Answer:
The specific heat capacity of the zinc metal measured in this experiment is 0.427 J/g.°C
Explanation:
From the experimental data, the water loses heat because its initial temperature is greater than the final temperature of the mixture. On the other hand, the zinc metal gains heat because its initial temperature is less than the final temperature of the mixture
Heat loss by water = Heat gain by zinc metal
m1C1(T1 - T3) = m2C2(T3 - T2)
m1 is mass of water = 55.4 g
C1 is specific heat capacity of water = 4.2 J/g.°C
m2 is mass of zinc metal = 23.4 g
C2 is specific heat capacity of zinc metal
T1 is the initial temperature of water = 99.61 °C
T2 is the initial temperature of zinc metal = 21.6 °C
T3 is the final temperature of the mixture = 96.4 °C
55.4×4.2(99.61 - 96.4) = 23.4×C2(96.4 - 21.6)
746.9028 = 1750.32C2
C2 = 746.9028/1750.32 = 0.427 J/g.°C
Answer:
a) Ep = 5886[J]; b) v = 14[m/s]; c) W = 5886[J]; d) F = 1763.4[N]
Explanation:
a)
The potential energy can be found using the following expression, we will take the ground level as the reference point where the potential energy is equal to zero.
![E_{p} =m*g*h\\where:\\m = mass = 60[kg]\\g = gravity = 9.81[m/s^2]\\h = elevation = 10 [m]\\E_{p}=60*9.81*10\\E_{p}=5886[J]](https://tex.z-dn.net/?f=E_%7Bp%7D%20%3Dm%2Ag%2Ah%5C%5Cwhere%3A%5C%5Cm%20%3D%20mass%20%3D%2060%5Bkg%5D%5C%5Cg%20%3D%20gravity%20%3D%209.81%5Bm%2Fs%5E2%5D%5C%5Ch%20%3D%20elevation%20%3D%2010%20%5Bm%5D%5C%5CE_%7Bp%7D%3D60%2A9.81%2A10%5C%5CE_%7Bp%7D%3D5886%5BJ%5D)
b)
Since energy is conserved, that is, potential energy is transformed into kinetic energy, the moment the harpsichord touches water, all potential energy is transformed into kinetic energy.
![E_{p} = E_{k} \\5886 =0.5*m*v^{2} \\v = \sqrt{\frac{5886}{0.5*60} }\\v = 14[m/s]](https://tex.z-dn.net/?f=E_%7Bp%7D%20%3D%20E_%7Bk%7D%20%5C%5C5886%20%3D0.5%2Am%2Av%5E%7B2%7D%20%5C%5Cv%20%3D%20%5Csqrt%7B%5Cfrac%7B5886%7D%7B0.5%2A60%7D%20%7D%5C%5Cv%20%3D%2014%5Bm%2Fs%5D)
c)
The work is equal to
W = 5886 [J]
d)
We need to use the following equation and find the deceleration of the diver at the moment when he stops his velocity is zero.
![v_{f} ^{2}= v_{o} ^{2}-2*a*d\\where:\\d = 2.5[m]\\v_{f}=0\\v_{o} =14[m/s]\\Therefore\\a = \frac{14^{2} }{2*2.5} \\a = 39.2[m/s^2]](https://tex.z-dn.net/?f=v_%7Bf%7D%20%5E%7B2%7D%3D%20v_%7Bo%7D%20%5E%7B2%7D-2%2Aa%2Ad%5C%5Cwhere%3A%5C%5Cd%20%3D%202.5%5Bm%5D%5C%5Cv_%7Bf%7D%3D0%5C%5Cv_%7Bo%7D%20%3D14%5Bm%2Fs%5D%5C%5CTherefore%5C%5Ca%20%3D%20%5Cfrac%7B14%5E%7B2%7D%20%7D%7B2%2A2.5%7D%20%5C%5Ca%20%3D%2039.2%5Bm%2Fs%5E2%5D)
By performing a sum of forces equal to the product of mass by acceleration (newton's second law), we can find the force that acts to reduce the speed of the diver to zero.
m*g - F = m*a
F = m*a - m*g
F = (60*39.2) - (60*9.81)
F = 1763.4 [N]
Answer:
Maximum height, h = 11.32 meters
Explanation:
It is given that,
The baseball is thrown directly upward at time, t = 0
Initial speed of the baseball, u = 14.9 m/s
Ignoring the resistance in this case and using a = g = 9.8 m/s²
We have to find the maximum height the ball reaches above where it leaves your hand. Let the maximum height is h. Using third equation of motion as :

At maximum height, v = 0
and a = -g = -9.8 m/s²


h = 11.32 meters
Hence, the maximum height of the baseball is 11.32 meters.