isetan:

Hi, another science question--


A basketball player lifts and throws a ball upwards (at an angle) and lands into the net. Is the potential energy greater when the ball leaves his hand or is it when the ball almost touches the net? Tks.

http://i62.tinypic.com/2rr87z4.png\">

The (gravitational) potential energy is greater when the ball touches the net. The higher an object is from the ground, the greater its gravitational potential energy relative to the ground.

The ball leaves his hand with a starting amount of kinetic energy, which decreases as the ball ascends and slows down. Some of the ball's kinetic energy is converted to gravitational potential energy.

The question is clear enough as it is written. Kinetic energy is energy an object has due to its motion. Kinetic energy is not a type of potential energy. There are many kinds of potential energy (chemical, elastic, ...) but it is clear from context that the question refers to gravitational potential energy.

cookieandmonster:

Please help with this question : Tom and Jerry were both going to Jurong Point from the same place. Jerry started the journey at 12 noon and took 4 minutes to reach Jurong Point. Tom started his journey a minute later and took 3 minutes to reach. At what time would Tom catch up with Jerry?

cookieandmonster:

A car and a coach travelled from Town A to Town B. The coach left Town A at 0648 and it took 5 hours to reach Town B. The car started 30 minutes later than the coach and it took 4 hours to reach Town B. At what time did the car catch up with the coach?

Their positions between the starting point and Jurong Point and between Towns A and B at various times are labelled below:

http://i58.tinypic.com/i2oqbl.png\">

Hence Tom catches up with Jerry at 12.04 pm (only when they both reach Jurong Point). The car catches up with the coach at 9.18 am.

[A more abstract way of thinking about the car and the coach: The problem is space- and time-symmetric around the midpoint of Towns A and B and the 5-hour duration of the problem. (Switch Towns A and B and reverse the paths of both, and you still have the same situation. The coach travels for 5 hours. The car waits 30 min, travels for 4 hours and waits 30 min.) This means that at the middle of the 5 hours, both car and coach must be at the midpoint of Towns A and B, so the car catches up with the coach at this point, at 9.18 am. If the vehicles were anywhere other than the midpoint, it would break the symmetry of the problem.]

genny:

The diagram below shows a rectangle STUV. ...

Please help to reply. Thanks. The answer is 33 cm2 but don't know how to get there.

Since ΔSTW takes up 1/4 of the area of rectangle STUV,
area of rectangle STUV = 120 cm²
area of ΔTWU = area of STUV − area of ΔSTW − area of ΔVWU − area of ΔSWV = 15 cm²
area of ΔXWU = 3/5 × 30 cm² (since XU is 3/5 of VU) = 18 cm²
shaded area = 15 + 18 = 33 cm²

Jamesbond:

The diagram below shows the path travelled by a pebble after a boy threw it from P. ...

My ans is 1. Given ans is 2. Can anybody explain?

As written, A, B and C are false, and D is true. A is false because the pebble has gravitational potential energy, relative to the ground, before it started moving. B is false. The pebble has kinetic energy, relative to the ground, everywhere along its trajectory in the air. The highest point of the pebble's trajectory in the air (near to Q) is the point of minimum kinetic energy along that trajectory. At this point, the pebble is not moving in the up/down direction, but is moving to the right, and so has nonzero kinetic energy. Relative to the ground, kinetic energy is only zero at the endpoints of the trajectory: before it started moving, and when it comes to a stop after it hits the ground.

Such a statement would be true: \"From P to R [or S], the amount of kinetic energy of the pebble decreased [before it reaches the highest point of the trajectory, near to Q], and then increased [after it passes the highest point of the trajectory].\"

Jamesbond:

There are identical wind turbines located at four different areas, P, Q, R and S. ...
My ans is 3. Given ans is 4. Can anybody explain?

There is camera distortion but it seems to me like the graph is consistent with the statement that the highest point at Area Q is in Jan. In such cases, it would be good to overlay a set square on the graph.