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elixir [45]
2 years ago
3

Suppose our sun had 4 times its present mass but the earth orbited it at the same distance as it presently does. What would be t

he length of the year on the earth under those conditions?A.1/2 as long as the present yearB. Twice as long as the present year.C. four times as long as the present year.D. the same as the present year.E. 1/4 as long as the present year
Physics
1 answer:
Volgvan2 years ago
6 0

M₁ = initial mass of sun = M

M₂ = final mass of sun = 4 M

T₁ = initial time period

T₂ = final time period

r₁ = initial distance from sun = r

r₂ = final distance from sun = r

Using Kepler's third law

T₁² = 4π²r₁³/(GM₁)                                    eq-1

Using Kepler's third law

T₂² = 4π²r₂³/(GM₂)                                 eq-2

Dividing eq-1 by eq-2

T₁² /T₂² = (4π²r₁³/(GM₁))/(4π²r₂³/(GM₂))

T₁² /T₂² = (M₂/M₁) (r₁³/r₂³)

T₁² /T₂² = (4M/M) (r³/r³)

T₁² /T₂² = 4

T²₁ = 4 T²₂

Taking square root both side

T₁ = 2 T₂

T₂ = T₁ /2

hence the correct choice is

A. 1/2 as long as the present year

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Brian and Jack decided to investigate which water fountain at school has the coldest water. The two boys take measurements using
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Answer:

b

Explanation:

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2 years ago
An aluminum rod and a nickel rod are both 5.00 m long at 20.0 degree Celsius. The temperature of each is raised to 70.0 degrees
vitfil [10]

Answer:

0.002925 m

Explanation:

Lt = LO(1 +α Δt ) here Lt is total length Lo is original length α is coefficient of linear expansion and Δt is change in temperature

<h2>for aluminium</h2>

α=25×10^-6

Lt = 5(1+25×10^-6×(70-20))

Lt = 5 (1+25×10^-6×50)

Lt = 5 ( 1+0.00125)

Lt = 5×1.00125

Lt =5.00625 m

<h2>for nickel </h2>

α=13.3×10^-6

Lt =5(1+13.3×10^-6×50)

Lt = 5(1+0.000665)

Lt =5.003325 m

hence difference in length =5.00625-5.003325

                                           = 0.002925 m

3 0
2 years ago
A moving sidewalk 95 m in length carries passengers at a speed of 0.53 m/s. One passenger has a normal walking speed of 1.24 m/s
Archy [21]

Answer:

a) t = 1.8 x 10² s

b) t = 54 s

c) t = 49 s

Explanation:

a) The equation for the position of an object moving in a straight line at constan speed is:

x = x0 + v * t

where

x = position at time t

x0 = initial position

v = velocity

t = time

In this case, the origin of our reference system is at the begining of the sidewalk.

a) To calculate the time the passenger travels on the sidewalk without wlaking, we can use the equation for the position, using as speed the speed of the sidewalk:

x = x0 + v * t

95 m = 0m + 0. 53 m/s * t

t = 95 m/ 0.53 m/s

t = 1.8 x 10² s

b) Now, the speed of the passenger will be her walking speed plus the speed of th sidewalk (0.53 m/s + 1.24 m/s = 1.77 m/s)

t = 95 m/ 1.77 m/s = 54 s

c) In this case, the passenger is located 95 m from the begining of the sidewalk, then, x0 = 95 m and the final position will be x = 0. She walks in an opposite direction to the movement of the sidewalk, towards the origin of the system of reference ( the begining of the sidewalk). Then, her speed will be negative ( v = 0.53 m/s - 2*(1.24 m/s) = -1.95 m/s. Then:

0 m = 95 m -1.95 m/s * t

t = -95 m / -1.95 m/s = 49 s

3 0
2 years ago
Astronomers have discovered a new planet called "Xandar" beyond the orbit of Pluto (No, not really but I need a fake planet for
Burka [1]

Answer:

m = 1.82E+23 kg

Explanation:

G = universal gravitational constant = 6.67E-11 N·m²/kg²

r = radius of orbit = 72,600 km = 7.26E+07 m

C = circumference of orbit = 2πr = 4.56E+08 m

P = period of orbit = 12.9 d = 1,114,560 s

v = orbital velocity of satellite Jim = C/P = 409 m/s

m = mass of Xandar = to be determined

v = √(Gm/r)

v² = [√(Gm/r)]²

v² = Gm/r

rv² = Gm

rv²/G = m

m = rv²/G

mG = universal gravitational constant = 6.67E-11 N·m²/kg²

r = radius of orbit = 72,600 km = 7.26E+07 m

C = circumference of orbit = 2πr = 4.56E+08 m

P = period of orbit = 12.9 d = 1,114,560 s

v = orbital velocity of satellite Jim = C/P = 409 m/s

m = mass of Xandar = to be determined

v = √(Gm/r)

v² = [√(Gm/r)]²

v² = Gm/r

rv² = Gm

rv²/G = m

m = rv²/G

m = 1.82E+23 kg

3 0
2 years ago
A material that has a fracture toughness of 33 MPa.m0.5 is to be made into a large panel that is 2000 mm long by 250 mm wide and
scoray [572]

Answer:

F_{allow} = 208.15kN

Explanation:

The word 'nun' for thickness, I will interpret in international units, that is, mm.

We will begin by defining the intensity factor for the steel through the relationship between the safety factor and the fracture resistance of the panel.

The equation is,

K_{allow} =\frac{K_c}{N}

We know that K_c is 33Mpa*m^{0.5} and our Safety factor is 2,

K_{allow} = \frac{33Mpa*m^{0.5}}{2} = 16.5MPa.m^{0.5}

Now we will need to find the average width of both the crack and the panel, these values are found by multiplying the measured values given by 1/2

<em>For the crack;</em>

\alpha = 0.5*L_c = 0.5*4mm = 2mm

<em>For the panel</em>

\gamma = 0.5*W = 0.5*250mm = 125mm

To find now the goemetry factor we need to use this equation

\beta = \sqrt{sec(\frac{\pi\alpha}{2\gamma})}\\\beta = \sqrt{sec(\frac{2\pi}{2*125mm})}\\\beta = 1

That allow us to determine the allowable nominal stress,

\sigma_{allow} = \frac{K_{allow}}{\beta \sqrt{\pi\alpha}}

\sigma_{allow} = \frac{16.5}{1*\sqrt{2*10^{-3} \pi}}

\sigma_{allow} = 208.15Mpa

So to get the force we need only to apply the equation of Force, where

F_{allow}=\sigma_{allow}*L_c*W

F_{allow} = 208.15*250*4

F_{allow} = 208.15kN

That is the maximum tensile load before a catastrophic failure.

4 0
2 years ago
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