The time constant of a simple RL circuit is defined as _______. We say that R is the resistance of the circuit and L is the inductance of the circuit.

Answer:

Explanation:

initial vertical velocity = 17.5 m/s

using g=-9.81 m/s^2

apply kinematics equation

v1^2-v0^2=2gS

solve for S with v1=0, v0=+17.5

S = (v1^2-v0^2)/2g

=(0-17.5^2)/(2*(-9.81))

= 15.61 m

Answer:

Explanation:

Given;

initial angular displacement, θ₁ = 0 radians

final angular displacement, θ₂ = 4.5 radians

Angular velocity is calculated as;

[tex]\omega = \frac{\delta \theta}{\delta t}= \frac{4.5 -0}{18} \\\\\omega = 0.25 \ rad/s[/tex]

Angular acceleration is calculated as;

[tex]\alpha = \frac{\delta \omega}{\delta t} = \frac{\omega _f - \omega_i}{t_2 -t_1}[/tex]

where;

[tex]\omega_f[/tex] is the final angular velocity = 0 rad/s

[tex]\omega _i[/tex] is the initial angular velocity = 0.25 rad/s

t₂ is the final time of the motion = 3 seconds

t₁ is the initial time of the motion = 18 seconds

[tex]\alpha =\frac{\omega _f - \omega_i}{t_2 -t_1} \\\\\alpha = \frac{0 - 0.25}{3-18} \\\\\alpha = 0.017 \ rad/s^2[/tex]

Answer:

the internal energy of the mixture at final state = 238kJ/kg

Explanation:

Given

V= 0.6m³

m=5kg

R=0.287kJ/kg.K

T=320 K

from ideal gas equation

PV = nRT

where P is pressure, V is volume, n is number of mole, R is ideal gas constant , T is the temperature.

Recall, mole = mass/molar mass

attached is calculation of the question.

Answer:

The answer is "[tex]\bold{7.30 \times 10^{-6}}[/tex]"

Explanation:

length of the copper wire:

L= 62.9 cm

r is the radius of the loop then:

[tex]r=\frac{L}{2 \pi}\\[/tex]

  [tex]=\frac{62.9}{2\times 3.14}\\\\=\frac{62.9}{6.28}\\\\=10.01\\[/tex]

area of the loop Is:

[tex]A_L= \pi r^2[/tex]

     [tex]=100.2001\times 3.14\\\\=314.628[/tex]

change in magnetic field is:

[tex]=\frac{dB}{dt} \\\\ = 0.01\ \frac{T}{s}[/tex]

then the induced emf is:  [tex]e = A_L \times \frac{dB}{dt}[/tex]

                                              [tex]=314.628 \times 0.01\\\\=3.14\times 10^{-5}V[/tex]

resistivity of the copper wire is: [tex]\rho =[/tex]  1.69 × 10-8Ω·m

diameter d = 1.15mm

radius (r) = 0.5mm

               [tex]= 0.5 \times 10^{-3} \ m[/tex]

hence the resistance of the wire is:

[tex]R=\frac{\rho L}{\pi r^2}\\[/tex]

   [tex]=\frac{1.69 \times 10^{-8}(62.9)}{3.14 \times (0.5 \times 10^{-3})^2}\\\\=\frac{1.69 \times 10^{-8}(62.9)}{3.14 \times 0.5 \times 0.5 \times 10^{-6}}\\\\=\frac{1.69 \times 10^{-8}(62.9)}{3.14 \times 0.25 \times 10^{-6}}\\\\=135.41 \times 10^{-2}\\=1.35\times 10^{-4}\\[/tex]

Power:

[tex]P=\frac{e^2}{R}[/tex]

[tex]=\frac{3.14\times 10^{-5}\times 3.14\times 10^{-5}}{1.35 \times 10^{-4}}\\\\=7.30 \times 10^{-6}[/tex]

The final answer is: [tex]\boxed{7.30 \times 10^{-6} \ W}[/tex]

Answer:

It is the mean of 460 swimmers

Explanation:

In this question, we are concerned with knowing the mean of the population w

Now, according to the question at hand, we have a total population of 460 swimmers and we have taken out 346 swimmers for a study

The population mean in this case is simply the mean of the swimming times of the 460 swimmers

There is another related thing here called the sample mean. For the sample mean, we only make a reference to the mean of the 346 swimmers who were taken out from the population to conduct a separate study

So conclusively, the population mean w is simply the mean of the total 460 swimmers

Find the acceleration, a .

