1. The resistance of an electric device is 40,000 microhms. Convert that measurement to ohms2. When an electric soldering iron is used in a 110 V circuit, the current flowing through the iron is
2 A. What is the resistance of the iron?
3. A current of 0.2 A flows through an electric bell having a resistance of 65 ohms. What must be
the applied voltage in the circuit?


Answer 1


(1) 0.04 ohms (2) 55 ohms (3) 13 volt


(1) The resistance of an electric device is 40,000 microhms.

We need to convert it into ohms.

1\ \mu \Omega =10^(-6)\ \Omega

To covert 40,000 microhms to ohms, multiply 40,000 and 10⁻⁶ as follows :

40000 \ \mu \Omega =40000 * 10^(-6)\ \Omega\n\n=0.04\ \Omega

(2) Voltage used, V = 110 V

Current, I = 2 A

We need to find the resistance of the iron. Using Ohms law to find it as follows :

V = IR, where R is resistance

R=(V)/(I)\n\nR=(110)/(2)\n\nR=55\ \Omega

(3) Current, I = 0.2 A

Resistance, R = 65 ohms

We need to find the applied voltage in the circuit. Using Ohms law to find it as follows :


V = 0.2 × 65

V = 13 volt

Answer 2


1. 0.04 Ohms

2. 55 Ohms

3. 13 Volts


Penn Foster

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The individual calcium atom has a positive, not negative, 2 charge.


Did the quiz also had it on the unit test on edgunity.

Hope this helps guys!

A 62.0 kg skier is moving at 6.90 m/s on a frictionless, horizontal, snow-covered plateau when she encounters a rough patch 4.50 m long. The coefficient of kinetic friction between this patch and her skis is 0.300. After crossing the rough patch and returning to friction-free snow, she skis down an icy, frictionless hill 2.50 m high.How fast is the skier moving when she gets to the bottom ofthe hill?



The given data is as follows.

      Mass, m = 62 kg,       Initial speed, v_(1) = 6.90 m/s

 Length of rough patch, L = 4.50 m,      coefficient of friction, \mu_(k) = 0.3

 Height of inclined plane, h = 2.50 m

According to energy conservation equation,

     (Final kinetic energy) + (Final potential energy) = Initial kinetic energy + Initial potential energy - work done by the friction

     K.E_(2) + U_(2) = K.E_(1) + U_(1) - W_(f)

    (1)/(2)mv^(2)_(2) + U_(2) = (1)/(2)mv^(2)_(1) + mgh - \mu_(k)mgL

Since, final potential energy is equal to zero. Therefore, the equation will be as follows.

    (1)/(2)mv^(2)_(2) = (1)/(2)mv^(2)_(1) + mgh - \mu_(k)mgL    

Cancelling the common terms in the above equation, we get

     (1)/(2)v^(2)_(2) = (1)/(2)v^(2)_(1) + gh - \mu_(k)gL

                         = (1)/(2)(6.90)^(2)_(1) + 9.8 m/s^(2) * 2.50 m - (0.3 * 9.8 * 4.50 m)

                         = 36.055 - 13.23

                         = 22.825

               v_(2) = √(2 * 22.825)

                           = 6.75 m/s

Thus, we can conclude that the skier is moving at a speed of 6.75 m/s when she gets to the bottom of the hill.



mass, m = 62 kg

initial velocity, u = 6.9 m/s

length, l = 4.5 m

height, h = 2.5 m

coefficient of friction, μ = 0.3

Final kinetic energy + final potential energy = initial kinetic energy + initial potential energy + wok done by friction

Let the final velocity is v.

0.5 mv² + 0 = 0.5 mu² + μmgl + mgh

0.5 v² = 0.5 x 6.9 x 6.9 + 0.3 x 9.8 x 4.5 + 9.8 x 2.5

0.5 v² = 23.805 + 13.23 + 24.5

v² = 123.07

v = 11.1 m/s  

While leaning out a window that is 6.0 m above the ground, you drop a 0.60-kg basketball to a friend at ground level. Your friend catches the ball at a height of 1.6 m above the ground. Determine the following.(a) the amount of work done by the force of gravity on the ball.(b) the gravitational potential energy of the ball-earth system, relative to the ground when it is released.(c) the gravitational potential energy of the ball-earth system, relative to the ground when it is caught.



a) W = 25.872 J

b) - 35.28 J

c) - 9.408


a) The amount of work done by the force of gravity on the ball = Change in potential energy between the two vertical points = - mg (H₂ - H₁)

F = - mg (gravity is acting downwards)

F = - 0.6 × 9.8 = - 5.88 N

(H₂ - H₁) = (1.6 - 6) = - 4.4 m

W = (-5.88)(-4.4) = 25.872 J

b) Gravitational-potential energy of the ball when it was released relative to the ground = (- mg) H₁ = (- 0.6 × 9.8) × 6 = - 35.28 J

c) Gravitational-potential energy of the ball when it is caught relative to the ground = (-mg)(H₂) = -0.6 × 9.8 × 1.6 = - 9.408 J

