Your new toaster has two separate toasting units, each of which consumes 600 watts of power when it is in use. When you operate one unit, a current of 5 amperes flowsthrough the wiring in your home and the wires waste about 1 watt of power handling that current. If you operate both toasting units at once, your toaster consumes 1200 watts and the current flowing through the wiring in your home doubles to 10 amperes. How much power will the wires in your home waste now

Answer:

173.75 km/hr in the NW direction.

Explanation:

Velocity is the time rate of change in displacement of a body. Mathematically:

v = d / t

where d = displacement

t = time

Therefore, the velocity of the car is:

v = 556 / 3.2 = 173.75 km/hr

The velocity of the car is 173.75 km/hr in the NW direction.

The velocity of a car will be "173.75 km/hr".

The velocity of something like a car moving northward on something like a prominent motorway as well as the velocity of something like a rocket launching towards spacecraft both might be determined or monitored.

Displacement, d = 556 km

Time, t = 3.2 hours

We know the relation,

→ Velocity = [tex]\frac{Displacement}{Time}[/tex]

or,

→ V = [tex]\frac{d}{t}[/tex]

By substituting the values, we get

      = [tex]\frac{556}{3.2}[/tex]

      = [tex]173.75[/tex] km/hr

Thus the response above is right.

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Answer:

= 0.0532 cm^3

Explanation:

The computation of volume of the sphere is shown below:-

[tex]Density = \frac{Mass}{Volume}[/tex]

Where,

Density = 13.6 g/cm^3

Mass of sphere = 0.723 g

now we will put the values into the above formula to reach volume of the sphere which is here below:-

[tex]Volume = \frac{0.723}{13.6}[/tex]

= 0.0532 cm^3

Therefore for computing the volume of the sphere we simply applied the above formula.

Answer:

a = -0.8 m/s²

Here, negative sign indicates that the acceleration has opposite direction to the direction of motion.

Explanation:

First we find the angular acceleration of the ball from the following formula:

α = (ωf - ωi)/t

where,

α = angular acceleration = ?

ωf = final angular velocity = 7 rad/s

ωi = initial angular velocity = 13 rad/s

t = Time taken = 15 s

Therefore,

α = (7 rad/s - 13 rad/s)/15 s

α = - 0.4 rad/s

negative sign shows that acceleration is in opposite direction to the direction of motion.

Now, for the linear acceleration, we use the formula:

a = rα

where,

a = linear acceleration = ?

r = radius of circular path = length of rope = 2 m

therefore,

a = (2 m)(- 0.4 rad/s²)

a = -0.8 m/s²

Here, negative sign indicates that the acceleration has opposite direction to the direction of motion.

Explanation:

The answer is B because Nuclear family mean a family with two kids and Mrs. Parker have two kids

Answer:

The average upward force exerted by the water is 988.2 N

Explanation:

Given;

mass of the diver, m = 60 kg

height of the board above the water, h = 10 m

time when her feet touched the water, t = 2.10 s

The final velocity of the diver, when she is under the influence of acceleration of free  fall.

V² = U² + 2gh

where;

V is the final velocity

U is the initial velocity = 0

g is acceleration due gravity

h is the height of fall

V² = U² + 2gh

V² = 0 + 2 x 9.8 x 10

V² = 196

V = √196

V = 14 m/s

Acceleration of the diver during 2.10 s before her feet touched the water.

14 m/s is her initial velocity at this sage,

her final velocity at this stage is zero (0)

V = U + at

0 = 14 + 2.1(a)

2.1a = -14

a = -14 / 2.1

a = -6.67 m/s²

The average upward force exerted by the water;

[tex]F_{on\ diver} = mg - F_{ \ water}\\\\ma = mg - F_{ \ water}\\\\F_{ \ water} = mg - ma\\\\F_{ \ water} = m(g-a)\\\\F_{ \ water} = 60[9.8-(-6.67)]\\\\F_{ \ water} = 60 (9.8+6.67)\\\\F_{ \ water} = 60(16.47)\\\\F_{ \ water} = 988.2 \ N[/tex]