Force = Mass × Acceleration

We know that ,

Force = Mass × Acceleration

★ Substituting the values in the above formula,we get:

⇒ 480 = 40 × Acceleration

⇒ Acceleration, a = 480/40

⇒ Acceleration,a = 12 m/s

Thus,the acceleration of a body is 12 metres per seconds.

Answer:

21,920 Pascals

Explanation:

P = ρgh

P = (1096 kg/m³) (10 m/s²) (2 m)

P = 21,920 Pa

Answer:

a=4m/s²

Explanation:

F=ma

0.4=0.1a

Answer:

a=4m/s

Explanation:

F=ma

0.4=0.1a

[tex] \frac{0.4}{0.1} = \frac{0.1}{0.1} [/tex]

a =4m/ s

Answer: Net electrostatic force on C is 24.2×[tex]10^{-2}[/tex] Newtons.

Explanation: Coulomb's Law is used to determine Electrostatic Force. Its formula is:

F = k.[tex]\frac{q_{0}.q_{1}}{r^{2}}[/tex]

where:

k is electrostatic constant (k = 8.987×[tex]10^{9}[/tex] Nm²/C²);

q is the charge of the object in Coulumb;

r is the distance between charges;

The net force is the sum of all the forces acting on C, so:

Force B on C:

They are both positive, so there is a relpusive force acting between them on the y-axis.

[tex]F_{BC} = 8,987.10^{9}.\frac{4.35.10^{-3}.9.67.10^{-4}}{(6.14.10^{2})^{2}}[/tex]

[tex]F_{BC} = 10.03.10^{-2}[/tex] N

Force D on C:

There is an atractive force between them on the x-axis.

[tex]F_{CD} = 8.987.10^{9}.\frac{9.67.10^{-4}.1.92.10^{-3}}{(1.42.10^{3})^{2}}[/tex]

[tex]F_{CD} = 13.64.10^{-4}[/tex] N

Force A on C:

First, find the distance between objects:

The distance is a diagonal line that divides the rectangle into a right triangle. Distance is square of the hypotenuse .

[tex]r^{2} = (6.14.10^2)^{2} + (1.42.10^{3})^{2}[/tex]

[tex]r^{2} = 37.72.10^{4}[/tex]

and hypotenuse: r = [tex]6.14.10^2[/tex]m

There is an atractive force between charges, but there are components of the force in x- and y-axis. So, because of that, force will be:

[tex]F_{CA} = F_{CA}[/tex].sinα + [tex]F_{CA}.[/tex]cosα

[tex]F_{CA} = 8.987.10^{9}.\frac{3.12.10^{-3}.9.67.10^{-4}}{37.72.10^{4}}[/tex]

[tex]F_{CA} = 7.2.10^{-2}[/tex]

The trigonometric relations is taken from the rectangle:

sinα = [tex]\frac{6.14.10^{2}}{6.14.10^{2}}[/tex]

cosα = [tex]\frac{1.42.10^{3}}{6.14.10^{2}}[/tex]

[tex]F_{CA}.[/tex]cosα = [tex]7.2.10^{-2}(\frac{1.42.10^{3}}{6.14.10^{2}})[/tex] = 0.17

[tex]F_{CA}.[/tex]sinα = [tex]7.2.10^{-2}.(\frac{6.14.10^{2}}{6.14.10^{2}} )[/tex] = 0.072

[tex]F_{CA} =[/tex] 0.17î + 0.072^j

Now, sum up all the terms in its respective axis:

X: [tex]13.64.10^{-4} + 0.17 =[/tex] 0.1714

Y: [tex]10.03.10^{-2} + 7.2.10^{-2}[/tex] = 0.1723

These forms another right triangle, whose hypotenuse is the net electrostatic force:

[tex]F_{net} = \sqrt{(0.1714)^{2} + (0.1723)^2}[/tex]

[tex]F_{net} = 24.3.10^{-2}[/tex] N

The net electrostatic force acting on C has magnitude [tex]F_{net} = 24.3.10^{-2}[/tex] N.

Answer:

B) The wavelength of both transverse and longitudinal waves is measured parallel to the direction of the travel of the wave

Answer:

B) The wavelength of both transverse and longitudinal waves is measured parallel to the direction of the travel of the wave.