(25) A grinding machine is supported on an isolator that has two springs, each with stiffness of k and one viscous damper with damping constant of c=1.8 kNs/m. The floor on which the machine is mounted is subjected to a harmonic disturbance due to the operation of an unbalanced engine in the vicinity of the grinding machine. The floor oscillates with amplitude Y=3 mm and frequency of 18 Hz. Because of other design constraints, the stiffness of each spring must be greater than 3.25 MN/m. What is the minimum required stiffness of each of the two springs to limit the grinding machine’s steady-state amplitude of oscillation to at most 10 mm? Assume that the grinding machine and the wheel are a rigid body of weight 4200 N and can move in only the vertical direction (the springs deflect the same amount).



k = 15.62 MN/m



- The viscous damping constant, c = 1.8 KNs/m

- The floor oscillation magnitude, Yo = 3 mm

- The frequency of floor oscillation, f = 18 Hz.

- The combined weight of the grinding machine and the wheel, W = 4200 N

- Two springs of identical stiffness k are attached in parallel arrangement.


- The stiffness k > 3.25 MN/m

- The grinding machine’s steady-state amplitude of oscillation to at most 10 mm. ( Xo ≤ 10 mm )


What is the minimum required stiffness of each of the two springs as per the constraints given.


- The floor experiences some harmonic excitation due to the unbalanced engine running in the vicinity of the grinding wheel. The amplitude "Yo" and the frequency "f" of the floor excitation is given

- The floor is excited with a harmonic displacement of the form:

                         y ( t ) = Y_o*sin ( w*t )


           Yo : The amplitude of excitation = 3 mm

           w : The excited frequency = 2*π*f = 2*π*18 = 36π

- The harmonic excitation of the floor takes the form:

                       y ( t ) = 3*sin ( 36\pi *t )                          

- The equation of motion for the floor excitation of mass-spring-damper system is given as follows:

                      m*(d^2x)/(dt^2) + c*(dx)/(dt) + k_e_q*x = k_e_q*y(t) + c*(dy)/(dt)\n\n(m)/(k_e_q)*(d^2x)/(dt^2) + (c)/(k_e_q)*(dx)/(dt) + x = y(t) + (c)/(k_e_q)*(dy)/(dt)


     m: The combined mass of the rigid body ( wheel + grinding wheel body)        c : The viscous damping coefficient

     k_eq: The equivalent spring stiffness of the system ( parallel )

     x : The absolute motion of mass ( free vibration + excitation )

- We will use the following substitutions to determine the general form of the equation of motion:

                                   w_n = \sqrt{(k_e_q)/(m) } , \n\np = (c)/(2√(k_e_q*m) ) =  (1800)/(2√(k_e_q*428.135) ) =  (43.49628)/(√(k_e_q) )


               w_n: The natural frequency

               p = ζ = damping ratio = c / cc , damping constant/critical constant


- The Equation of motion becomes:

                         (1)/(w^2_n)*(d^2x)/(dt^2) + (2*p)/(w_n)*(dx)/(dt) + x = y(t) + (2*p)/(w_n)*(dy)/(dt)


- The steady solution of a damped mass-spring system is assumed to be take the form of harmonic excitation of floor i.e:

                         X_s_s = X_o*sin ( wt + \alpha  )       


              X_o : The amplitude of the steady-state vibration.

              α: The phase angle ( α )

- The steady state solution is independent from system's initial conditions and only depends on the system parameters and the base excitation conditions.

- The general amplitude ( X_o ) for a damped system is given by the relation:

                        X_o = Y_o*\sqrt{(1+ ( 2*p*r)^2)/(( 1 - r^2)^2 + ( 2*p*r)^2) }              


                r = Frequency ratio =  (w)/(w_n) =  \frac{36*\pi }{\sqrt{(k_e_q*g)/(W) } } = \frac{36*\pi }{\sqrt{(k_e_q)/(428.135) } } = (36*\pi*√(428.135)  )/(√(k_e_q) )


- We will use the one of the constraints given to limit the amplitude of steady state oscillation ( Xo ≤ 10 mm ):

- We will use the expression for steady state amplitude of oscillation ( Xo ) and determine a function of frequency ratio ( r ) and damping ratio ( ζ ):


                    ((X_o )/(Y_o))^2 \geq (1+ ( 2*p*r)^2)/(( 1 - r^2)^2 + ( 2*p*r)^2)\n\n((X_o )/(Y_o))^2 \geq (1+ ( 2*(43.49628)/(√(k_e_q) )*(36*\pi*√(428.135)  )/(√(k_e_q) ))^2)/(( 1 - ((36*\pi*√(428.135)  )/(√(k_e_q) ))^2)^2 + ( 2*(43.49628)/(√(k_e_q) )*(36*\pi*√(428.135)  )/(√(k_e_q) ))^2)\n\n

