Earth's radius is 638 x 10^6 m, and it completes one revolution every day. What is...
The earth (of radius of 6371km) rotates on itself at a rate of one revolution every 24 hours and is titled by an angle of 23.5°. Let's assume that the earth is a perfect sphere, which allows us to consider a uniform circular motion 23.5 R equator Figure 1: Earth rotation Everybody on earth share the same time of rotation, but based on their location on the earth, they wil have different velocities. A person located in Hawaii, at an...
Earth Mass/Radius: 1 M= 5.97 x 10^24 kg 1 R= 6.38 x 10^6 m Compare th centripetal force of s 75 kg person standing on the equator of the earth to that of the gravitational force due to the earths rotation. In other words, compute the ratio ( F gravitational/ F centripetal). How many revolutions per day would the earth have to turn to make these forces equal to one another (to makes this ratio exactly 1)?
What is the (magnitude of the) centripetal acceleration (as a multiple of g=9.8 m/s2g=9.8 m/s2) towards the Earth's axis of a person standing on the surface of the Earth at a latitude of 78.9∘78.9∘? Recall that latitude is measured from the equator, and assume the Earth's radius is 6,370 km6,370 km.
The earth (of radius of 6371km) rotates on itself at a rate of one revolution every 24 hours and is titled by an angle of 23.5°. Let's assume that the earth is a perfect sphere, which allows us to consider a uniform circular motion 23.5 R equator Figure 1: Earth rotation Everybody on earth share the same time of rotation, but based on their location on the earth, they wil have different velocities. A person located in Hawaii, at an...
What is the (magnitude of the) centripetal acceleration (as a multiple of g=9.8 m/s2) towards the Earth's axis of a person standing on the surface of the Earth at a latitude of 62.2∘? Recall that latitude is measured from the equator, and assume the Earth's radius is 6,370 km.
2. The earth has a radius of 6*106 m and makes a full revolution around its axis in 24 h. (i) Calculate the angular speed of the earth. ω = 7.3*10-5 s-1 (ii) Calculate the centripetal acceleration of one point on the surface of the earth (at the equator). ac = 0.03 m/s2 (for the records, about 3% of g) THE ANSWERS ARE IN BOLD, COULD YOU PLEASE SHOW THE WORKING NEEDED TO ACHIEVE THESE ANSWERS.
A large merry-go-round completes one revolution every 15.5 s. Compute the acceleration of a child seated on it, a distance of 5.95 m from its center. magnitude
A large merry-go-round completes one revolution every 25.5 s. Compute the acceleration of a child seated on it, a distance of 5.15 m from its center. magnitude ? direction?
A satellite of mass m completes one revolution around the earth at a constant speed s and radius r. Which of the following statements are false? The centripetal force is directed towards the center and is nonzero. The net force is directed inwards towards the earth. The satellite is accelerating. The magnitude of the work done by the earth on the satellite is nonzero. All of the above are true.
Q3) A car moving with a constant speed of 35.0 m/s completes one lap around a circular track of radius 250 m. What is the magnitude of the centripetal acceleration of the car? When the car is at the point 'A', what is its direction of centripetal acceleration (show by making an arrow on the figure)