Explanation: hope this helps ;)

Answer:

(a) The power wasted for 0.289 cm wire diameter is 15.93 W

(b) The power wasted for 0.417 cm wire diameter is 7.61 W

Explanation:

Given;

diameter of the wire, d = 0.289 cm = 0.00289 m

voltage of the wire, V = 120 V

Power drawn, P = 1850 W

The resistivity of the wire, ρ = 1.68 x 10⁻⁸ Ω⋅m

Area of the wire;

A = πd²/4

A = (π x 0.00289²) / 4

A = 6.561 x 10⁻⁶ m²

(a) At 26 m of this wire, the resistance of the is

R = ρL / A

R = (1.68 x 10⁻⁸  x 26) / 6.561 x 10⁻⁶

R = 0.067 Ω

Current in the wire is calculated as;

P = IV

I = P / V

I = 1850 / 120

I = 15.417 A

Power wasted = I²R

Power wasted = (15.417²)(0.067)

Power wasted = 15.93 W

(b) when a diameter of 0.417 cm is used instead;

d = 0.417 cm = 0.00417 m

A = πd²/4

A = (π x 0.00417²) / 4

A = 1.366 x 10⁻⁵ m²

Resistance of the wire at 26 m length of wire and  1.366 x 10⁻⁵ m² area;

R = ρL / A

R = (1.68 x 10⁻⁸  x 26) / 1.366 x 10⁻⁵

R = 0.032 Ω

Power wasted = I²R

Power wasted = (15.417²)(0.032)

Power wasted = 7.61 W

Answer:

The  z-component of the force is  [tex]\= F_z = 0.00141 \ N[/tex]    

Explanation:

From the question we are told that

          The charge on the particle is [tex]q = 7.76 *0^{-8} \ C[/tex]    

           The magnitude of the magnetic field is  [tex]B = 0.700\r i \ T[/tex]

            The  velocity of the particle toward the x-direction is  [tex]v_x = -1.68*10^{4}\r i \ m/s[/tex]

           The  velocity of the particle toward the y-direction is

[tex]v_y = -2.61*10^{4}\ \r j \ m/s[/tex]

           The  velocity of the particle toward the z-direction is

[tex]v_y = -5.85*10^{4}\ \r k \ m/s[/tex]

Generally the force on this particle is mathematically represented as

          [tex]\= F = q (\= v X \= B )[/tex]

So  we have    

          [tex]\= F = q ( v_x \r i + v_y \r j + v_z \r k ) \ \ X \ ( \= B i)[/tex]

         [tex]\= F = q (v_y B(-\r k) + v_z B\r j)[/tex]      

  substituting values

       [tex]\= F = (7.7 *10^{-8})([ (-2.61*10^{4}) (0.700)](-\r z) + [(5.58*10^{4}) (0.700)]\r y)[/tex]    

      [tex]\= F= 0.00303\ \r j +0.00141\ \r k[/tex]                  

So the z-component of the force is  [tex]\= F_z = 0.00141 \ N[/tex]    

Note :  The  cross-multiplication template of unit vectors is  shown on the first uploaded image  ( From Wikibooks ).

Answer:

it is desired that the lenses stop this ray, its polarization must be vertical

Explanation:

To answer this exercise, let's analyze the rays of light reflected on a horizontal surface, when the incident light that we consider non-polarized is reflected on a surface, the electric field of light moves the electrons on the surface horizontally and this re-emits the radiation same shape, that is horizontal.

The other vertical direction the atoms have a lot of movement restricted by the attraction on the surface, so for the reflected ray this polarization is attenuated, this does not stop the transmitted ray where the two polarizations are transmitted.

Total polarizations only for one angle, but in general as we approach dominant polarization it horizontal. Specifically the angle for full polarization is

                        n = tan teaP

Now we can analyze what polarization the lenses have, if the ray that comes is polarized horizontally and it is desired that the lenses stop this ray, its polarization must be vertical

Answer:

Fx = 2.5 N

Fy = 5 N

|F| = 5.59 N

Explanation:

Given:-

- The mass of puck, m = 4.0 kg

- The initial velocity of puck, u = 3.00 i m/s

- The final velocity of puck, v = ( 8.00 i + 10.00 j ) m/s

- The time interval for the duration of force, Δt = 8 seconds

Find:-

the components of the force and (b) its magnitude.

Solution:-

- We will set up a coordinate system ( x - y ) plane. With unit vectors i and j along x and y axes respectively.

- To model the situation we will seek help from Newton's second law of motion. Defined by the rate of change of linear momentum of the system.