Therefore, the average upward force exerted by the water is 988.2 N

Answer:

The angle of refraction of sheet 3 when sheet 1 is on top of it is [tex]\theta_{r_s } = 23.1 ^o[/tex]

Explanation:

From the question we are told that

     The angle of incidence is  [tex]\theta _i = 26.50 ^o[/tex]

      The angle of refraction angle for  sheet 1 is  [tex]\theta _{r_1}} = 31.70 ^o[/tex]

       The angle of refraction for sheet 3 is  [tex]\theta _{r_3}} = 36.70 ^o[/tex]

According to Snell's  law  

       [tex]\frac{n_2}{n_1} = \frac{sin (\theta_1)}{sin (\theta_{r_1})}[/tex]

Where  [tex]n_1 \ and \ n_2[/tex]  are refractive index of sheet 1  and  sheet 2  

       =>   [tex]n_2 = n_1 \frac{sin(\theta_i)}{sin (\theta _{r_1})}[/tex]

Also  when sheet 3 in on top of sheet 2

       [tex]\frac{n_2}{n_3} = \frac{sin \theta_i}{sin \theta_{r_3}}[/tex]

substituting for  [tex]n_2[/tex]

      [tex]n_1 \frac{sin(\theta_i)}{sin (\theta _{r_1})} = n_3 \frac{sin \theta_i}{sin \theta_{r_3}}[/tex]

      [tex]n_1 \frac{sin(\theta_i)}{sin (\theta _{r_1})} = n_3 \frac{sin \theta_i}{sin \theta_{r_3}}[/tex]

=>    [tex]n_3 = n_1 * \frac{sin(\theta_{r_3})}{sin(\theta_{r_1})}[/tex]

when sheet 1 in on top of sheet 3

        [tex]\frac{n_3}{n_1} = \frac{sin(\theta_i)}{\theta_{r_s}}[/tex]

where [tex]r_s[/tex] is the angle of refraction when sheet 1 is on top of sheet 3

substituting for  [tex]n_3[/tex]

         [tex]\frac{ n_1 * \frac{sin(\theta_{r_3})}{sin(\theta_{r_1})}}{n_1} = \frac{sin(\theta_i)}{\theta_{r_s}}[/tex]

=>   [tex]sin (\theta _{r_s}) = n_1 * sin (\theta_i) * \frac{sin (\theta_{r_1})}{ n_1 * sin(\theta_{r_3})}[/tex]

substituting values

      [tex]sin (\theta _{r_s}) = n_1 * sin (26.50) * \frac{sin (31.70)}{ n_1 * sin(36.70)}[/tex]

=>     [tex]\theta_{r_s } = sin^{-1} (0.3923)[/tex]

=>   [tex]\theta_{r_s } = 23.1 ^o[/tex]

Answer:

The width of the slit will be ".946 mm".

Explanation:

The given values are:

Wavelength = 610 × 10⁻⁹

Length, L = 3 m

As we know,

⇒  [tex]\frac{y}{L} = \frac{m(wavelength)}{a}[/tex]

On putting the estimated values, we get

⇒  [tex]\frac{2\times 10^{-3}}{3.1} = \frac{(1)(610 X 10^{-9})}{a}[/tex]

On applying cross-multiplication, we get

⇒  [tex]a=9.46\times 10^{-4}[/tex]

⇒  [tex]a = .946 mm[/tex]

Question:

A particle moving along the x-axis has a position given by x=(24t - 2.0t³)m, where t is measured in s. What is the magnitude of the acceleration of the particle at the instant when its velocity is zero

Answer:

24 m/s

Explanation:

Given:

x=(24t - 2.0t³)m

First find velocity function v(t):

v(t) = ẋ(t) = 24 - 2*3t²

v(t) = ẋ(t) = 24 - 6t²

Find the acceleration function a(t):

a(t) = Ẍ(t) = V(t) = -6*2t

a(t) = Ẍ(t) = V(t) = -12t

At acceleration = 0, take time as T in velocity function.

