The picture to the right shows which wave behavior?

Explanation:

I have a lot to say it was very nice to meet my parents are u doing well I dont want too its been so much I love you so I was like u know I am not a man but you are the auditions I have been in a long long long life is a triangle and a chair for me and my parents think about the way I

Answer:

1.93×10²⁸ s

Explanation:

From the question given above, the following data were obtained:

Number of electron (e) = 2×10²⁴

Current (I) = 10 A

Time (t) =?

Next, we shall determine the quantity of electricity flowing through pasing through the point. This can be obtained as follow:

1 e = 96500 C

Therefore,

2×10²⁴ e = 2×10²⁴ e × 96500 / 1 e

2×10²⁴ e = 1.93×10²⁹ C

Thus, 1.93×10²⁹ C of electricity is passing through the point.

Finally, we shall determine the time. This can be obtained as follow:

Current (I) = 10 A

Quantity of electricity = 1.93×10²⁹ C

Time (t) =?

Q = it

1.93×10²⁹ = 10 × t

Divide both side by 10

t = 1.93×10²⁹ / 10

t = 1.93×10²⁸ s

Thus, it took 1.93×10²⁸ s for 2×10²⁴ electrons to pass through the point

Answer:

jjnn ok jjjmkkmmkijnnkko

Answer:

2.03 x 0.05= 0.1015

.........

Answer:

Explanation:

From chapter 2 of the book: Medical Assisting: Administrative Skills, Miguel Perez is an administrative assistant. The duties of an administrative assistant in an healthcare professional setting range from performing medical clerical services like keeping patient's files organized, scheduling appointments and answering calls.

So, when a patient has medical concerns and calls the doctor's office, Miguel should answer the phone, know the patient's concerns, put the patient on hold and hand the phone over to a medical personnel - either a nurse or a physician. This is because Miguel is not a trained healthcare professional and cannot offer medical advice or assistance.

Answer:

Oscillation whose amplitude reduce with time are called damped oscillation. This happen because of the friction. In oscillation if its amplitude doesn't change with time then they are called Undamped oscillation

Damped and undamped vibration refer to two different types of vibrations. The main difference between damped and undamped vibration is that undamped vibration refer to vibrations where energy of the vibrating object does not get dissipated to surroundings over time, whereas damped vibration refers to vibrations where the vibrating object loses its energy to the surroundings.

Answer:

a) the total electric potential is 2282000 V

b) the total electric potential (in V) at the point with coordinates (0, 1.50 cm) is 1330769.23 V

Explanation:

Given the data in the question and as illustrated in the image below;

a) Determine the total electric potential (in V) at the origin.

We know that; electric potential due to multiple charges is equal to sum of electric potentials due to individual charges

so

Electric potential at p in the diagram 1 below is;

Vp = V1 + V2

Vp = kq1/r1 + kq2/r2

we know that; Coulomb constant, k = 9 × 10⁹ C

q1 = 4.60 uC = 4.60 × 10⁻⁶ C

r1 = 1.25 cm = 0.0125 m

q2 = -2.06 uC = -2.06 × 10⁻⁶ C

location x2 = −1.80 cm; so r2 = 1.80 cm = 0.018 m

so we substitute

Vp = ( 9 × 10⁹ × 4.60 × 10⁻⁶/ 0.0125 ) + ( 9 × 10⁹ × -2.06 × 10⁻⁶ / 0.018 )

Vp = (3312000) + ( -1030000 )

Vp = 3312000 -1030000

Vp = 2282000 V

Therefore, the total electric potential is 2282000 V

b)

the total electric potential (in V) at the point with coordinates (0, 1.50 cm).

As illustrated in the second image;

r1² = 0.015² + 0.0125²

r1 = √[ 0.015² + 0.0125² ]

r1 = √0.00038125

r1 = 0.0195

Also

r2² = 0.015² + 0.018²

r2 = √[ 0.015² + 0.018² ]

r2 = √0.000549

r2 = 0.0234

Now, Electric Potential at P in the second image below will be;

Vp = V1 + V2

Vp = kq1/r1 + kq2/r2

we substitute

Vp = ( 9 × 10⁹ × 4.60 × 10⁻⁶/ 0.0195 ) + ( 9 × 10⁹ × -2.06 × 10⁻⁶ / 0.0234 )

Vp = 2123076.923 + ( -762962.962 )

Vp = 2123076.923 -792307.692

Vp =  1330769.23 V

Therefore, the total electric potential (in V) at the point with coordinates (0, 1.50 cm) is 1330769.23 V

a) The total electric potential is 2282000 V

b) The total electric potential (in V) at the point with coordinates (0, 1.50 cm) is 1330769.23 V

The electric potential is defined as the amount of work energy needed to move a unit of electric charge from a reference point to a specific point in an electric field.