                                [tex]F_net = \frac{m*( v - u ) }{dt}[/tex]

Where,

                   Fnet: The net force that acts on the puck-rocket system

- Here we will assume that the mass of rocket is negligible compared to the mass of the puck. The only force ( F ) acting on the puck is due to the thrust produced of the rocket. The dry and air frictions are both neglected for the analysis.

- We will apply the newton's second law of motion in component forms. And determine the components of force F, as  ( Fx ) and ( Fy ) as follows:

                         [tex]F_x = \frac{m* ( v_x - u_x)}{dt} \\\\F_x = \frac{4* ( 8 - 3)}{8} \\\\F_x = 2.5 N\\\\F_y = \frac{m* ( v_y - u_y)}{dt} \\\\F_y = \frac{4* ( 10 - 0)}{8} \\\\F_y = 5 N\\\\[/tex]

- We will apply the Pythagorean theorem and determine the magnitude of the thrust force produced by the rocket with which the puck accelerated:

                          [tex]| F | = \sqrt{( F_x)^2 + ( F_y)^2} \\\\| F | = \sqrt{( 2.5)^2 + ( 5)^2} \\\\| F | = \sqrt{31.25} \\\\| F | = 5.590[/tex]            

Answer: the magnitude of the thrust force is F = 5.59 N

Answer:

The length of the rod for the condition on the question to be met is [tex]L = 1.5077 \ m[/tex]

Explanation:

The  Diagram for this  question is  gotten from the first uploaded image  

From the question we are told that

          The mass of the rod is [tex]M = 3.41 \ kg[/tex]

           The mass of each small bodies is  [tex]m = 0.249 \ kg[/tex]

           The moment of inertia of the three-body system with respect to the described axis is   [tex]I = 0.929 \ kg \cdot m^2[/tex]

             The length of the rod is  L  

     Generally the moment of inertia of this three-body system with respect to the described axis can be mathematically represented as

        [tex]I = I_r + 2 I_m[/tex]

Where  [tex]I_r[/tex] is the moment of inertia of the rod about the describe axis which is mathematically represented as  

        [tex]I_r = \frac{ML^2 }{12}[/tex]

And   [tex]I_m[/tex] the  moment of inertia of the two small bodies which (from the diagram can be assumed as two small spheres) can be mathematically represented  as

           [tex]I_m = m * [\frac{L} {2} ]^2 = m* \frac{L^2}{4}[/tex]

Thus  [tex]2 * I_m = 2 * m \frac{L^2}{4} = m * \frac{L^2}{2}[/tex]

Hence

       [tex]I = M * \frac{L^2}{12} + m * \frac{L^2}{2}[/tex]

=>   [tex]I = [\frac{M}{12} + \frac{m}{2}] L^2[/tex]

substituting vales  we have  

        [tex]0.929 = [\frac{3.41}{12} + \frac{0.249}{2}] L^2[/tex]

       [tex]L = \sqrt{\frac{0.929}{0.40867} }[/tex]

      [tex]L = 1.5077 \ m[/tex]

Answer:

Momentum of the car = [tex]1.575\times 10^{-2}[/tex] kg meter per second

Explanation:

Julie is playing with a car which has mass = 6.3 g = [tex]6.3\times 10^{-3}[/tex] kg

Velocity of the car is 2.5 meter per second

Since formula to calculate the momentum of an object is,

p = mv

Where, p = momentum of the object

m = mass of the object

v = velocity of the object

By substituting these values in the formula,

p = [tex](6.3\times 10^{-3})\times 2.5[/tex]

  = [tex]1.575\times 10^{-2}[/tex] Kg meter per second

Therefore, momentum of the car will be [tex]1.575\times 10^{-2}[/tex] Kg meter per second.

Answer:

16200 km/s

270 km/min

4.5 km/h

Explanation:

Acceleration Formula: Average Acceleration = Δv/Δt (change in velocity over change in time)

Simply plug in our known variables and solve:

a = (45.0 - 0)/10

a = 45.0/10

a = 4.5 km/h

Answer:

[tex]\boxed{\mathrm{4.5 \: kmph/s \: or \: 1.25 \: m/s^2 }}[/tex]

Explanation:

[tex]\displaystyle \mathrm{acceleration = \frac{change \: in \: velocity}{time \: taken}}[/tex]

[tex]\displaystyle \mathrm{a = \frac{v - u}{t}}[/tex]