0 =v(T) = 24 - 6T²

Solve for T

[tex] T = \sqrt{\frac{-24}{6}} = \sqrt{-4} = -2 [/tex]

Substitute -2 for t in acceleration function:

a(t) = a(T) = a(-2) = -12(-2) = 24 m/s

Acceleration = 24m/s

Answer:

A and B is positive charge

C_negative

Explanation:

because when an ebonite is rubbed with fur produce negative charge due to law of electrostatic like charge repel and unlike attract

Answer:

La motocicleta recorre 25 metros en 1 segundo si circula a una velocidad de 90 km/h

Explanation:

La velocidad es una magnitud que expresa el desplazamiento que realiza un objeto en una unidad determinada de tiempo, esto es, relaciona el cambio de posición (o desplazamiento) con el tiempo.

Siendo la velocidad es el espacio recorrido en un período de tiempo determinado, entonces 90 km/h indica que en 1 hora la motocicleta recorre 90 km. Entonces, siendo 1 h= 3600 segundos (1 h=60 minutos y 1 minuto=60 segundos) podes aplicar la siguiente regla de tres: si en 3600 segundos (1 hora) la motocicleta recorre 90 km, entonces en 1 segundo ¿cuánta distancia recorrerá?

[tex]distancia=\frac{1 segundo*90 km}{3600 segundos}[/tex]

distancia= 0.025 km

Por otro lado, aplicas la siguiente regla de tres: si 1 km es igual a 1,000 metros, ¿0.025 km cuántos metros son?

[tex]distancia=\frac{0.025 km*1,000 metros}{1 km}[/tex]

distancia= 25 metros

La motocicleta recorre 25 metros en 1 segundo si circula a una velocidad de 90 km/h

La cantidad de metros que recorre una motocicleta en un segundo si viaja a una velocidad de 90 km / h es de 25 m / s.

Para obtener la velocidad de la motocicleta en un segundo, necesitaremos convertir 90 km / h en metros por segundo

Usando la tasa de conversión;

1000 m = 1 km

1 hora = 3600 segundos

[tex]\frac{90km}{hr} = \frac{90km \times 1000 m}{1km \times 3600s} \\\\\frac{90km}{hr} = \frac{90,000 m}{3600s} =25m/s \\[/tex]

Esto muestra que la cantidad de metros que recorre una motocicleta en un segundo si viaja a una velocidad de 90 km / h es de 25 m / s.

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Answer:

89.11kg

Explanation:

Note an object weighs less when in a fluid and the weight of the volume of the fluid displaced is known as the upthrust.

Now, the person is going to displace the volume 89/1025 =0.087m3 { from density D = mass(M)/volume(V)}

The weight of the fluid displaced is the density of the fluid × volume of fluid displaced.

The weight of the fluid=0.087m3× 1kg/me = 0.087kg

Now the weight of the fluid displaced is referred to as the upthrust.

Now the real weight - the apparent weight = the upthrust.

Hence the apparent weight = real weight - upthrust

Apparent weight = 89.2-0.087 = 89.11kg

Answer:

4.8 × 10^15 N

Explanation:

Electric Field is defined as Force per unit Charge.

This is expressed mathematically as;

E= F/Q

Where E- Electric Field

F- Force

Q- charge

From the expression above by change of subject of formula for F, we have;

F=E×Q

= 1.2 * 10^6 ×4.0 * 10^9

= 4.8 × 10^15 N

Answer:

The curve should be banked at an angle of 13 degrees.

Explanation:

We have,

Radius of a highway curve is 274 m

Speed of car on this curve is 25 m/s

Let [tex]\theta[/tex] is the banking angle. On a banked curve, the angle of safe diving is given by following expression.

[tex]\tan\theta=\dfrac{v^2}{Rg}[/tex]

g = 10 m/s²

Plugging all the values in above formula,

[tex]\tan\theta=\dfrac{(25)^2}{274\times 9.8}\\\\\theta=\tan^{-1}\left(\dfrac{(25)^{2}}{274\times9.8}\right)\\\\\theta=13^{\circ}[/tex]

So, the curve should be banked at an angle of 13 degrees.

