Find the centripetal acceleration of a person standing on the Earth equator. Consider the radius of the Earth to be 6400000 m and the period of rotation to be 1 day O0 0.043 m/sec2 O0 0.34 m/sec2 0.43 m/sec2
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- The International Space Station orbits the Earth in a circle with a radius of approximately 6688185 m (which is only a few hundred miles above Earth's surface). At that distance, the acceleration due to gravity is only 8.87m / (s ^ 2) what is the velocity of the space station in this orbit? Your answer should be in units of m / sA 5 kg mass is spun around in a circle of radius 2m with a period of 7 s. What equation would you use to calculate the centripetal acceleration? Calculate the centripetal acceleration. Possible Formulas that can be used to answer the question: v=(2πr)/T ac=v2/r ac=(4π2r)/T2 Fc=mac Fg=mg F=(Gm1m2)/d2 g=Gm/r2 T2=(4π2/Gm)r3 v=√(Gm)/r g=9.80m/s2 G=6.67x10-11 (N∙m2)/kg2The earth rotates once per day about an axis passing through the north and south poles, an axis that is perpendicular to the plane of the equator. Assuming the earth is a sphere with a radius of 6.38 x 106 m, determine the speed and centripetal acceleration of a person situated (a) at the equator and (b) at a latitude of 50.0° north of the equator. (a) v = i (b) ac= V = ac = i i i Units Units Units Units 8 Re
- An object in a circular orbit is moving at a constant speed. This is called centripetal acceleration. Yet the speed is constant. How is it accelerating?The earth rotates once per day about an axis passing through the north and south poles, an axis that is perpendicular to the plane of the equator. Assuming the earth is a sphere with a radius of 6.38 x 10 m, determine the speed and centripetal acceleration of a person situated (a) at the equator and (b) at a latitude of 65.0 ° north of the equatorA 0.85 kg ball experiences a centripetal acceleration of 6.5 m/s2 as it goes around a 1.2 m radius circle. How fast was it travelling?
- it's urgentA tennis ball connected to a string of radius 0.80 m is spun around in a vertical, circular path at a uniform speed of 2.5 m/s. When the ball is at the bottom of the circle, what is the magnitude of its acceleration? Your answer needs to have 2 significant figures, including the negative sign in your answer if needed. Do not include the positive sign if the answer is positive. No unit is needed in your answer, it is already given in the question statement.