Why vacuum flask is known as thermos flak?

Asubmarine has a "crush depth" (that is, the depth at which water pressure will crush the submarine) of 250 m. what is the approximate pressure (water plus atmospheric) at this depth? (recall that the density of seawater is 1025 kg/m3, g = 9.81 m/s2, and 1 kg/(ms2) = 1 pa = 9.8692 10-6 atm.) a. 34.8 atm b. 24.8 atm c. 25.8 atm d. 7.8 atm

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

Aleaky 10-kg bucket is lifted from the ground to a height of 15 m at a constant speed with a rope that weighs 0.7 kg/m. initially the bucket contains 45 kg of water, but the water leaks at a constant rate and finishes draining just as the bucket reaches the 15-m level. find the work done. (use 9.8 m/s2 for g.) show how to approximate the required work by a riemann sum. (let x be the height in meters above the ground. enter xi* as xi.) lim n → ∞ n (9.8)(45−3 xi) i = 1 δx express the work as an integral. 0 dx evaluate the integral. (round your answer to the nearest integer.) j