Answer:

14.04 m/s

Explanation:

To find the velocity of the first car after the collision, we can use the equation of conservation of momentum:

m1v1 + m2v2 = m1'v1' + m2'v2'

We have the following data:

m1 = m1' = 328,

m2 = m2' = 790,

v1 = 19.1,

v2 = 13,

v2' = 15.1.

Using this data, we can find v1' (final velocity of the first car):

328 * 19.1 + 790 * 13 = 328 * v1' + 790 * 15.1

16534.8 = 328 * v1' + 11929

328 * v1' = 4605.8

v1' = 14.04 m/s

Answer:

a) 5.143 m/s^2

b) T = 15.43 N

c) Fr = 10.29 N

d) 5.143 m/s^2 , T = 15.43 N

Explanation:

Given:-

- The mass of left most block, m1 = 1.0 kg

- The mass of center block, m2 = 2.0 kg

- The mass of right most block, m3 = 4.0 kg

- A force that acts on the right most block, F = 36 N

Solution:-

a)

- For the first part we will consider the three blocks with masses ( m1 , m2 , and m3 ) as one system on which a force of F = 36 N is acted upon. The masses m1 and m3 are connected with a string with tension ( T ) and the m1 and m2 are in contact.

- We apply the Newton's second law of motion to the system with acceleration ( a ) and the combined mass ( M ) of the three blocks as follows:

                       [tex]F = M*a\\\\36 = ( 1 + 2 + 4 )*a\\\\a = \frac{36}{7}\\\\a = 5.143 \frac{m}{s^2}[/tex]

Answer: The system moves in the direction of external force ( F ) i.e to the right with an acceleration of 5.143 m/s^2

b)

- The blocks with mass ( m1 and m3 ) are connected with a string with tension ( T ) with a combined acceleration of ( a ).

- We will isolate the massive block ( m3 ) and notice that two opposing forces ( F and T ) act on the block.

- We will again apply the Newton's 2nd law of motion for the block m3 as follows:

                       [tex]F_n_e_t = m_3 * a\\\\F - T = m_3 * a\\\\36 - T = 4*5.143\\\\T = 36 - 20.5714\\\\T = 15.43 N[/tex]

Answer:- A tension of T = 15.43 Newtons acts on both blocks ( m1 and m3 )

c)

- We will now isolate the left most block ( m1 ) and draw a free body diagram. This block experiences two forces that is due to tension ( T ) and a reaction force ( Fr ) exerted by block ( m2 ) onto ( m3 ).

- Again we will apply the the Newton's 2nd law of motion for the block m3 as follows:

                         [tex]F_n_e_t = m_1*a\\\\T - F_r = m_1*a\\\\15.43 - F_r = 1*5.143\\\\F_r = 15.43 - 5.143\\\\F_r = 10.29 N[/tex]

- The reaction force ( Fr ) is contact between masses ( m1 and m2 ) exists as a pair of equal magnitude and opposite direction acting on both the masses. ( Newton's Third Law of motion )

Answer: The block m2 experiences a contact force of ( Fr = 10.29 N ) to the right.

d)

- If we were to stack the block ( m2 ) on-top of block ( m1 ) such that block ( m2 ) does not slip we the initial system would remain the same and move with the same acceleration calculated in part a) i.e 5.143 m/s^2

- We will check to see if the tension ( T ) differs or not as the two block ( m1 and m2 ) both experience the same Tension force ( T ) as a sub-system. with a combined mass of ( m1 + m2 ).

- We apply the Newton's 2nd law of motion for the block m3 as follows:

                            [tex]T = ( m_1 + m_2 ) *a\\\\T = ( 1 + 2 ) * 5.143\\\\T = 15.43 N[/tex]

Answer: The acceleration of the whole system remains the same at a = 5.143 m/s^2 and the tension T = 15.43 N also remains the same.

