Solution
distance travelled by Chris
\Delta t=\frac{1}{3600}hr.
X_{c}= [(\frac{21+0}{2})+(\frac{33+21}{2})+(\frac{55+47}{2})+(\frac{63+55}{2})+(\frac{70+63}{2})+(\frac{76+70}{2})+(\frac{82+76}{2})+(\frac{87+82}{2})+(\frac{91+87}{2})]\times\frac{1}{3600}
=\frac{579.5}{3600}=0.161miles
Kelly,
\Delta t=\frac{1}{3600}hr.
X_{k}=[(\frac{24+0}{2})+(\frac{3+24}{2})+(\frac{55+39}{2})+(\frac{62+55}{2})+(\frac{71+62}{2})+(\frac{79+71}{2})+(\frac{85+79}{2})+(\frac{85+92}{2})+(\frac{99+92}{2})+(\frac{103+99}{2})]\times\frac{1}{3600}
=\frac{657.5}{3600}
\Delta X=X_{k}-X_{C}=0.021miles
Answer:
<em>The K.E from A to B won't increase...</em>
Explanation:
That's because the P.E from A to B is increasing. The K.E will increase if charge moves from a higher potential to a lower potential i.e., from B to A.
That is the reason there is no effect on net K.E when moving from a potential to same potential over and over (A to C).
What you do is, multiply 16.0 and 12.4 together. then multiply that by 40a
Answer:
d. 37 °C
Explanation:
= mass of lump of metal = 250 g
= specific heat of lump of metal = 0.25 cal/g°C
= Initial temperature of lump of metal = 70 °C
= mass of water = 75 g
= specific heat of water = 1 cal/g°C
= Initial temperature of water = 20 °C
= mass of calorimeter = 500 g
= specific heat of calorimeter = 0.10 cal/g°C
= Initial temperature of calorimeter = 20 °C
= Final equilibrium temperature
Using conservation of heat
Heat lost by lump of metal = heat gained by water + heat gained by calorimeter
The answer is True since thats what usually goes on