Given the data in the question and as illustrated in the image below;

a) Determine the total electric potential (in V) at the origin.

We know that; electric potential due to multiple charges is equal to sum of electric potentials due to individual charges

Electric potential at p in diagram 1 below is;

[tex]V_P=V_1+V_2[/tex]

[tex]Vp = \dfrac{kq_1}{r_1} + \dfrac{kq_2}{r_2}[/tex]

we know that; the Coulomb constant, k = 9 × 10⁹ C

q1 = 4.60 uC = 4.60 × 10⁻⁶ C

r1 = 1.25 cm = 0.0125 m

q2 = -2.06 uC = -2.06 × 10⁻⁶ C

location x2 = −1.80 cm; so r2 = 1.80 cm = 0.018 m

so we substitute

Vp = ( 9 × 10⁹ × 4.60 × 10⁻⁶/ 0.0125 ) + ( 9 × 10⁹ × -2.06 × 10⁻⁶ / 0.018 )

Vp = (3312000) + ( -1030000 )

Vp = 3312000 -1030000

Vp = 2282000 V

Therefore, the total electric potential is 2282000 V

b)The total electric potential (in V) at the point with coordinates (0, 1.50 cm).

As illustrated in the second image;

[tex]r_1^2=0.015^2+0.0125^2[/tex]

[tex]r_1 = \sqrt{[ 0.015^2 + 0.0125^2 ][/tex]

[tex]r_1 = \sqrt{0.00038125}[/tex]

[tex]r_1 = 0.0195[/tex]

Also

[tex]r_2^2 = 0.015^2 + 0.018^2[/tex]

[tex]r_2 = \sqrt{0.015^2 + 0.018^2}[/tex]

[tex]r_2 = \sqrt{0.000549[/tex]

[tex]r_2 = 0.0234[/tex]

Now, Electric Potential at P in the second image below will be;

Vp = V1 + V2

[tex]Vp = \dfrac{kq_1}{r_1} + \dfrac{kq_2}{r_2}[/tex]

we substitute

Vp = ( 9 × 10⁹ × 4.60 × 10⁻⁶/ 0.0195 ) + ( 9 × 10⁹ × -2.06 × 10⁻⁶ / 0.0234 )

Vp = 2123076.923 + ( -762962.962 )

Vp = 2123076.923 -792307.692

Vp =  1330769.23 V

Therefore, the total electric potential (in V) at the point with coordinates (0, 1.50 cm) is 1330769.23 V

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it's up in Gogle trust me

[tex] \\ [/tex]

Frequency, f, is how many cycles of an oscillation occur per second and is measured in cycles per second or hertz (Hz). The period of a wave, T, is the amount of time it takes a wave to vibrate one full cycle. These two terms are inversely proportional to each other: f = 1/T and T = 1/f.

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Hope It Helps!

Answer:

Inverse

Explanation:

Frequency is the number of cycles in a second. Frequency is the inverse of a period

frequency = 1 / period

Answer:

.50 M

Explanation:

5*.50=2.5 + 2*.25=.5 = 3n

6*.50= 3N

Final answer is .50M

Answer: The answer is D. KNO3

Explanation:

The graph shows that the KN03 going straight up from the temperature sign so you reversed that so that it will make it to 90°C to 30°C

To solve this we must be knowing each and every concept related to solubility. Therefore, the correct option is option D among all the given options.

The greatest amount of one material that may be dissolved in the other is referred to as its solubility. It is the most solute that may be dissolved into a solvent near equilibrium, resulting in a saturated solution.

When specific circumstances are satisfied, more solute can be dissolved further than the solubility limit point, resulting in a supersaturated solution. Adding extra solute after saturation or supersaturation does not enhance the concentration in the solution. Rather, the excess solute begins to precipitated out of solution. KNO[tex]_3[/tex] is the compound that would have the greatest percentage recovered after cooling a saturated solution of that compound from 90°C to 30°C.

Therefore, the correct option is option D.

