Answer: The correct option is (B) 0.875 Ω.
Explanation:
Equivalent resistance: It is defined as the sum of the all resistance present in the circuit.
There are two conditions:
(1) Resistors in series
Equivalent resistance, ![R_s=R_1+R_2+R_3+R_4........R_n](https://tex.z-dn.net/?f=R_s%3DR_1%2BR_2%2BR_3%2BR_4........R_n)
(2) Resistors in parallel
Equivalent resistance, ![R_p=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}+\frac{1}{R_4}........\frac{1}{R_n}](https://tex.z-dn.net/?f=R_p%3D%5Cfrac%7B1%7D%7BR_1%7D%2B%5Cfrac%7B1%7D%7BR_2%7D%2B%5Cfrac%7B1%7D%7BR_3%7D%2B%5Cfrac%7B1%7D%7BR_4%7D........%5Cfrac%7B1%7D%7BR_n%7D)
According to the given circuit, the resistors are in parallel arrangement.
Given:
![R_A=2.00\Omega\\\\R_B=4.00\Omega\\\\R_C=8.00\Omega](https://tex.z-dn.net/?f=R_A%3D2.00%5COmega%5C%5C%5C%5CR_B%3D4.00%5COmega%5C%5C%5C%5CR_C%3D8.00%5COmega)
Thus,
Equivalent resistance will be:
![R_p=\frac{1}{R_A}+\frac{1}{R_B}+\frac{1}{R_C}\\\\R_p=\frac{1}{2.00}+\frac{1}{4.00}+\frac{1}{8.00}\\\\R_p=0.875\Omega](https://tex.z-dn.net/?f=R_p%3D%5Cfrac%7B1%7D%7BR_A%7D%2B%5Cfrac%7B1%7D%7BR_B%7D%2B%5Cfrac%7B1%7D%7BR_C%7D%5C%5C%5C%5CR_p%3D%5Cfrac%7B1%7D%7B2.00%7D%2B%5Cfrac%7B1%7D%7B4.00%7D%2B%5Cfrac%7B1%7D%7B8.00%7D%5C%5C%5C%5CR_p%3D0.875%5COmega)
Therefore, the equivalent resistance of the circuit is 0.875 Ω.
Answer:
a)801KJoules
b)The woman must climb stairs for an equivalent of 1.57km
Number of stairs=7850 stairs
Explanation:
a)We use the conversion between Joules and Kcal:
4184Joules=1Kcal
540Kcal*(1484Joules/1Kcal)=801KJoules
b)The physical Work must be equal to the food energy
![W_{physical}=E_{food}\\ mgh=E_{food}\\h=E_{food}/(mg)=801KJoules/(52Kg*9.81m/s^{2})=1.57Km\\](https://tex.z-dn.net/?f=W_%7Bphysical%7D%3DE_%7Bfood%7D%5C%5C%20mgh%3DE_%7Bfood%7D%5C%5Ch%3DE_%7Bfood%7D%2F%28mg%29%3D801KJoules%2F%2852Kg%2A9.81m%2Fs%5E%7B2%7D%29%3D1.57Km%5C%5C)
The woman must climb stairs for an equivalent of 1.57Km
For example if an stair measure 20cm=0.2m
Number of stairs=1.57km/0.2m=7850 stairs
Answer:
a) Qh= 6750 kJ
b) e = 0.296
c)
= 390.625° C
Explanation:
Given:
Work done, W = 2000 kJ
Heat flow, Q = 4750 kJ
Temperature at which heat flows out,
= 275° C
a) Now, the heat flow through the engine (Qh)
Qh = W + Q
or
Qh = 2000 + 4750
or
Qh= 6750 kJ
b) The efficiency (e) is given as:
![e=\frac{W}{Q_h}](https://tex.z-dn.net/?f=e%3D%5Cfrac%7BW%7D%7BQ_h%7D)
on substituting the values, we get
![e=\frac{2000}{6750}](https://tex.z-dn.net/?f=e%3D%5Cfrac%7B2000%7D%7B6750%7D)
or
e = 0.296
c) ![e=1-\frac{T_c}{T_H}](https://tex.z-dn.net/?f=e%3D1-%5Cfrac%7BT_c%7D%7BT_H%7D)
where,
is the temperature at which heat flow
on substituting the values, we get
![0.296=1-\frac{275}{T_H}](https://tex.z-dn.net/?f=0.296%3D1-%5Cfrac%7B275%7D%7BT_H%7D)
or
= 390.625° C
Thrust - Friction = mass * acceleration
Friction = coefficient * mass * gravity
Friction = 0.10 * 75 * 9.8 = 73.5 N
Substituting:
200N - 73.5 = 75a
a = 1.69 m/s^2
Obtaining final velocity at the end of 41 seconds:
Vf = at
Vf = 1.69 (41)
Vf = 69.15 m/s
Using the formula for displacement for linear motion problems:
s = Vo*t + (1/2)a*t^2
s = 0 + (1/2)(1.69)(41^2)
s = 1420.445 meters
After he runs out of fuel, we use the following formula to find the distance that he coasts:
F = ma = coefficient*m*g
a = 0.10(9.8)
a = -0.98
Vf^2 = Vo^2 + 2as
0 = 69.15^2 + 2(-0.98)s
s = 2439. 65
Adding the 2 displacements together to obtain the total distance:
Total distance = 2439.65 + <span>1420.45 meters
Total distance = 3860.1 meters</span>