[tex]\displaystyle \mathrm{v=final \: velocity}\\\displaystyle \mathrm{u=initial \: velocity}[/tex]

[tex]\displaystyle \mathrm{a = \frac{45- 0}{10}}[/tex]

[tex]\displaystyle \mathrm{a = \frac{45}{10}}[/tex]

[tex]\displaystyle \mathrm{a = 4.5}[/tex]

[tex]\mathrm{4.5 \: kmph/s = 1.25 \: m/s^2 }[/tex]

Answer:

(a) Fn = 0.33 N

(b) Pn = 263.22 x 10³ Pa = 263.22 KPa

Explanation:

(a)

First, we need to calculate the total force exerted by all nails on the magician. This force must be equal to the weight of magician:

F = W

where,

F = Total Force exerted by all nails = ?

W = Weight of magician = mg = (64.3 kg)(9.8 m/s²) = 630.14 N

Therefore,

F = 630.14 N

Now, we calculate the force exerted by each nail:

Fn = F/n

where,

Fn = force exerted by each nail = ?

n = Total no. of nails = 1900

Therefore,

Fn = 630.14 N/1900

Fn = 0.33 N

(b)

The pressure exerted by each nail is given as:

Pn = Fn/An

where,

Pn = Pressure exerted by each nail = ?

An = Area of contact for each nail = 1.26 x 10⁻⁶ m²

Therefore,

Pn = 0.33 N/1.26 x 10⁻⁶ m²

Pn = 263.22 x 10³ Pa = 263.22 KPa

Answer:

The speed is  [tex]v =10.27 *10^{7} \ m/s[/tex]

Explanation:

From the question we are told that

      The  voltage  is  [tex]V = 30 kV = 30*10^{3} V[/tex]

      The  initial velocity of the electron is  [tex]u = 0 \ m/s[/tex]

Generally according to the law of energy conservation

    Electric potential Energy  =  Kinetic energy of the electron

So  

      [tex]PE = KE[/tex]

Where  

      [tex]KE = \frac{1}{2} * m* v^2[/tex]

Here  m is the mass of the electron with a value of  [tex]m = 9.11 *10^{-31} \ kg[/tex]

     and  

         [tex]PE = e * V[/tex]

      Here  e is the charge on the electron with a value  [tex]e = 1.60 *10^{-19} \ C[/tex]

=>    [tex]e * V = \frac{1}{2} * m * v^2[/tex]

=>      [tex]v = \sqrt{ \frac{2 * e * V}{m} }[/tex]

substituting values  

           [tex]v = \sqrt{ \frac{2 * (1.60*10^{-19}) * 30*10^{3}}{9.11 *10^{-31}} }[/tex]

          [tex]v =10.27 *10^{7} \ m/s[/tex]

Answer: lower

There are a number of factors that can affect the coefficient of friction, including surface conditions.

Values of the coefficient of sliding friction can be a good reference for specific combinations of materials. The frictional force and normal reaction are directly proportion but an increase or decrease in coefficient of friction will cause an increase or decrease in the resistance of sliding respectively

Answer:

n = 1.6*10^9 capillaries

Explanation:

In order to calculate the number of capillaries, you take into account that the following relation must be accomplished:

[tex]A_1v_1=nA_2v_2[/tex]               (1)

A1: area of the aorta

v1: speed of the blood in the aorta = 40cm/s

n: number of capillaries = ?

A2: area of each capillary

v2: speed of the blood in each capillary

For the calculation of A1 and A2 you use the formula for the cross sectional area of a cylinder, that is, the area of a circle:

[tex]A=\pi r^2\\\\A_1=\pi r_1^2=\pi(1.0cm)^2=3.1415 cm^2\\\\A_2=\pi r_2^2=\pi (5*10^{-4}cm)^2=7.85*10^{-7}cm^2[/tex]

Where you have used the values of the radius for the aorta and the capillaries.

Next, you solve the equation (1) for n, and replace the values of all parameters:

[tex]n=\frac{A_1v_1}{A_2v_2}=\frac{(3.1415cm^2)(40cm/s)}{(7.85*10^{-7}cm^2)(0.10cm/s)}=1.6*10^9[/tex]

Then, the number of capillaries is 1.6*10^9

Answer:

a.   q2 = 16.4μC, positive charge

b.   F = 0.900N

c.   downward

Explanation:

a. In order to calculate the charge of the unknown charge you use the following formula, for the electric force between two charges:

[tex]F_e=k\frac{q_1q_2}{r^2}[/tex]            (1)

k: Coulomb's constant = 8.98*10^9Nm^2/C^2

r: distance between the charges = 0.300m

q1: charge 1 = -0.550 μC = 0.550*10^-6C

q2: charge 2 = ?

