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postnew [5]
3 years ago
15

Tidal friction is slowing the rotation of the Earth. As a result, the orbit of the Moon is increasing in radius at a rate of app

roximately 4.0 cm/yr. Assuming this to be a constant rate, how many years will pass before the radius of the Moon's orbit increases by 3.84x 10^7 m (10%)?
Physics
1 answer:
melomori [17]3 years ago
7 0

Answer:

967500000 years

Explanation:

The Speed at which the radius of the orbit of the Moon is increasing is 4 cm/yr

Converting to m

1 m = 100 cm

1\ cm=\frac{1}{100}\ m

4\ cm\y=\frac{4}{100}=0.04\ m/yr

The distance by which the radius increases is 3.84×10⁷ m

Time = Distance / Speed

\text{Time}=\frac{3.87\times 10^7}{0.04}\\\Rightarrow \text{Time}=967500000\ yr

967500000 years will pass before the radius of the orbit increases by 10%.

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Determine the boundary work done by a gas during an expansion process if the pressure and volume values at various states are me
astraxan [27]

Answer:

W=0.217kJ

Explanation:

#Boundary Work is energy expended when a force acts through a displacement

The boundary of work can be calculated as the area under the P-V curve for all the 4 stages measured:

W= \sum_{n=1}^{5} \frac{P_{n+1}+P_{n+2}+P_{n+3}+P_{n+4}+P_{n+5}}{2}\\\\=0.5((290+300)(1.1-1)+(270+290)(1.2-1.1)+(250+270)(1.4-1.3)+(220+250)(1.7-1.4)+(200+220)(2-1.7))\times 10^-^3kJ\\\\=0.217kJ

Hence the boundary of work done by a gas during this 4-stage process is 0.217kJ

7 0
3 years ago
How many electrons must be removed from each of two 4.85-kg copper spheres to make the electric force of repulsion between them
VladimirAG [237]

Answer:

n=2.611*10^{9}electrons

Explanation:

Given data

mass of copper m=4.85 kg

charge of electron qe= -1.6×10⁻¹⁹C

To find

Number of electron n must be removed

Solution

Equate the magnitude of electric force Fe of repulsion between two two spheres to the magnitude of gravitational force of attraction between them

So

F_{e}=F_{g}\\ k\frac{q_{e}q_{e}}{r^{2} }=G\frac{m^{2} }{r^{2} }\\  k(q_{e})^{2}=Gm^{2}\\

where q is charge of each sphere which is equal to number n of removed electrons multiplied by each charge qe

So

k(nq_{e})^{2}=Gm^{2}\\n=\sqrt{\frac{Gm^{2}}{k(q_{e})^{2}} }\\ n=\sqrt{\frac{(6.67*10^{-11}N.m^{2}/kg^{2})(4.85kg)^{2}}{(8.99*10^{9}N.m^{2}/C^{2}  )(1.6*10^{-16}C)^{2}} }\\n=2.611*10^{9}electrons

3 0
3 years ago
what happens to the specific heat capacity of a material if it changes state? i.e. from solid to liquid
Zigmanuir [339]

Answer:

A solid substance at its melting point has less energy than the same mass of the substance when it is a liquid at the same temperature. ... This heat energy allows the change of state to happen, and the temperature remains constant during the process.

Explanation:

6 0
3 years ago
A boulder on the mythical planet mongo drops off a cliff and falls from rest 1000 m in 10.0 s. (A) what's the initial speed of t
Amiraneli [1.4K]
At rest, initial speed zero

x=v(initial) t+ 1/2 at^2
-1000m=0(10) + 1/2 a 10^2
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6 0
3 years ago
A 25 kg child plays on a swing having support ropes that are 2.20 m long. A friend pulls her back until the ropes are 42◦ from t
Semmy [17]

Answer:

A) P.E = 138.44 J

B) The velocity of swing at bottom, v = 3.33 m/s

C) The work done, W = -138.44 J

Explanation:

Given,

The mass of the child, m = 25 Kg

The length of the swing rope, L = 2.2 m

The angle of the swing to the vertical position, ∅ = 42°

A) The potential energy at the initial position ∅ = 42° is given by the relation

                                P.E = mgh joule

Considering h  = 0 for the vertical position

The h at ∅ = 42° is  h = L (1 - cos∅)

                               P.E = mgL (1 - cos∅)

Substituting the given values in the above equation

                               P.E = 25 x 9.8 x 2.2 (1 - cos42°)

                                      = 138.44 J

The potential energy for the child just as she is released, compared to the potential energy at the bottom of the swing is, P.E = 138.44 J

B) The velocity of the swing at the bottom.

At bottom of the swing the P.E is completely transformed into the K.E

                  ∴                 K.E = P.E

                                     1/2 mv² = 138.44

                                     1/2 x 25 x v² 138.44

                                            v² = 11.0752

                                             v = 3.33 m/s

The velocity of the swing at the bottom is, v = 3.33 m/s

C) The work done by the tension in the rope from initial position to the bottom

             Tension on string, T = Force acting on the swing, F

                      W=L\int\limits^0_\phi{F} \, d \phi

                             =L\int\limits^0_\phi{mg.sin \phi} \, d \phi

                            = -Lmg[cos\phi]_{42}^{0}

                            = - 2.2 x 25 x 9.8 [cos0 - cos 42°]

                            = - 138.44 J

The negative sign in the in energy is that the work done is towards the gravitational force of attraction.

The work done by the tension in the ropes as the child swings from the initial position to the bottom of the swing, W = - 138.44 J

3 0
3 years ago
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