I'm pretty sure that there should be an options to choose. Anyway, I've seen this question before and I know that this is an example of <span>the phi phenomenon.</span>
Answer:
Distance of the object from eye is approx 4.52 m
Explanation:
As we know that the object subtend a small angle on both the eyes which is given as

now we know that the distance between two eyes is given as
d = 7.50 cm
so we have

so here the radius is same as the distance from eye while arc is the distance between two eyes
so we have


Answer:
24) W = 75 [J]; 25) W = 1794[J]; 26) n = 8.8 (times) or 9 (times)
Explanation:
24) This problem can be solved by means of the following equation.

where:
DU = internal energy difference [J]
Q = Heat transfer [J]
W = work [J]
Since there are no temperature changes the internal energy change is equal to zero
DU = 0
therefore:

The work is equal to the heat transfered, W = 75 [J].
25) The heat transfer can be calculated by means of the following equation.
![Q = m*c_{p}*DT\\where:\\m = mass = 0.4[kg]\\c_{p} = specific heat = 897[J/kg*K]\\DT= 5 [C]](https://tex.z-dn.net/?f=Q%20%3D%20m%2Ac_%7Bp%7D%2ADT%5C%5Cwhere%3A%5C%5Cm%20%3D%20mass%20%3D%200.4%5Bkg%5D%5C%5Cc_%7Bp%7D%20%3D%20specific%20heat%20%3D%20897%5BJ%2Fkg%2AK%5D%5C%5CDT%3D%205%20%5BC%5D)
Q = 0.4*897*5 = 1794[J]
Work is equal to heat transfer, W = 1794[J]
26) Each time the bag falls the potential energy is transformed into heat energy, which is released into the environment. In this way the potential energy is equal to the developed heat.

where:
m = mass = 0.5[kg]
g = gravity = 9.81[m/s^2]
h = 1.5 [m]
![E_{p}=0.5*9.81*1.5\\E_{p}=7.36[J]](https://tex.z-dn.net/?f=E_%7Bp%7D%3D0.5%2A9.81%2A1.5%5C%5CE_%7Bp%7D%3D7.36%5BJ%5D)
The heat developed can be calculated by means of the following equation.
![Q=m*c_{p}*DT\\Q=0.5*130*1\\Q=65[J]](https://tex.z-dn.net/?f=Q%3Dm%2Ac_%7Bp%7D%2ADT%5C%5CQ%3D0.5%2A130%2A1%5C%5CQ%3D65%5BJ%5D)
The number of times will be calculated as follows
n = 65/7.36
n = 8.8 (times) or 9 (times)
A: decreases in specific heat capacity
Answer:
a)906.5 Nm^2/C
b) 0
c) 742.56132 N•m^2/C
Explanation:
a) The plane is parallel to the yz-plane.
We know that
flux ∅= EAcosθ
3.7×1000×0.350×0.700=906.5 N•m^2/C
(b) The plane is parallel to the xy-plane.
here theta = 90 degree
therefore,
0 N•m^2/C
(c) The plane contains the y-axis, and its normal makes an angle of 35.0° with the x-axis.
therefore, applying the flux formula we get
3.7×1000×0.3500×0.700×cos35°= 742.56132 N•m^2/C