Two students are balancing on a 10 m seesaw . the seesaw is designed so that each side of the seesaw is 5 m long. the student on the left weighs 60 kg and is sitting three meters away from from the fulcrum at the center. the student on the right weighs 45 kg. the seesaw is parallel to the ground. the Mass of the board is evenly distributed so that is center of mass is over the fulcrum. imagine that two students are sitting on the seesaw so that the torque is 0 N*m. which of the following changes will alter the torque of the seesaw? a. another student stands perfectly on the center of the seesaw.
b. both students move toward the center by one meter.
c. the heavier student moves forward 1 m, while the lighter student moves forward 1.33 m.
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initial balance

60 * 3 = 45 x

x = 4

so the student on the right is 4 meters from the fulcrum.

a. There is no torque resulting from more mass at the fulcrum.

b.

60 * (3-1) = 120

45 * (4-1) = 135, the student on the right hits the ground

c,

60 * (3-1) = 120

45 (4-1.33) = 120.15 pretty close to balanced.

Explanation: It's b

a high pressure system is a dense air mass that is usually cooler and drier than the surrounding air. a low pressure system is a less dense air mass that is usually wetter and warmer than the surrounding air.

in general, areas that experience high atmospheric pressure also experience fair weather. low pressure systems can cause the formation of clouds and storms. air usually flows from areas of high pressure to areas of low pressure..

im srry but ur wrong the answer is d a smooth rock, some red dirt, and an empty bird's nest.