Answer:

A lens

Explanation:

A lens forms images when light passes Through it bending the rays of in the process.A phenomena called refraction and the speed of light changes in the process because it enters a medium since it's wavelength is reduced.

The type of device that forms images by changing the speed at which light travels is the lens.

A lens bends a light beam at an aimed perspective and converges or diffuses bundles of rays by taking benefit of refraction taking vicinity while the mild travels from air into glass or plastic. For that purpose, the aspect geometry of a lens has a spherical parent, which can be kind of divided into sorts.

A lens bends a mild beam at an aimed perspective and converges or diffuses bundles of rays through taking gain of refraction taking area whilst the mild travels from air into glass or plastic. For that motive, the facet geometry of a lens has a round parent, which may be kind of divided into sorts.

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Answer:

v = 0.059 m/s

Explanation:

To find the final speed of Olaf and the ball you use the conservation momentum law. The momentum of Olaf and the ball before catches the ball is the same of the momentum of Olaf and the ball after. Then, you have:

[tex]mv_{1i}+Mv_{2i}=(m+M)v[/tex]  (1)

m: mass of the ball = 0.400kg

M: mass of Olaf = 75.0 kg

v1i: initial velocity of the ball = 11.3m/s

v2i: initial velocity of Olaf = 0m/s

v: final velocity of Olaf and the ball

You solve the equation (1) for v and replace the values of all variables:

[tex]v=\frac{mv_{1i}}{m+M}=\frac{(0.400kg)(11.3m/s)}{0.400kg+75.0kg}=0.059\frac{m}{s}[/tex]

Hence, after Olaf catches the ball, the velocity of Olaf and the ball is 0.059m/s

Answer:

decreases, decreases

Explanation:

Answer:

1. c

2. e

3. d

Explanation:

1.

From Einstein's Photoelectric Equation, we know that:

Energy given up by photon = Work Function + K.E of Electron

hc/λ = φ + K.E

where,

h = Plank's Constant = 6.626 x 10⁻³⁴ J.s

c = speed of light = 3 x 10⁸ m/s

λ = wavelength of light source = 200 nm = 2 x 10⁻⁷ m

φ = (5.1 eV)(1.6 x 10⁻¹⁹ J/eV) = 8.16 x 10⁻¹⁹ J

Therefore,

(6.626 x 10⁻³⁴ J.s)(3 x 10⁸ m/s)/(2 x 10⁻⁷ m) - 8.16 x 10⁻¹⁹ = K.E

K.E = (9.939 - 8.16) x 10⁻¹⁹ J

K.E = 1.778 x 10⁻¹⁹ J

The positive answer shows that electrons will be emitted. Since it is clear from the equation the the K.E of electron decreases with the increase in work function. Therefore:

c. less energetic photo-electrons (on average)

2.

The increase in light sources means an increase in the intensity of light. The no. of photons are increased, due to increase of intensity. Thus, more photons hit the metal and they eject greater no. of electrons. Therefore,

e. more photo-electrons ejected

3.

X-rays have smaller wavelength and greater energy than ultraviolet rays. Thus, the photons with greater energy will strike the metal and as a result, electrons with higher energy will be ejected.

d. more energetic photo-electrons (on average)

Answer:

If you pull a permanent magnet rapidly away from a tank circuit, what is likely to happen in that circuit?

Charge will oscillate in the tank's capacitor and inductor.

Explanation:

Answer:

25 m/s

Explanation:

Data provided in the question

20 m/s for 10 minutes

And, the 30 m/s for another 10 minutes

Based on the above information, the average speed is

As we know that

[tex]Average\ speed = \frac{Total\ distance}{Total\ time}[/tex]

[tex]= \frac{20\times10\times60 + 30\times10\times60 }{20\times60}[/tex]

= 25 m/s

1 hour = 60 minutes

1 minute = 60 seconds

Hence, the average speed is 25 m/s

In the question,  there are miles is given but instead of this we use the minutes as we have to find out the average speed and time should not be in miles it should be in minutes, hour or seconds

Therefore we considered the same

Answer:

The terminal velocity is [tex]v_t =17.5 \ m/s[/tex]

