A small immersion heater is rated at 285 W . The specific heat of water is 4186 J/kg⋅C∘. Estimate how long it will take to heat a cup of soup (assume this is 250 mL of water) from 20 ∘C to 64 ∘C. Ignore the heat loss to the surrounding environment.

The primary benefit of taking into account a child's interests while selecting exercise activities is that the child will a. play more willing with others b. succeed at their favorite activity c. allow parents to relax afterwards d. gain motivation to exercise select the best answer from the choices provided.

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

What is the electric force acting between two charges of -0.0045 c and -0.0025 c that are 0.0060 m apart? use fe=kq1q2/r^2 and k = 9.00 x 10^9 n*m^2/c^2 a. 1.7 x 10^7 n b. -1.7 x 10^7 n c. -2.8 x 10^9 n d. 2.8 x 10^9 n