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Oxana [17]
3 years ago
5

What force is required to accelerate a 6 kg bowling ball at 2 m/s/s forward?

Physics
1 answer:
Ber [7]3 years ago
6 0

Answer:

12N

Explanation:

Force = mass * acceleration.

6 * 2 = 12

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What is the weight of a 4.2 kg bowling ball on Mars?
Nataliya [291]

What is the weight of a 4.2 kg bowling ball on Mars?

Answer:

1.59 kg

Explanation:

The formula is:

<u>F = G((Mm)/r2) </u>

F is the gravitational force between two objects,

G is the Gravitational Constant (6.674×10-11 Newtons x meters2 / kilograms2),

M is the planet's mass (kg),

m is your mass (kg), and

r is the distance (m) between the centers of the two masses (the planet's radius).

Hope this helps

--Jay

8 0
3 years ago
What is coulomb law​
MrRa [10]

Answer:Coulomb's law states that: The magnitude of the electrostatic force of attraction or repulsion between two point charges is directly proportional to the product of the magnitudes of charges and inversely proportional to the square of the distance between them.

Explanation:Coulomb's law, or Coulomb's inverse-square law, is an experimental law of physics that quantifies the amount of force between two stationary, electrically charged particles. The electric force between charged bodies at rest is conventionally called electrostatic force or Coulomb force

6 0
3 years ago
Use the ratio version of Kepler’s third law and the orbital information of Mars to determine Earth’s distance from the Sun. Mars
zhuklara [117]

Kepler's third law is used to determine the relationship between the orbital period of a planet and the radius of the planet.

The distance of the earth from the sun is 1.50 \times 10^{11}\;\rm m.

<h3>What is Kepler's third law?</h3>

Kepler's Third Law states that the square of the orbital period of a planet is directly proportional to the cube of the radius of their orbits. It means that the period for a planet to orbit the Sun increases rapidly with the radius of its orbit.

T^2 \propto R^3

Given that Mars’s orbital period T is 687 days, and Mars’s distance from the Sun R is 2.279 × 10^11 m.

By using Kepler's third law, this can be written as,

T^2 \propto R^3

T^2 = kR^3

Substituting the values, we get the value of constant k for mars.

687^2 = k\times (2.279 \times 10^{11})^3

k = 3.92 \times 10^{-29}

The value of constant k is the same for Earth as well, also we know that the orbital period for Earth is 365 days. So the R is calculated as given below.

365^3 = 3.92\times 10^{-29} R^3

R^3 = 3.39 \times 10^{33}

R= 1.50 \times 10^{11}\;\rm m

Hence we can conclude that the distance of the earth from the sun is 1.50 \times 10^{11}\;\rm m.

To know more about Kepler's third law, follow the link given below.

brainly.com/question/7783290.

6 0
2 years ago
Los huracanes y tornados se producen por las corrientes de convencion
Montano1993 [528]

Answer:

correct!

Explanation:

4 0
3 years ago
ou are out stargazing with your 13.4-cm telescope. You point your telescope at an interesting formation in the sky, which you th
Alinara [238K]

Answer:

θ = 4.716 10⁻⁶ rad

Explanation:

In order for the releases to be considered separate, they must meet the Rayleigh criterion that establishes that the maximum diffraction of one star must coincide with the first minimum of the diffraction pattern of the second star.

We use the diffraction equation for a slit

            a sin θ = m λ

The minimum occurs at m = 1

             sin θ = λ / a

Since the angles in these systems are very small, we can approximate the sine to its angle in radians

             θ = λ / a

The telescope has a circular aperture whereby polar cords should be used, which introduces a constant number

           θ = 1.22 λ / a

Let's calculate

          θ = 1.22 518 10⁻⁹ / 13.4 10⁻²

          θ = 4.716 10⁻⁶ rad

8 0
2 years ago
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