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Svetllana [295]
3 years ago
7

The planet Krypton has a mass of 7.6 × 1023 kg and radius of 1.7 × 106 m. What is the acceleration of an object in free fall nea

r the surface of Krypton? The gravitational constant is 6.6726 × 10−11 N · m2 /kg2 . Answer in units of m/s 2 .
Physics
1 answer:
olga55 [171]3 years ago
5 0

Answer:

17.55 m/s²

Explanation:

Parameters given:

Mass of Krypton, M = 7.6 * 10^23 kg

Radius, R = 1.7 * 10^6 m

Gravitational constant, G = 6.6726 * 10^(-11) Nm²/kg²

Acceleration due to gravity of planet of mass M is given as:

g = GM/R²

Since the object is close to the surface of Krypton, we can say that the distance from the Centre of Krypton is the radius of the planet Krypton.

Therefore,

g = (6.6726 * 10^(-11) * 7.6 * 10^23)/(1.7 * 10^6)²

g = 17.55 m/s²

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Superman is flying 54.5 m/s when he sees
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348.34 m/s. When Superman reaches the train, his final velocity will be 348.34 m/s.

To solve this problem, we are going to use the kinematics equations for constant aceleration. The key for this problem are the equations d=v_{0} t+\frac{at^{2} }{2} and v_{f} =v_{0} +at where d is distance, v_{0} is the initial velocity, v_{f} is the final velocity, t is time, and a is aceleration.

Superman's initial velocity is v_{0}=54.5\frac{m}{s}, and he will have to cover a distance d = 850m in a time t = 4.22s. Since we know d, v_{0} and t, we have to find the aceleration a in order to find v_{f}.

From the equation d=v_{0} t+\frac{at^{2} }{2} we have to clear a, getting the equation as follows: a=\frac{2(d-v_{0}t) }{t^{2} }.

Substituting the values:

a=\frac{2(850m-54.5\frac{m}{s}.4.22s) }{(4.22s)^{2}}=69.63\frac{m}{s^{2}}

To find v_{f} we use the equation v_{f} =v_{0} +at.

Substituting the values:

v_{f} =54.5\frac{m}{s} +(69.63\frac{m}{s^{2}}.4.22s)=348.34\frac{m}{s}

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3 years ago
A diagram of a closed circuit with power source on the left labeled 45 V, two resistors on the top labeled 2.0 Ohms and 3.0 Ohms
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A water wheel rotates with a period of 2.26 s. If the water wheel has a radius of l.94 m,
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Answer:

The correct answer is B)

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When a wheel rotates without sliding, the straight-line distance covered by the wheel's center-of-mass is exactly equal to the rotational distance covered by a point on the edge of the wheel.  So given that the distances and times are same, the translational speed of the center of the wheel amounts to or becomes the same as the rotational speed of a point on the edge of the wheel.

The formula for calculating the velocity of a point on the edge of the wheel is given as

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The answer is left in Meters/Seconds so we will work with our information as is given in the question.

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Cheers!

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