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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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The energy stored in a capacitor is given by
U= \frac{1}{2} CV^2
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Let's calculate the energy of the first capacitor:
U_1 =  \frac{1}{2} (25\cdot 10^{-6}F)(120 V)=1.5 \cdot 10^{-3}J

And now the energy of the second capacitor:
U_2 =  \frac{1}{2} (5 \cdot 10^{-6}F)(120 V)=3 \cdot 10^{-4}J

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U=U_1 +U_2 = 1.8 \cdot 10^{-3}J
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An 870 N firefighter, F_ff, stands on a ladder that is 8.00 m long and has a weight of 355 N, F_ladder. The weight of the ladder
castortr0y [4]

Answer:

Explanation:

Weight of the ladder is 355N

WL = 355N

The weight of the ladder acts at center of the ladder I.e at 4m from the bottom

Weight of firefighters is 870N

Wf = 870N

The fire fighter is at 6.3m from the bottom of the ladder.

N1 is the normal force exerted  by the wall

N2 is the normal force exerted  by the ground

Using Newton law

Check attachment

ΣFy = 0, since body is in equilibrium

N₂ - WL - Wf = 0

N₂ = WL + Wf

N₂ = 870 + 355

N₂ = 1225 N

This the normal force exerted by the ground on the wall.

Now to get N₁, let take moment about point A

So, before we take the moment we need to make sure that the forces are perpendicular to the plane(ladder), we need to resolve the weight of the ladder, firefighter and the normal of the wall to be perpendicular to the plane.

ΣMa = 0

Clockwise moment is equal to anti-clockwise moment

Moment Is the produce of force and perpendicular distance.

M = F×r

So,

WL•Cos50 × 4 + Wf•Cos50 × 6.3 —N₁•Sin50 × 8 = 0

355•Cos50 × 4 + 870•Cos50 × 6.3 =

N₁•Sin50 × 8

1825.52 + 3523.12 = 6.13N₁

6.13N₁ = 5348.64

N₁ = 5348.64/6.13

N₁ = 872.54 N.

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Determine the energy required to accelerate an electron from (a) 0.500c to 0.900c and (b) 0.900c to 0.990c
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Answer:

0.582 MeV

2.45 MeV

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E_r=0.511\ MeV=Electron rest energy

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W=\left(\frac{1}{\sqrt{1-\frac{v_f^2}{c^2}}}-\frac{1}{\sqrt{1-\frac{v_i^2}{c^2}}}\right)E_r\\\Rightarrow W=\left(\frac{1}{\sqrt{1-0.9^2}}-\frac{1}{\sqrt{1-0.5^2}}\right)0.511\\\Rightarrow W=0.582\ MeV

Energy required is 0.582 MeV

(b) 0.900c to 0.990c

W=\left(\frac{1}{\sqrt{1-\frac{v_f^2}{c^2}}}-\frac{1}{\sqrt{1-\frac{v_i^2}{c^2}}}\right)E_r\\\Rightarrow W=\left(\frac{1}{\sqrt{1-0.99^2}}-\frac{1}{\sqrt{1-0.9^2}}\right)0.511\\\Rightarrow W=2.45\ MeV

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