by 2.20 cm. If the surface charg
bekas [8.4K]

Answer:

5.3\times 10^3 N/C

Explanation:

We are given that

Distance between plates=d=2.2 cm=2.2\times 10^{-2} m

1 cm=10^{-2} m

\sigma=47nC/m^2=47\times 10^{-9}C/m^2

Using 1 nC=10^{-9} C

We have to find the magnitude of E in the region between the plates.

We know that the electric field for parallel plates

E=\frac{\sigma}{2\epsilon_0}

E_1=\frac{\sigma}{2\epsilon_0}

E_2=\frac{\sigma}{2\epsilon_0}

E=E_1+E_2

E=\frac{\sigma}{2\epsilon_0}+\frac{\sigma}{2\epsilon_0}=\frac{\sigma}{\epsilon_0}

Where \epsilon_0=8.85\times 10^{-12}C^2/Nm^2

Substitute the values

E=\frac{47\times 10^{-9}}{8.85\times 10^{-12}}

E=5.3\times 10^3 N/C

Hence, the magnitude of E in the region between the plates=5.3\times 10^3 N/C

5 0
3 years ago
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