                    ((X_o )/(Y_o))^2 \geq ( 1 + (41442858448.85813)/(k_e_q^2 ))/([ 1 - ((5476277.91201  )/(k_e_q) )]^2 +  (41442858448.85813)/(k_e_q^2 )  )}\n\n((X_o )/(Y_o))^2 \geq ( (k_e_q^2 + 41442858448.85813)/(k^2_e_q ))/([ ((k_e_q - 5476277.91201)^2  )/(k_e_q^2) ] +  (41442858448.85813)/(k_e_q^2 )  )}\n

                   ((X_o )/(Y_o))^2 \geq ( k_e_q^2 + 41442858448.85813)/( (k_e_q - 5476277.91201)^2 +41442858448.85813 )}\n\n((10 )/(3))^2 \geq ( k_e_q^2 + 41442858448.85813)/( k^2_e_q  -10952555.82402*k_e_q +3.00311*10^1^3 )}\n\n\n10.11111*k^2_e_q  -121695064.71133*k_e_q +3.33637*10^1^4 \geq 0

- Solve the inequality (  quadratic ):

       k1_e_q \geq  7811740.790197058  (N)/(m)  \n\nk2_e_q \leq   4224034.972855095 (N)/(m)

- The equivalent stiffness of the system is due to the parallel arrangement of the identical springs:

                k_e_q = (k^2)/(2k) = (k)/(2)

- Therefore,

                  k1 \geq  7811740.790197058*2 = 15.62  (MN)/(m)  \n\nk2 \leq   4224034.972855095*2 = 8.448 (MN)/(m)  

- The minimum stiffness of spring is minimum of the two values:

                k = 15.62 MN/m

What If? Fluoride ions (which have the same charge as an electron) are initially moving with the same speed as the electrons from part (a) through a different uniform electric field. The ions come to a stop in the same distance d. Let the mass of an ion be M and the mass of an electron be m. Find the ratio of the magnitude of electric field the ions travel through to the magnitude of the electric field found in part (a). (Use the following as necessary: d, K, m, M, and e for the charge of the electron.)



E₁ / E₂ = M / m


Let the electric field be E₁ and E₂ for ions and electrons respectively .

Force on ions = E₁ e where e is charge on ions .

Acceleration on ions a = E₁ e / M . Let initial velocity of both be u . Final velocity v = 0

v² = u² - 2as

0 = u² - 2 x E₁ e d  / M  

u² = 2 x E₁ e d  / M

Similarly for electrons

u² = 2 x E₂ e d  / m


2 x E₁ e d  / M =  2 x E₂ e d  / m

E₁ / E₂ = M / m

Final answer:

The ratio of the magnitude of the electric field the ions travel through to the magnitude of the electric field found in part (a) is M/m.


The ratio of the magnitude of the electric field the ions travel through to the magnitude of the electric field found in part (a) can be determined using the concept of mechanical energy conservation. Since the ions come to a stop, their initial kinetic energy must be equal to the work done by the electric field on them. The work done is given by the equation:

Work = Change in kinetic energy

The change in kinetic energy can be calculated using the formula:

Change in kinetic energy = (1/2)Mv2 - (1/2)mv2

where M and m are the masses of the ions and electrons respectively, and v is their initial speed. Solving for the ratio, we get:

Ratio = (1/2)M/(1/2)m = M/m

So, the ratio of the magnitude of electric field the ions travel through to the magnitude of the electric field found in part (a) is M/m.

Learn more about Electric field here:



A businessperson took a small airplane for a quick flight up the coast for a lunch meeting and then returned home. The plane flew a total of 4 hours, and each way the trip was 200 miles. What was the speed of the wind that affected the plane, which was flying at a speed of 120mph? Round your answer to the nearest whole number.



Speed of the wind is 48.989 mph


We have given each trip is of 200 miles

So total distance = 200 +200 = 400 miles

Speed of the airplane = 120 mph

Let the speed of the wind = x mph

So the speed of the airplane with wind = 120+x

So time taken by airplane with wind = (200)/(120+x)

Speed of the airplane against the wind = 120 - x

So time taken by the airplane against the wind =(200)/(120-x)

Total time is given as t= 4 hour

So (200)/(120+x)+(200)/(120-x)=4




x = 48.989 mph



  Type                           Distance             Rate         Time

Headwind 200 120-r   200/120-r

Tailwind     200  120 - r  200/120 - r

We know the times add to 4, so we write the equation:

200/120−r +   200/120 + r = 4  

We multiply both sides by the LCD and simplify to get:

(120−r)(120+r) ((200/120 -r ) + 200/120+r) = 4(120 -r) (120 +r)


Factor the 200 and simplify inside the parentheses to find:





−2,400= −r^2

49 ≈ r

The speed of the wind is 49mph.