Fe: electric force = 0.900N

The force exerted in the second charge points upward, then, the sign of the second charge is positive because this charge is getting closer to the first one.

You solve the equation (1) for the second charge ans replace the values of the other parameters:

[tex]q_2=\frac{r^2F_e}{kq_1}=\frac{(0.300m)^2(0.900N)}{(8.98*10^9Nm^2/C^2)(0.550*10^{-6}C)}\\\\q_2=1.64*10^{-5}C\\\\q_2=16.4*10^{-6}C=16.4*10\mu C[/tex]

The values of the second charge is 1.64 μC

b. By the third Newton Law, you have that the force exerted in the second charge is equal to the force exerted by the first charge on the second one.

The force exerted on the first charge is 0.900N

c. The charges are attracting between them, then, the force exerted on the first charge is pointing downward.

Answer:

Object 2 has the larger drag coefficient

Explanation:

The drag force, D, is given by the equation:

[tex]D = 0.5 c \rho A v^2[/tex]

Object 1 has twice the diameter of object 2.

If [tex]d_2 = d[/tex]

[tex]d_1 = 2d[/tex]

Area of object 2, [tex]A_2 = \frac{\pi d^2 }{4}[/tex]

Area of object 1:

[tex]A_1 = \frac{\pi (2d)^2 }{4}\\A_1 = \pi d^2[/tex]

Since all other parameters are still the same except the drag coefficient:

For object 1:

[tex]D = 0.5 c_1 \rho A_1 v^2\\D = 0.5 c_1 \rho (\pi d^2) v^2[/tex]

For object 2:

[tex]D = 0.5 c_2 \rho A_2 v^2\\D = 0.5 c_2 \rho (\pi d^2/4) v^2[/tex]

Since the drag force for the two objects are the same:

[tex]0.5 c_1 \rho (\pi d^2) v^2 = 0.5 c_2 \rho (\pi d^2/4) v^2\\4c_1 = c_2[/tex]

Obviously from the equation above, c₂ is larger than c₁, this means that object 2 has the larger drag coefficient

Answer:

The current and the applied emf can be in phase if either of the two changes are made.

1) The inductance of the inductor is doubled, with everything else remaining constant.

2) The capacitance of the capacitor is doubled, with everything else remaining constant.

Explanation:

The current and applied emf for this type of circuit would be in phase when there is no phase difference between the two quantities. That is, Φ = 0°.

The phase difference between current and applied emf is given as

Φ = tan⁻¹ [(XL - Xc)/R]

XL = Impedance due to the inductor

Xc = Impedance due to the capacitor

R = Resistance of the resistor.

For Φ to be 0°, tan⁻¹ [(XL - Xc)/R] = 0

But only tan⁻¹ 0 = 0 rad

So, for the phase difference to be 0,

[(XL - Xc)/R] = 0

Meaning

XL = Xc

But for this question,

XL = 100 Ω, Xc = 200 Ω

For them to be equal, we have to find a way to increase the impedance of the inductor or reduce the impedance of the capacitor.

The impedance are given as

XL = 2πfL

Xc = (1/2πfC)

f = Frequency

L = Inductance of the inductor

C = capacitance of the capacitor

The impedance of the inductor can be increased from 100 Ω to 200 Ω by doubling the inductance of the inductor.

And the impedance of the capacitor can be reduced from 200 Ω to 100 Ω by also doubling the capacitance of the capacitor.

So, these are either of the two ways to make the current and applied emf to be in phase.

Hope this Helps!!!