Explanation:

From the question we are told that

       The mass of the squirrel is  [tex]m_s = 50\ g = \frac{50}{1000} = 0.05 \ kg[/tex]

      The surface area is   [tex]A_s = 935 cm^2 = \frac{935}{10000} = 0.0935 \ m^2[/tex]

       The height of fall is  h =4.8 m

        The length of the prism is [tex]l = 23.2 = 0.232 \ m[/tex]

          The width of the prism is [tex]w = 11.6 = 0.116 \ m[/tex]

The terminal velocity is mathematically represented as

       [tex]v_t = \sqrt{\frac{2 * m_s * g }{\dho_s * C * A } }[/tex]

Where [tex]\rho[/tex]  is the density of a rectangular prism with a constant values of [tex]\rho = 1.21 \ kg/m^3[/tex]

            [tex]C[/tex] is the drag coefficient for a horizontal skydiver with a value = 1

            A  is the area of the prism the squirrel is assumed to be which is mathematically represented as

      [tex]A = 0.116 * 0.232[/tex]

       [tex]A = 0.026912 \ m^2[/tex]

 substituting values

      [tex]v_t = \sqrt{\frac{2 * 0.510 * 9.8 }{1.21 * 1 * 0.026912 } }[/tex]

     [tex]v_t =17.5 \ m/s[/tex]

1)  100 ° C

2) 323 K

hope it helps youuuuuu

Answer:

(A) 4* 6 ^ ⁻6 T m² (B) 2 * 10 ^ ⁻6 v

Explanation:

Solution

Given that:

A refrigerator magnet about = 2 mm

The estimated magnetic field strength of the magnet is = 5 m T

The Area = 8 cm²

Now,

(A) The magnetic flux ΦB = BA

Thus,

ΦB  = (5 * 10^⁻ 3) ( 4 * 10 ^⁻2) * ( 2 * 10^ ⁻2) Tm²

So,

ΦB =  4* 6 ^ ⁻6 T m²

(B)By applying Faraday's Law we have the following formula given below:

Ε = Bℓυ

Here,

ℓ = 2 cm the same as 2 * 10 ^⁻2 m

B = 5 m T = 5 * 10 ^ ⁻3 T

υ = 2 cm/s  = 2 * 10 ^ ⁻2 m/s

Thus,

Ε = (5 * 10 ^ ⁻3 T) *  (2 * 10 ^ ⁻2) (2 * 10 ^ ⁻2) v

E =2 * 10 ^ ⁻6 v

A) The magnetic flux ΦB into the refrigerator door behind the magnet :

B) The estimated emf induced under the rectangle on the door's surface ;

Given data :

magnetic field strength of magnet ( B )  = 5 mT

size of refrigerator magnet = 2 mm

Area of magnet ( A )  = 4 * 2 = 8 cm²

where ; ΦB  = BA

ΦB = ( 5 * 10⁻³ ) * ( 4 * 10⁻² ) * ( 2 * 10⁻² ) Tm²

      =  4 * 6⁻⁶ Tm²

To determine the estimated emf we will apply Faraday's law

Ε = Bℓυ ---- ( 2 )

where : B = 5 * 10⁻³ T,  ℓ = 2 * 10⁻² m,  υ = 2 * 10⁻² m/s

insert values into equation 2

E = ( 5 * 10⁻³ ) * ( 2 * 10⁻² ) * ( 2 * 10⁻² )

  = 2 * 10⁻⁶ v

Hence we can conclude that The magnetic flux ΦB is 4 * 6⁻⁶ Tm² and The estimated emf induced is  2 * 10⁻⁶ v

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Answer:

V = 0.0283 m³ = 28300 cm³

T₂ = 1200 K

Explanation:

The volume of the gas can be determined by using General Gas Equation:

PV = nRT

where,

P = Pressure of Gas = (72 cm of Hg)(1333.2239 Pa/cm of Hg) = 95992.12 Pa

V = Volume of Gas = ?

