(b) Fig. 3.2 shows the speed and direction of motion of an object at a point in time, 150.0 m/s
object
Fig. 3.2
The resultant force on the object is zero for 10 seconds.
Deduce the speed and direction of motion after 5 seconds. Indicate the speed and direction
of the object by drawing a labelled arrow next to the object in Fig. 3.3.

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

An architect plans to use solar energy to heat the next house he designs. what principle of absorption and infrared energy can be applied to the design of the new house? how could she apply to those principals?

Air conditioners not only cool air, but dry it as well. suppose that a room in a home measures 5.0m×9.0m×2.4m . if the outdoor temperature is 30 ∘c and the vapor pressure of water in the air is 85 % of the vapor pressure of water at this temperature, what mass of water must be removed from the air each time the volume of air in the room is cycled through the air conditioner? the vapor pressure for water at 30 ∘c is 31.8 torr.