Answer:

a = -5.10 m/s^2

her acceleration on the rough ice is -5.10 m/s^2

Explanation:

The distance travelled on the rough ice is equal to the width of the rough ice.

distance d = 5.0 m

Initial speed u = 9.2 m/s

Final speed v = 5.8 m/s

The time taken to move through the rough ice can be calculated using the equation of motion;

d = 0.5(u+v)t

time t = 2d/(u+v)

Substituting the given values;

t = 2(5)/(9.2+5.8)

t = 2/3 = 0.66667 second

The acceleration is the change in velocity per unit time;

acceleration a = ∆v/t

a = (v-u)/t

Substituting the values;

a = (5.8-9.2)/0.66667

a = -5.099974500127

a = -5.10 m/s^2

her acceleration on the rough ice is -5.10 m/s^2

Answer:

The power dissipated by each bulb is  [tex]P = 10.0 \ W[/tex]

Explanation:

From the question we are told that

    The  power rating of both bulbs is  [tex]P = 40 \ W[/tex]

     The voltage  rating of both bulb is  [tex]V = 120 \ V[/tex]

     The  both bulbs are connected a voltage of  [tex]V_C = 120 V[/tex]

The amount of power rating of each  bulb is mathematically represented as

       [tex]P = \frac{V^2}{R }[/tex]

=>    [tex]R = \frac{V^2}{P}[/tex]

substituting values

       [tex]R = \frac{ (120)^2}{40}[/tex]

      [tex]R = 360 \Omega[/tex]

Now given that the bulbs are connected is  series, the equivalent resistance is  evaluated as

          [tex]R_{eq } = R +R[/tex]

substituting values

          [tex]R_{eq } = 360 + 360[/tex]

         [tex]R_{eq } =720 \ \Omega[/tex]

The  current flowing through the bulbs is mathematically evaluated as

         [tex]I =\frac{V_C}{R_{eq}}[/tex]

substituting values

      [tex]I =\frac{120}{720}[/tex]

      [tex]I = 0.1667 \ A[/tex]

Now  the power dissipated by both bulbs is mathematically represented as

           [tex]P = I ^2 * R[/tex]

substituting values      

         [tex]P = 0.1668^2 * 360[/tex]

         [tex]P = 10.0 \ W[/tex]

The power that should be dissipated by each bulb is P = 10.0 W.

Since

The power rating of both bulbs is P = 40 W.

The voltage rating of both bulbs is V = 120 V.

And, both bulks that should be connected a voltage of Vc = 120V

Now the amount of power that should be rated of each bulb should be

P = V^2/R

So, R = V^2/P

= 120^2/40

= 360Ω

The equivalent resistance should be

I = Vc/Req

= 120/720

= 0.1667 A

Now the power is = 0.1668^2 * 360

= 10.0 W

Learn more about power here:  https://brainly.com/question/14089891

Answer:

Do u have a picture of the graph?

Explanation:

I can solve it with refraction

Answer:

The new voltage between the plates of the capacitor is 18 V

Explanation:

The charge on parallel plate capacitor is calculated as;

q = CV

Where;

V is the battery voltage

C is the capacitance of the capacitor, calculated as;

[tex]C = \frac{\epsilon _0A}{d} \\\\q =CV = (\frac{\epsilon _0A}{d})V = \frac{\epsilon _0A V}{d}[/tex]

[tex]q = \frac{\epsilon _0A V}{d}[/tex]

where;

ε₀ is permittivity of free space

A is the area of the capacitor

d is the space between the parallel plate capacitors

If only the space between the capacitors is doubled and every other parameter is kept constant, the new voltage will be calculated as;

[tex]q = \frac{\epsilon _0A V}{d} \\\\\frac{\epsilon _0A V}{d} = \frac{\epsilon _0A V}{d} \\\\\frac{V_1}{d_1} = \frac{V_2}{d_2} \\\\V_2 = \frac{V_1d_2}{d_1} \\\\(d_2 = 2d_1)\\\\V_2 = \frac{V_1*2d_1}{d_1} \\\\(V_1 = 9V)\\\\V_2 = \frac{9*2d_1}{d_1} \\\\V_2 = 9*2\\\\V_2 = 18 \ V[/tex]

Therefore, the new voltage between the plates of the capacitor is 18 V

Answer:

kinetic energy (K.E) = 5.28 ×10⁻¹⁷            

Explanation:

Given:

Mass of  α particle (m) = 6.50 × 10⁻²⁷ kg

Charge of  α particle (q) = 3.20 × 10⁻¹⁹ C

Potential difference ΔV = 165 V

Find:

kinetic energy (K.E)

Computation:

kinetic energy (K.E) = (ΔV)(q)

kinetic energy (K.E) = (165)(3.20×10⁻¹⁹)

kinetic energy (K.E) = 528 (10⁻¹⁹)

kinetic energy (K.E) = 5.28 ×10⁻¹⁷              

There are six statements on the list.

The first 2 are true, and the last 2 are true.

The 2 in the middle aren't true.  They are false.