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IRINA_888 [86]
4 years ago
8

The existence of the dwarf planet Pluto was proposed based on irregularities in Neptune's orbit. Pluto was subsequently discover

ed near its predicted position. But it now appears the discovery was fortuitous, because Pluto is small and the irregularities in Neptune's orbit were not well known. To illustrate that Pluto has a minor effect on the orbit of Neptune, calculate the acceleration due to gravity at Neptune, in m/s2, due to Pluto when they are 4.50x1012 m apart, as they are at present. The mass of Pluto is 1.4x1022 kg.
Group of answer choices
4.6x10-14
5.7x10-15
6.8x10-16
7.9x10-17
Physics
1 answer:
givi [52]4 years ago
4 0

Answer:

Acceleration due to gravity, a=4.61\times 10^{-14}\ m/s^2

Explanation:

It is given that,

Mass of Pluto, m=1.4\times 10^{22}\ kg

Distance between Neptune and Pluto, r=4.5\times 10^{12}\ m

The force of gravity is balanced by the gravitational force between Neptune and Pluto. It is given by :

a=\dfrac{Gm}{r^2}

a=\dfrac{6.67\times 10^{-11}\times 1.4\times 10^{22}}{(4.5\times 10^{12})^2}

a=4.61\times 10^{-14}\ m/s^2

So, the acceleration due to gravity at Neptune due to Pluto is 4.61\times 10^{-14}\ m/s^2. Hence, this is the required solution.  

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Answer:

1) The force Christian can exert on his bicycle before picking up the the cargo is 529.74 N

2) The force Christian can exert on his bicycle after picking up the the cargo is 647.46 N

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Explanation:

The given parameters are;

The mass of Christian and his bicycle = 54 kg

The mass of the cargo = 12 kg

1) The force Christian can exert on his bicycle before picking up the the cargo = Mass of Christian and his bicycle × Acceleration due to gravity

∴ The force Christian can exert on his bicycle before picking up the the cargo = 54 kg × 9.81 m/s² = 529.74 N

2) The force Christian can exert on his bicycle after picking up the the cargo = (54 + 12) kg × 9.81 m/s² = 647.46 N

Therefore, Christian has to exert more force on his bike after picking up the cargo.

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A uniform solid disk rolls without slipping down an incline making an angle θ with the horizontal. What is its acceleration? (En
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Answer:

aCM = (2/3)*g*Sin θ

Explanation:

Consider a uniform solid disk having mass M,  radius R and rotational inertia I  about its center of mass, rolling without  slipping down an inclined plane.

In order to get the linear acceleration of the object’s center of mass, aCM ,

down the incline,  we analyze this as follows:

The force of gravity (W = Mg) acting straight down  is resolved into components parallel and  perpendicular to the incline.

Since the object rolls without  slipping there is a force of  friction (Ff) acting on the object,  at it’s point of contact with the  incline, in the direction up  the incline.

Newton’s 2nd Law gives then for acceleration down the incline

∑Fx' = m*aCM   ⇒    m*g*Sin θ - Ff = m*aCM

The force of friction also causes a torque around the center of mass

having lever arm R so we can also write

τ = R*Ff = I*α

Solving for the friction,    Ff = I*α / R

This is used in the expression  derived from the 2nd Law:

m*g*Sin θ - Ff = m*g*Sin θ - (I*α / R) = m*aCM

The objects angular acceleration is related to the linear acceleration  of the edge that contacts the incline by

a = R*α

Since the object rolls without  slipping this has the same  magnitude as aCM so we have  that

α = aCM / R

Using this in

m*g*Sin θ - (I*α / R) = m*g*Sin θ - (I*(aCM / R) / R) = m*aCM

⇒  aCM = (m*g*Sin θ*R²) / (I + m*R²)

if I = (1/2)*m*R²   (for a uniform solid disk)

we get

aCM = (2/3)*g*Sin θ

6 0
3 years ago
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