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MissTica
3 years ago
7

A new planet is discovered beyond Pluto at a mean distance to the sun of 4004 million miles. Using Kepler's third law, determine

an estimate for the time T to travel around the sun in an orbit.
Physics
1 answer:
AVprozaik [17]3 years ago
7 0

Answer:

103239.89 days

Explanation:

Kepler's third law states that the square of the orbital period of a planet is directly proportional to the cube of the semi-major axis of its orbit.

                               a³ / T² = 7.496 × 10⁻⁶  (a.u.³/days²)

where,

a is the distance of the semi-major axis in a.u

T is the orbit time in days

Converting the mean distance of the new planet to astronomical unit (a.u.)

                       1 a.u = 9.296 × 10⁷ miles        

                                      \frac{4004 * 10^{6}}{9.296 * 10^{7}}  =  43.07\ a.u.

Substituting the values into Kepler's third law equation;

                                    \frac{(43.07)^{3}}{T^{2}}  =  7.496 * 10^{-6}  

                                    T^{2} = \frac{(43.07)^{3}}{7.496 * 10^{-6}} (days)²

                                    T^{2} = \sqrt{\frac{(43.07)^{3}}{7.496 * 10^{-6}}}

                                    T = 103239.89 days

An estimate time T for the new planet to travel around the sun in an orbit is 103239.89 days

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An archer fires an arrow at a castle 230 m away. The arrow is in flight for 6 seconds, then hits the wall of the castle and stic
Vika [28.1K]

Answer:

(a). The initial speed of the arrow is 49.96 m/s.

(b). The angle is 39.90°.

Explanation:

Given that,

Horizontal distance = 230 m

Time t = 6 sec

Vertical distance = 16 m

We need to calculate the horizontal component

Using formula of horizontal component

R =u\cos\theta t

Put the value into the formula

\dfrac{230}{6} = u\cos\theta

u\cos\theta=38.33.....(I)

We need to calculate the height

Using vertical component

H=u\sin\theta t-\dfrac{1}{2}gt^2

Put the value in the equation

16 =u\sin\theta\times6-\dfrac{1}{2}\times9.8\times6^2

u\sin\theta=\dfrac{16+9.8\times18}{6}

u\sin\theta=32.06.....(II)

Dividing equation (II) and (I)

\dfrac{u\sin\theta}{u\cos\theta}=\dfrac{32.06}{38.33}

\tan\theta=0.8364

\theta=\tan^{-1}0.8364

\theta=39.90^{\circ}

(a). We need to calculate the initial speed

Using equation (I)

u\cos\theta\times t=38.33

Put the value into the formula

u =\dfrac{230}{6\times\cos39.90}

u=49.96\ m/s

(b). We have already calculate the angle.

Hence, (a). The initial speed of the arrow is 49.96 m/s.

(b). The angle is 39.90°.

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3 years ago
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kherson [118]

Answer:

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Explanation:

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3 years ago
A pitcher throws a 0.140 kg baseball, and it approaches the bat at a speed of 35.0 m/s. The bat does Wnc = 75.0 J of work on the
Eva8 [605]

Answer:

The speed of the ball is 42.5 m/s

Explanation:

The initial kinetic energy of the ball is:

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Using kinematic equation we can find the speed of the ball after being 25 m above the point of collision:

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V_f=42.5m/s

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