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VladimirAG [237]
3 years ago
10

. A newly discovered planet has three times the mass and five times the radius of Earth. What is the ratio of the acceleration d

ue to gravity at the surface of the new planet to the acceleration due to gravity at the surface of Earth
Physics
1 answer:
NikAS [45]3 years ago
3 0

Answer:

0.12

Explanation:

The acceleration due to gravity of a planet with mass M and radius R is given as:

g = (G*M) / R²

Where G is gravitational constant.

The mass of the planet M = 3 times the mass of earth = 3 * 5.972 * 10^24 kg

The radius of the planet R = 5 times the radius of earth = 5 * 6.371 * 10^6 m

Therefore:

g(planet) = (6.67 * 10^(-11) * 3 * 5.972 * 10^24) / (5 * 6.371 * 10^6)²

g(planet) = 1.18 m/s²

Therefore ratio of acceleration due to gravity on the surface of the planet, g(planet) to acceleration due to gravity on the surface of the planet, g(earth) is:

g(planet)/g(earth) = 1.18/9.8 = 0.12

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In the context of the loop and junction rules for electrical circuits, a junction is: Group of answer choices where a wire is be
vitfil [10]

Answer:

In the context of the loop and junction rules for electrical circuits, a junction is where three or more wires are joined.

Explanation:

A point where at least three circuit paths meet i.e wires, is referred to as a junction.

Kirchhoff’s circuit laws are two(2) equations first published by Gustav Kirchhoff in 1845. Fundamentally, they address conservation of energy and charge in the context of electrical circuits. One of the laws known as Kirchoff's Current Law deals with the principle of application of conserved energy in electrical circuits. Kirchoff's Current Law states that the sum of all currents entering a junction must equal the sum of all currents leaving the junction.

This basically means, the algebraic sum of currents in a network of conductors(wires) meeting at a point is equal to zero

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3 years ago
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A purse at radius 2.00 m and a wallet at radius 3.00 m travel in uniform circular motion on the floor of a merry-go-round as the
djverab [1.8K]

Complete Question:

A purse at radius 2.00 m and a wallet at radius 3.00 m travel in uniform circular motion on the floor of a merry-go-round as the ride turns.

They are on the same radial line. At one instant, the acceleration of the purse is (2.00 m/s2 ) i + (4.00 m/s2 ) j .At that instant and in unit-vector notation, what is the acceleration of the wallet

Answer:

aw = 3 i + 6 j m/s2

Explanation:

  • Since both objects travel in uniform circular motion, the only acceleration that they suffer is the centripetal one, that keeps them rotating.
  • It can be showed that the centripetal acceleration is directly proportional to the square of the angular velocity, as follows:

       a_{c} = \omega^{2} * r (1)

  • Since both objects are located on the same radial line, and they travel in uniform circular motion, by definition of angular velocity, both have the same angular velocity ω.

       ∴ ωp = ωw (2)

           ⇒ a_{p} = \omega_{p} ^{2} * r_{p} (3)

               a_{w} = \omega_{w}^{2} * r_{w} (4)

  • Dividing (4) by (3), from (2), we have:

        \frac{a_{w} }{a_{p}} = \frac{r_{w} }{r_{p}}

  • Solving for aw, we get:

        a_{w} = a_{p} *\frac{r_{w} }{r_{p} } = (2.0 i + 4.0 j) m/s2 * 1.5 = 3 i +6j m/s2

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Bill Nye- Static Electricity Answer Key?
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A block is sent up a frictionless ramp along which an x axis extends upward. The figure below gives the kinetic energy of the bl
ss7ja [257]
Kinetic energy =1/2 mv^2 

<span>m=2ke/v^2 </span>

<span>m=2(34)/3.6^2 </span>

<span>m=5.24 </span>

<span>force normal = mg </span>
<span>=5.24 x 9.8 </span>
<span>force normal = 51.4N

Thank you for posting your question here at brainly. I hope the answer will help you. Feel free to ask more questions here.


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3 years ago
Two blocks of masses m and M are connected by a string and pass over a frictionless pulley. Mass m hangs vertically, and mass M
horsena [70]

Answer:

sin\theta - \mu_k cos\theta = \frac{m}{M}

sin\theta - \mu_k cos\theta = 1

Explanation:

Force of friction on M mass so that it will move down the inclined plane is given as

F_f = \mu Mgcos\theta

now if it is moving down the inclined plane at constant speed

so we will have

Mgsin\theta - T - \mu mgcos\theta = 0

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so we have

T - mg = 0

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so we have

sin\theta - \mu_k cos\theta = \frac{m}{M}

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sin\theta - \mu_k cos\theta = 1

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3 years ago
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