In the circuit shown in the figure above, suppose that the value of R1 is 100 k ohms and the value of R2 is 470 k ohms. At which of the following locations in the circuit would you measure the highest voltage with your meter? A. Between points A and B
B. Between points A and C
C. Between points B and E
D. Between points B and C​
in the circuit shown in the figure above, suppose that - 1


Answer 1




voltage between A and C is equal battery's voltage.

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The wind-chill index is modeled by the function W = 13.12 + 0.6215T − 11.37v0.16 + 0.3965Tv0.16 where T is the temperature (°C) and v is the wind speed (km/h). When T = 12°C and v = 18 km/h, by how much would you expect the apparent temperature W to drop if the actual temperature decreases by 1°C? (Round your answers to two decimal places.)


This question involves the concepts of derivative, apparent temperature, actual temperature,and wind speed.

The drop in apparent temperature will be "1.25°C".

The apparent temperature (W) is given in terms of actual temperature (T) and wind speed (v) is given by the following function:

W = 13.12 + 0.6215\ T-11.37\ v^(0.16)+0.3965\ Tv^(0.16)

Taking the derivative with respect to actual temperature, we get:

(dW)/(dT)=0.6215+0.3965\ v^(0.16)\n\n


dW = drop in apparent temperatures = ?

dT = drop in actual temperature = - 1°C

v = wind speed = 18 km/h



dW = - 1.25°C

Learn more about derivatives here:




Given that

w=13.12 +0.6215 T-11.37 v^(0.16)+0.3965 T v^(0.16)

Given that T= 12°C and v=19 km/h

Now to find the drop in the apparent temperature w

(dw)/(dT)=0.6215 +0.3965v^(0.16)


(\Delta w)/(\Delta T)= 0.6215 +0.3965 v^(0.16)

Now by putting the values v=19 km/hr and ΔT=1

(\Delta w)/(1)=0.6215 +0.3965* 18^(0.16)


So we can say that when temperature is decrease by 1°C then apparent temperature will decrease by 1.25°C at given velocity.

A vertical straight wire carrying an upward 24-A current exerts an attractive force per unit length of 88 X 104N/m on a second parallel wire 7.0 cm away. What current (magnitude and direction) flows in the second wire?



The current flows in the second wire is 1.3*10^(10)\ A


Given that,

Upward current = 24 A

Force per unit length(F)/(l) =88*10^(4)\ N/m

Distance = 7.0 cm

We need to calculate the current in second wire

Using formula of magnetic force


(F)/(l)=(\mu I_(1)I_(2))/(2\pi r)


(F)/(l)=force per unit length

I₁= current in first wire

I₂=current in second wire

r = distance between the wires

Put the value into the formula

88*10^(4)=(4\pi*10^(-7)*24* I_(2))/(2\pi *7*10^(-2))


I_(2)=1.3*10^(10)\ A

Hence, The current flows in the second wire is 1.3*10^(10)\ A

Two identical particles, each with a mass of 4.5 mg and a charge of 30 nC, are moving directly toward each other with equal speeds of 4.0 m/s at an instant when the distance separating the two is equal to 25 cm. How far apart will they be when closest to one another?



   r₁ = 20.5 cm


In this exercise we can use the conservation of energy

the gravitational power energy is always attractive, the electrical power energy is repulsive if the charges are of the same sign

starting point.

        Em₀ = U_g + U_e + K = -G (m_1m_2)/(r) +k (q_1q_2)/(r) - 2 ( (1)/(2)  m v^2)

the two in the kinetic energy is because they are two particles

final point. When it is detained

        Em_f = U_g + U_e = -G (m_1m_2)/(r_1) + k (q_1q_2)/(r_1)

the energy is conserved

        Em₀ = em_f

the charges and masses of the two particles are equal

         -G (m^2)/(r) + k (q^2)/(r) + m v^2 = - G (m^2)/(r_1) + k (q^2)/(r_1)        


sustitute the values

-6.67-11 (4.5 10-3) ² / 0.25 - 9, 109 (30 10-9) ² / 0.25 + 4.5 10-3 4² = - 6.67 10- 11 (4.5 10-3) ² / r1 -9 109 (30 10-9) ² / r1

    -5.4 10⁻¹⁵ + 3.24 10⁻⁵ - 7.2 10⁻⁵ = -1.35 10⁻¹⁵ / r₁  + 8.1 10⁻⁶ / r₁

We can see that the terms that correspond to the gravitational potential energy are much smaller than the terms of the electric power, which is why we depress them.

      3.24 10⁻⁵ - 7.2 10⁻⁵ =  8.1 10⁻⁶ / r₁

      -3.96 10⁻⁵ = 8.1 10⁻⁶ / r₁

      r₁ = 8.1 10⁻⁶ /3.96 10⁻⁵

      r₁ = 2.045 10⁻¹ m

      r₁ = 20.5 cm

A nearsighted person has a far point of 40cm. What power spectacle lens is needed if the lens is 2cm from the eye



The value is p =   - 2.63 \ Diopters


From the question we are told that  

      The value of the far point is  a =  40 \ cm  =  0.4 \  m

      The distance of the lens to the eye is  b =  2 \ cm = 0.02


        1 Diopter = >  1 m^(-1)

Generally the power spectacle lens needed is mathematically represented as

           p = (1)/(d_o )  + (1)/(d_i)

Here d_o is the object distance which for a near sighted person is d_o =  \infty

And  d_i is the image distance which is evaluated as

        d_i =  b - a

=>     d_i =  0.02 - 0.4

=>     d_i = -0.38 \  m


         p = (1)/(\infty )  + (1)/(-0.38)

=>      p = 0   - 2.63

=>      p =   - 2.63 \ Diopters

a mass of .4 kg is raised by a vertical distance of .450 m in the earth's gravitational field. what is the change in its gravitational potential energy



E = 1.76 J


Given that,

Mass of an object, m = 0.4 kg

It moves by a vertical distance of 0.45 m in the Earth's gravitational field.

We need to find the change in its gravitational potential energy. It can be given by the formula as follow :

E=mgh\n\nE=0.4* 9.8* 0.45\n\nE=1.76\ J

So, the change in its gravitational potential energy is 1.76 J.

The distance in the x direction between two control points on a vertical aerial photograph is 4.5". If the distance between these same two points is 3.6" on another photograph having a scale of 1:24,000, determine the scale of the first vertical aerial photograph. Of the focal length of the camera is 6"and the average elevation at these points is 100 ft, determine the flying height from which each photograph was taken



Use proportions to find the scale of the first photo, then use that scale and other given information to fill in the equation



S = scale of the photo

f = focal length of the camera (in feet)

H = flying height

h = average elevation