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Answer:

P₂ = 4098.96 Pa

Explanation:

For this exercise let's use Bernoulli's equation

Let's use the subscript 1 for the point of the liquid surface and the subscript 2 for the ends (point A)

         P₁ + ½ ρ v₁² + ρ g y₁ = P₂ + ½ ρ v₂² + ρ g y₂

the velocity at the end of the tank

        v₂ = w r₂

the velocity at the surface of the liquid is

        v₁ - w r₁

where r₂ = 1.5 m and r₁ = 1 m

the tank pressure is P₁ = P₀ = 0.1 10³ Pa

         P₂ = P₁ + ½ ρ [w² (r₁² - r₂²)] + ρ g (y₁ -y₂)

We must remember that the pressure measurements the distances are measured from the lowest part to the surface that has zero height

let's reduce the magnitudes to the SI system

         w = 3 rev / min (2π rad / 1rev) (1 min / 60 s) = 0.314159 rad / s

let's calculate

        P₂ = 0.1 10³ + ½ 800 0.314159² |(1² -1.5²)| + 800 9.8 |(1-1.5)|

        P₂ = 0.1 103 +78.96 + 3920

        P₂ = 4098.96 Pa

Answer:

A. 240 J

B. 60 J

C. 90 J

D. 0 J

E. 50 J

Explanation:

A. Determination of the potential energy of the coconut while it is still in the tree

Mass (m) = 2 Kg

Acceleration due to gravity (g) = 10 m/s²

Height (h) = 12 m

Potential energy (PE) =.?

PE = mgh

PE = 2 × 10 × 12

PE = 240 J

B. Determination of the potential energy of the coconut when it hits the bystander on the head,

Mass (m) = 2 Kg

Acceleration due to gravity (g) = 10 m/s²

Height (h) = 3 m

Potential energy (PE) =.?

PE = mgh

PE = 2 × 10 × 3

PE = 60 J

C. Determination of the potential energy of the coconut when it bounces up to its maximum height,

Mass (m) = 2 Kg

Acceleration due to gravity (g) = 10 m/s²

Height (h) = 3 + 1.5 = 4.5 m

Potential energy (PE) =.?

PE = mgh

PE = 2 × 10 × 4.5

PE = 90 J

D. Determination of the potential energy of the coconut when it lands on the ground,

Mass (m) = 2 Kg

Acceleration due to gravity (g) = 10 m/s²

Height (h) = 0 m

Potential energy (PE) =.?

PE = mgh

PE = 2 × 10 × 0

PE = 0 J

E. Determination of the potential energy of the coconut when it rolls into a ground hole, and falls 2.50 m to the bottom of the hole.

Mass (m) = 2 Kg

Acceleration due to gravity (g) = 10 m/s²

Height (h) = 2.50 m

Potential energy (PE) =.?

PE = mgh

PE = 2 × 10 × 2.50

PE = 50 J

(a) The potential energy of the coconut relative to the ground while it is still in the tree is 235.2 J.

(b) The potential energy of the coconut relative to the ground when it hits the bystander on the head is 58.8 J.

(c) The potential energy of the coconut relative to the ground when it bounces up to its maximum height is 88.2 J.

(d) The potential energy of the coconut relative to the ground when it lands on the ground is 0 J.

(e) The potential energy of the coconut when it rolls into a groundhog hole, and falls 2.50 m to the bottom of the hole is 49 J.

The given parameters;

The potential energy of the coconut relative to the ground while it is still in the tree;

[tex]P.E = mgh\\\\P.E = 2 \times 9.8 \times 12\\\\P.E = 235.2 \ J[/tex]

The potential energy of the coconut relative to the ground when it hits the bystander on the head;

[tex]P.E = 2 \times 9.8 \times 3 \\\\P.E = 58.8 \ J[/tex]

The potential energy of the coconut relative to the ground when it bounces up to its maximum height;

[tex]P.E = 2 \times 9.8 (1.5 + 3)\\\\P.E = 88.2 \ J[/tex]

The potential energy of the coconut relative to the ground when it lands on the ground;

[tex]P.E = 2 \times 9.8 \times 0\\\\P.E = 0 \ J[/tex]

The potential energy of the coconut relative to the ground when it rolls into a groundhog hole, and falls 2.50 m to the bottom of the hole;

[tex]P.E = 2\times 9.8 \times 2.5 \\\\P.E = 49 \ J[/tex]

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Answer:

[tex]c=1\ J/g^\circ C[/tex]

Explanation:

Given that,

Heat required, Q = 8000 J

Mass, m = 400 g

The change in temperature, [tex]\Delta T = 20^{\circ}[/tex]

The heat required due to change in temperature is given by :

[tex]Q=mc\Delta T\\\\c=\dfrac{Q}{m\Delta T}\\\\c=\dfrac{8000 }{400\times 20}\\\\c=1\ J/g^\circ C[/tex]

So, the specific heat of the substance is [tex]1\ J/g^\circ C[/tex]

Answer:

25 m/s

Explanation:

200/8 = 25

Answer:

Personal Health is the ability to take charge of your health by making conscious decisions to be healthy.