n = no. of moles = mass/molar mass = (35 g)/(32 g/mol) = 1.09 mol

R = General Gas Constant = 8.314 J/ mol.k

T = Temperature of Gas = 27°C + 273 = 300 k

Therefore,

(95992.12 Pa)(V) = (1.09 mol)(8.314 J/mol.k)(300 k)

V = 2718.678 J/95992.12 Pa

V = 0.0283 m³ = 28300 cm³

The Kinetic Energy of gas molecule is given as:

K.E = (3/2)(KT)

Also,

K.E = (1/2)(mv²)

Comparing both equations, we get:

(3/2)(KT) = (1/2)(mv²)

v² = 3KT/m

v = √(3KT/m)

where,

v = r.m.s velocity

K = Boltzamn Constant

T = Absolute Temperature

m = mass of gas molecule

At T₁ = 300 K, v = v₁

v₁ = √(3K*300/m)

v₁ = √(900 K/m)

Now, for v₂ = 2v₁ (double r.m.s velocity), T₂ = ?

v₂ = 2v₁ = √(3KT₂/m)

using value of v₁:

2√(900 K/m) = √(3KT₂/m)

4(900) = 3 T₂

T₂ = 1200 K

Answer with Explanation:

Concepts and reason

The concept to solve this problem is that if a capacitor is connected in a RC circuit then it allows the flow of charge through circuit only till it gets fully charged. Once the capacitor is charged it will not allow any charge or current to flow.

Opposite is the case with inductor in the RL circuit. According to Faraday's law an inductor develops an emf to oppose the voltage applied but once the flux change stops then the inductor behaves just like a normal wire as if no inductor is there.

In attached figure, resistor is connected in series to the capacitor.

As we considered [tex]V_{C}[/tex] the voltage across the capacitor and [tex]V_{s}[/tex] the voltage across the source.

Voltage across a resistor In RC circuit.

[tex]V_{R}=V_S\left ( e^{-\frac{t}{RC}} \right )[/tex]

Voltage across a resistor In RL circuit.

[tex]V_{R}=V_S\left (1- e^{-\frac{Rt}{L}} \right )[/tex]

The sketch of [tex]\mathbf{v_R(t)}[/tex] as a function of t for each of the circuits can be seen in the diagram attached below.

For the Pre-Laboratory exercise, based on the assumption that the RC circuit has a capacitor and a sensing resistor while the RL circuit has a sensing resistor and an inductor.

The input voltage for both circuits is regarded as the square wave and if the square wave is much larger than the time constant for each.

Therefore, we can conclude that the below diagram shows an appropriate sketch of  [tex]\mathbf{v_R(t)}[/tex] as a function of t for each of the circuits.

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Answer:

Its initial position was 471 m.

Explanation:

We have,

Final position of the object is 327 m

Displacement of the object is -144 m

It is required to find its initial position. The difference of final and initial position is equal to the displacement of the object. So,

[tex]d=\text{final position}-\text{initial position}\\\\-144=327-\text{initial position}\\\\\text{initial position}=327+144\\\\\text{initial position}=471\ m[/tex]

So, its initial position was 471 m.

Answer:

[tex]E=3.69*10^{-11}\frac{V}{m}[/tex]

Explanation:

To solve this problem you use the following formula, for the calculation of the electric field along the axis of the dipole.

[tex]E=\frac{p}{2\pi \epsilon_ox^3}[/tex]   (1)

p: dipole moment = 6.16*10^-30 Cm

x: distance to the center of mass of the dipole = 3.00*10^-9m

eo: dielectric permittivity of vacuum = 8.85*10^-12 C^2/Nm^2

You replace the values of the variables in the equation (1):

[tex]E=\frac{6.16*10^{-30}Cm}{2\pi(8.85*10^{-12}C^2/Nm^2)(3.00*10^{-9}m)^3}\\\\E=3.69*10^{-11}\frac{V}{m}[/tex]

Answer :the resultant of two vectors can be found using either the parallelogram method or the triangle method. don't know if this was helpful ?

Explanation:

Answer:

Analytical method.