1ST CORRECT ANSWER GETS BRAINLIEST ASAP Will can jog/walk the mile in 12 minutes and 40 seconds. He is motivated to improve his time by 25%. When he succeeds with this fitness goal he will be able to run the mile in what length of time?

9 minutes and 30 seconds

9 minutes and 50 seconds

10 minutes and 20 seconds

10 minutes and 30 seconds

Aplane flies a distance d from west to east at a constant speed vp , with respect to the air in a time twe . on the return trip, the plane flies a distance d from east to west at a constant speed vp , with respect to the air in a time tew . on both trips the wind blows from west to east at a constant speed va , with respect to the ground 1) what is tew in terms of vp, va, and d, as needed? tew=dvp tew=dvp+va tew=dva tew=dvpâ’va tew=12dvp tew=12dvp+va tew=12dvpâ’va your submissions: b submitted: saturday, january 26 at 6: 49 pm feedback: feedback will be given once all questions have been attempted and the grade cluster button has been pressed. 2) what is twe in terms of vp, va, and d, as needed? twe=dvp twe=dvp+va twe=dva twe=dvpâ’va twe=12dvp twe=12dvp+va twe=12dvpâ’va your submissions: d submitted: saturday, january 26 at 6: 49 pm feedback: feedback will be given once all questions have been attempted and the grade cluster button has been pressed. 3) assuming d = 2300 miles, vp = 400 miles/hr, and twe = 4 hours, what is va, the speed of the wind with respect to the ground? va = 100 miles/hr va = 575 miles/hr va = 200 miles/hr va = 175 miles/hr va = 975 miles/hr 4) once again, assuming d = 2300 miles, vp = 400 miles/hr, and twe = 4 hours, what is tew, the time it takes the plane to fly a distance d miles from east to west? tew = 5.75 hours tew = 11.5 hours tew = 7.67 hours tew = 8 hours tew = 10.2 hours

In a heat engine if 1000 j of heat enters the system the piston does 500 j of work, what is the final internal energy of the system if the initial energy was 2000 j? 1. write the equation 2.list out your known variables 3.plug the numbers into the equations 4.solve 5.write your solution statement that includes initial energy and final

Assume you measured the mass of the cart to be (500 ± 1) g and the mass of the additional mass you put on the cart to be (500 ± 1) g as well. since the scale you are using in the lab cannot measure objects heavier than 600g you will have to sum up individual pieces and propagate the error. so what would be the mass and the standard error of the cart and the mass