Answer:Personal Health is the ability to take charge of your health by making conscious decisions to be healthy. It not only refers to the physical well being of an individual but it also comprises the wellness of emotional, intellect, social, economical, spiritual and other areas of life.

Explanation:

Answer:

    q = 2 10⁻⁸ C

Explanation:

For this exercise we use the translational equilibrium equation

                    F_e -A =

                    F_e = W

the electric force is given by Coulomb's law

                    F_e = [tex]k \frac{q_1q_2}{r^2}[/tex]

in this case they indicate that the loads on the tapes are equal

                    F_e = k q² / r²

we substitute

                    k q² / r² = m g

                    q = [tex]\sqrt{ \frac{mg r^2}{k} }[/tex]

calculate  

                     q = [tex]\sqrt { \frac{ 12 \ 10^{-3} \ 9.8 (0.55 \ 10^{-2})^2 }{9 \ 10^9} }[/tex]    

                     q = [tex]\sqrt{ 3.9526 \ 10^{-16}[/tex]

                     q = 1,999 10⁻⁸ C

                     q = 2 10⁻⁸ C

Answer:

the answers, the correct one is D,   Rb₂ = 29.97 ohm

Explanation:

For this exercise we use ohm's law and the equivalent resistance ratio for series and parallel circuits.

Serial circuit

         (Ra + Rb) is = V

         (Ra + Rb) 0.111 = 10

         (Ra + Rb) = 10 / 0.111 = 90.09

parallel circuit

         [tex]\frac{1}{R} = \frac{1}{Ra} + \frac{1}{Rb}[/tex]

           R = [tex]\frac{Ra \ Rb}{Ra + Rb}[/tex]

          \frac{Ra \ Rb}{Ra + Rb} i_p = V

         \frac{Ra \ Rb}{Ra + Rb} 0.5 = 10

         \frac{Ra \ Rb}{Ra + Rb} = 10 / 0.5 = 20

we write and solve our system of equations

         Ra + Rb = 90.09

         \frac{Ra \ Rb}{Ra + Rb} = 20

we solve for Ra in the first equation

          Ra = 90.09 - Rb

           RaRb = 20 (Ra + Rb)

we substitute Ra in the second equation

           (90.09-Rb) Rb = 20 [(90.09-Rb) + Rb]

            90.09 Rb - Rb² = 20 90.09

            Rb² - 90.09 Rb + 1801.8 = 0

we solve the quadratic equation

            Rb = [90.09 ±[tex]\sqrt{90.09^2 - 4 \ 1801.8}[/tex] ] / 2

            Rb = [90.09 ± 30.15] / 2

            Rb₁ = 60.12 ohm

            Rb₂ = 29.97 ohm

the smallest value is Rb = 30 ohm

When checking the answers, the correct one is D

Answer:

the speed of the wave =1m/s

at least two charged interacting parts

Answer:

a)  v₀ₓ = 9.9 m / s, b)  x_woman = 32.7 m

Explanation:

A) In this exercise, the movement of the bagels is parabolic, we find the time it takes to reach the floor.

          y = y₀ + v_{oy} t - ½ g t²

          0 = y₀ + 0 - ½ gt²

          t = [tex]\sqrt{2y_o/g}[/tex]

let's calculate

          t = [tex]\sqrt{2 \ 52.9/9.8}[/tex]

          t = 3,286 s

Now we can analyze how long Henrieta has walked, she has a walking time before the bagel movement begins (t₀ = 4.50 s)

          t_woman = t₀ + t

           t_woman = 4.50 + 3.286

           t_woman = 7.786 s

The distance traveled in this time is

           x_{woman} = v_woman t_woman

            x_{woman} = 4.20 7.786

          x_{woman} = 32.7 m

For her to grab the bagel, the two of them must be at this point

          x_bagel = x_woman

          x_bael = vox t

          v₀ₓ = x_bagel / t

          v₀ₓ = 32.7 / 3,286

           v₀ₓ = 9.9 m / s

b) when catching the bagels this point x_woman = 32.7 m

Answer:

