Recent content by MaryBarnes

awesome! thanks for the help!

i attempted the answer: Q total=mcwΔt+msLf =8.0kg*4200j/kg*(100°C-25°C)+2.0kg*2.3*106 =7.12*106J

I see now. Thanks!

its 20, but my answer has a negative. i was going to edit it but don't know how to

=0.25kg*2100J/(kg °c)*((-10°C) - (-30) °C) =525*20°C =10500J P=10500J/150s =70w is this ok?

im going to retry my answer. will post shortly

3.3*105 is the latent heat of fusion for water that I've been given, now i see where i may have messed up.

q=mLv

Homework Statement a 0.25 kg piece of ice at -30°C is warmed by an electric heater and the following graph of temperature is produced. -use the information on the graph to determine the power output of the heater the graph shows that the ice went from -30degC to -10 degC in 150s Homework...

Homework Statement how much energy must be added to the water to create 2.0 kg of steam? Homework EquationsThe Attempt at a Solution I have no idea how I am supposed to figure the answer out. what formulas do I use?

Ok, thanks for mentioning that. Thank you for all of your help! Now I actually understand how work with these equations!

I tried the third answer using the sin and cosine law. Is this right? fnet^2=2N^2+17N^2 -2(2N)(17N)cos45 Fnet=square root of:244.9 Fnet=15.6N or 16N sin45/15.6N=sin/2N sin=2Nsin45/15.6N sin-1(0.0906) =5.2deg 45deg-5.2deg=40

if i subtract 24deg from 24 deg i get 0? is this really the answer? for the third question I'm just going to break it down into components because that way makes way more sense to me. maybe that's not the expectation, but i don't think id loose any marks for doing so. Thank you for all your...

so now I've tried tried what you suggested and I got 24.48 deg. Is this right?

"In your third answer, the magnitude of the net force ##F_{net}=24N## is correct, but the angle is incorrect because you have used the Sine Rule in some triangle that doesn't exist. I suggest you to draw a sketch for each problem with a completed parallelogram and angles properly annotated so...