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  1. M

    A container holds 8.0kg of water at 25degC...

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

    Check my answer for power output question

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

    Check my answer for power output question

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

    Check my answer for power output question

    im going to retry my answer. will post shortly
  5. M

    Check my answer for power output question

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

    Check my answer for power output question

    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...
  7. M

    A container holds 8.0kg of water at 25degC...

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

    Calculate the Net Force

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

    Calculate the Net Force

    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
  10. M

    Calculate the Net Force

    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...
  11. M

    Calculate the Net Force

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

    Calculate the Net Force

    "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...
  13. M

    Calculate the Net Force

    is it sin 55/33=sin/32 sin-1(o.7943) =52.59 or 53deg you said that the B= A+35deg so that would mean that A =B-35 deg? 53-35=18 deg also, how do my other answers look?
  14. M

    Calculate the Net Force

    my third answer is : 180deg-56deg=124 Fnet^2= F1^2+F^2 -2(F1)(F2) cosC Fnet^2=15^2=12N^2-2(15N)(12N)cos124 =24N sin/12=sin56/24 =sin-1(0.4145) =24.5 24N [W 24deg S]
  15. M

    Calculate the Net Force

    i've tried the second answer: 10N=8.0N[N] 10N-8.0=2N Fnet^2= F1^2+F^2 -2(F1)(F2) cosC Fnet^2=2n^2+17n^2-2*2n*17Ncos45 Fnet^2=244.9 square root of 244.9=15.6N or 16N sin/17N=sin45/15.6 =12/15.6 sin-1(o.77) =50deg 50deg-90deg=40deg 16N [W 40deg S]
  16. M

    Calculate the Net Force

    I don't know how to write down the symbols for the angles, how do you do that? anyways, after doing sin 55/33=sin/32 I got 52.59deg. is this right?
  17. M

    Calculate the Net Force

    angles are my weakness! would i do sin55/33=sin/32 ?
  18. M

    Calculate the Net Force

    I knew that the angle part couldn't be right! so this would be my answer now that i have changed the angle to 55deg: 90deg-35=55deg Fnet^2= F1^2+F^2 -2(F1)(F2) cosC Fnet^2=32N^2+38^2 -2(32N)(38N)cos55 Fnet^2= 1073 square root of 1073 = 32.7N or 33N sin 55/33N=sin/38 sin=38N sin55/33N...
  19. M

    Calculate the Net Force

    this is what i get by using the method you suggested to me: 180deg-35deg=145deg Fnet^2= F1^2+F^2 -2(F1)(F2) cosC Fnet^2=32N^2+38^2 -2(32N)(38N)cos145 Fnet^2= 4460.17 square root of 4460.17 = 66.78 or 67N sin 145/67N=sin/38 sin=38N sin145/67N sin=21.79/67N sin-1(0.3253) =18.9 or 19deg...
  20. M

    Calculate the Net Force

    The way that it is shown to me is using sinA/A=sinB/B=sinC/C or tan=opp/adj so what i did was 62.2N/sin125=32.0N/sin =24.8 or 25 deg
  21. M

    Calculate the Net Force

    Also, can anyone help me with the other two diagrams? I uploaded them in my initial post.
  22. M

    Calculate the Net Force

    i need to find the angle in order to figure out the direction. what formula do i use?
  23. M

    Calculate the Net Force

    I left the vectors out of the equation, but i do know that i need to put them into my final answer. we haven't been shown how to break things down into components in this unit, so we wouldn't be required to use it. The only examples shown to us how to find net force is this formula...
  24. M

    Calculate the Net Force

    Homework Statement Homework Equations The Attempt at a Solution 38) c^2=a^2=b^2-2abcosC c=([32.0N]^2=[38.0N]^2-2[32N]*[38N]cos125)1/2 c=62.15N sin law 32/sinA=62.2N/sin125 =24.8 or 25 i have no idea how to do the other two questions. please help ASAP!!!
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