1

Explanation:

I couldn't upload the complete pic because I'm browsing from phone

pls give me brainliest

Answer:

please find attached pdf

Explanation:

Answer: Throughout the scientific world where measurements are nearly always made in SI units, thermodynamic temperature is measured in kelvins (symbol: K). The Rankine scale uses the degree Rankine (symbol: °R) as its unit, which is the same magnitude as the degree Fahrenheit (symbol: °F).

Explanation:

Please mark me as the Brainiest if I got it right

what's the dimension symbol for thermodynamic temperature

Answer:

°R

Answer:

hello your question is incomplete attached below is the complete and the required circuit diagrams

answer :

Ai) This proves that when the current across the resistor is doubled the value of the voltage across the resistor doubles as well

B) It is better to take more measurements in the range 20mA < I < 40mA because of the amount of temperature reached by the bulb and the change in resistance is affected by the temperature

hence At 0 mA current, there won't be any noticeable change

Explanation:

Ai) The voltage across the resistor will double when you double the current through the resistor

Given that : V = I*R.  

lets assume : I = 2 amperes , R = 3 ohms

V = 2*3 = 6 v

secondly lets assume double the value of  (I)   i.e. I = 4 amperes

hence : V = 4*3 = 12 volts

This proves that when the current across the resistor is doubled the value of the voltage across the resistor doubles as well

Aii) Showing the two data points from simulation

I1                    V1            I2= 2I1         V2=2V1       V1/ I1 =V2/I2

0.9*10^3     9 * 10^3     1.8*10^3       18*10^3          10 ohms

1.6 * 10^3    16 * 10^3    3.2*10^3     32*10^3         10 ohms

B) It is better to take more measurements in the range 20mA < I < 40mA because of the amount of temperature reached by the bulb and the change in resistance is affected by the temperature

hence At 0 mA current, there won't be any noticeable change

Answer:

D.

Explanation:

Answer:

120 W lightbulb

Explanation:

Let the two lightbulb be A and B respectively.

Given the following data;

Power A = 120W

Power B = 90W

Voltage = 120V

To find the current flowing through each lightbulb;

a. For lightbulb A

Power = current * voltage

120 = current * 120

Current = 120/120

Current = 1 Ampere.

b. For lightbulb B

Current = power/voltage

Current = 90/120

Current = 0.75 Amperes

Therefore, the lightbulb that carries more current is A with 1 Ampere.

The  bulb that  carries more current is :

- A with 1 Ampere.

Let the two lightbulb be A and B respectively.

To find the current flowing through each lightbulb;

Thus, the lightbulb that carries more current is A with 1 Ampere.

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Answer:

_s = 37.77 m / s

Explanation:

This is an exercise of the Doppler effect that the change in the frequency of the sound due to the relative speed of the source and the observer, in this case the observer is still and the source is the one that moves closer to the observer, for which relation that describes the process is

                    f ’= f₀  [tex]\frac{v}{v - v_s}[/tex]

where d ’= 530 Make

when the ambulance passes away from the observer the relationship is

                    f ’’ = f₀ [tex]\frac{v}{v + v_s}[/tex]

where d ’’ = 424 beam

let's write the two expressions

               f ’ (v-v_s) = fo v

               f ’’  (v + v_s) = fo v

let's solve the system, subtract the two equations

                v (f ’- f’ ’) - v_s (f’ + f ’’) = 0

                v_s = v [tex]\frac{ f' - f''}{ f' + f''}[/tex]

the speed of sound is v = 340 m / s

let's calculate

                 v_s = 340 [tex](\frac{ 530 -424}{530+424} )[/tex]

                 v_s = 340 [tex](\frac{106}{954}[/tex])

                  v_s = 37.77 m / s

Answer:

zero

Explanation:

The computation of the normal force is shown below:

As we know that

F_c = mg - N

F_c = mv^2 ÷ r

N = mg - mv^2 ÷ r

N = m(g - v^2 ÷ r)

Assume that

The mass of the car is 1200 kg

And, r = 10 m

So,

= 1200 (9.8 - 15^2 ÷ 10)

= -15240 N

Since it comes in negative so the normal force would be zero