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Ierofanga [76]
3 years ago
8

In 1996, astronomers discovered an icy object beyond pluto that was given the designation 1996 tl 66. it has a semimajor axis of

84 au. what is its orbital period in years according to kepler's third law? (round your answer to the nearest whole number; do not include units)
Physics
1 answer:
antoniya [11.8K]3 years ago
7 0

Answer : 2446 years.

Explanation :

Length of semi major axis is, a=84\ au= 1.496\times 10^{11}\ m

According to Kepler's third law, square of time period of an orbit is directly proportional to the cube of the semi major axis.

i.e T^2=\dfrac{4\pi^2}{GM}a^3

where G is gravitational constant

M is  mass of sun, M=1.98\times 10^{30}\ Kg

So, T^2=\dfrac{4\times (3.14)^2}{6.6\times 10^{-11}Nm^2/Kg\times 1.98\times 10^{30}\Kg}

T^2=3\times 10^{-19}\times(84\times 1.496\times 10^{11})^3

T^2=3\times 10^{-19}\times 1984415.6\times 10^{33}

T^2=59532469.8\times 10^{14}\ s

T=7715.7\times 10^7\ s

since, 1\ sec=3.17\times 10^{-8}\ years

So, orbital period is approximately 2446 years.

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Answer:

The inside Pressure of the tank is 4499.12 lb/ft^{2}

Solution:

As per the question:

Volume of tank, V = 0.25 ft^{3}

The capacity of tank, V' = 50ft^{3}

Temperature, T' = 80^{\circ}F = 299.8 K

Temperature, T = 59^{\circ}F = 288.2 K

Now, from the eqn:

PV = nRT                      (1)

Volume of the gas in the container is constant.

V = V'

Similarly,

P'V' = n'RT'                       (2)

Also,

The amount of gas is double of the first case in the cylinder then:

n' = 2n

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n and n' are the no. of moles

Now, from eqn (1) and (2):

\frac{PV}{P'V'} = \frac{nRT}{n'RT'}

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A projectile is launched at an angle of 36.7 degrees above the horizontal with an initial speed of 175 m/s and lands at the same
Softa [21]

Answer:

a) The maximum height reached by the projectile is 558 m.

b) The projectile was 21.3 s in the air.

Explanation:

The position and velocity of the projectile at any time "t" is given by the following vectors:

r = (x0 + v0 · t · cos α, y0 + v0 · t · sin α + 1/2 · g · t²)

v = (v0 · cos α, v0 · sin α + g · t)

Where:

r = position vector at time "t"

x0 = initial horizontal position

v0 = initial velocity

t = time

α = launching angle

y0 = initial vertical position

g = acceleration due to gravity (-9.80 m/s² considering the upward direction as positive).

v = velocity vector at time t

a) Notice in the figure that at maximum height the velocity vector is horizontal. That means that the y-component of the velocity (vy) at that time is 0. Using this, we can find the time at which the projectile is at maximum height:

vy = v0 · sin α + g · t

0 = 175 m/s · sin 36.7° - 9.80 m/s² · t

-  175 m/s · sin 36.7° /  - 9.80 m/s² = t

t = 10.7 s

Now, we have to find the magnitude of the y-component of the vector position at that time to obtain the maximum height (In the figure, the vector position at t = 10.7 s is r1 and its y-component is r1y).

Notice in the figure that the frame of reference is located at the launching point, so that y0 = 0.

y = y0 + v0 · t · sin α + 1/2 · g · t²

y = 175 m/s · 10.7 s · sin 36.7° - 1/2 · 9.8 m/s² · (10.7 s)²

y = 558 m

The maximum height reached by the projectile is 558 m

b) Since the motion of the projectile is parabolic and the acceleration is the same during all the trajectory, the time of flight will be twice the time it takes the projectile to reach the maximum height. Then, the time of flight of the projectile will be (2 · 10.7 s) 21.4 s. However, let´s calculate it using the equation for the position of the projectile.

We know that at final time the y-component of the vector position (r final in the figure) is 0 (because the vector is horizontal, see figure). Then:

y = y0 + v0 · t · sin α + 1/2 · g · t²

0 = 175 m/s · t · sin 36.7° - 1/2 · 9.8 m/s² · t²

0 = t (175 m/s ·  sin 36.7 - 1/2 · 9.8 m/s² · t)

0 = 175 m/s ·  sin 36.7 - 1/2 · 9.8 m/s² · t

-  175 m/s ·  sin 36.7 / -(1/2 · 9.8 m/s²) = t

t = 21.3 s

The projectile was 21.3 s in the air.

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