PLEASE HELP I’LL MARK YOU BRAINLIEST!!!!

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:

The definition of that same given problem is outlined in the following section on the clarification.

Explanation:

The Q seems to be endless (hardly any R on the circuit). So energy equations to describe and forth through the inducer as well as the condenser.  

Presently take a gander at the energy stored in your condensers while charging is Q.

⇒  [tex]U =\frac{Qmax^2}{C}[/tex]

So conclude C doesn't change substantially as well as,

When,

⇒  [tex]Q=\frac{Qmax}{2}[/tex]

⇒  [tex]Q^2=\frac{Qmax^2}{4}[/tex]

And therefore only half of the population power generation remains in the condenser that tends to leave this same inductor energy at 3/4 U.

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.

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.

Answer:

t = 3.4 s

The box will come to rest in 3.4 s

Explanation:

For the block to come to rest, the friction force must become equal to the unbalanced force. Therefore:

Unbalanced Force = Frictional Force

but,

Unbalanced Force = ma

Frictional Force = μR = μW = μmg

Therefore,

ma = μmg

a = μg

where,

a = acceleration of box = ?

μ = coefficient of sliding friction = 0.45

g = 9.8 m/s²

Therefore,

a = (0.45)(9.8 m/s²)

a = -4.41 m/s²  (negative sign due to deceleration)

Now, for the time to stop, we use first equation of motion:

Vf = Vi + at

where,

Vf = Final Speed = 0 m/s (since box stops at last)

Vi = Initial Speed = 15 m/s

t = time to stop = ?

Therefore,

0 m/s = 15 m/s + (-4.41 m/s²)t

(-15 m/s)/(-4.41 m/s²) = t

t = 3.4 s

The box will come to rest in 3.4 s

Answer:

b. only ray that reflects back in the same direction it came from

Explanation:

C-rays can be said to be a ray that comes from the center of the curvature. It is known that any ray that comes from the center of the curvature reflects back in the same direction it came from, this is because the line joining from the center of the curvature to any point in the mirror is perpendicular to the mirror.

Correct answer is option B.

C-ray is the only ray that reflects back in the same direction it came from.

Option A is incorrect because for other rays, angle of incidence = angle of reflection. This is not a property of c-ray.

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:

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.   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.

The  lens is designed to work in the visible, near-infrared, and near-ultraviolet. The best resolution of this lens from a diffraction standpoint is: in the near-ultraviolet.

The act of bending light around corners such that it spreads out and illuminates regions where a shadow is anticipated is known as diffraction of light. In general, since both occur simultaneously, it is challenging to distinguish between diffraction and interference. The diffraction of light is what causes the silver lining we see in the sky. A silver lining appears in the sky when the sunlight penetrates or strikes the cloud.

Longer wavelengths of light are diffracted at a greater angle than shorter ones, with the amount of diffraction being dependent on the wavelength of the light. Hence, among the light waves of  the visible, near-infrared, and near-ultraviolet range, near-ultraviolet waves have the shortest wavelengths. So,  The best resolution of this lens from a diffraction standpoint is in the near-ultraviolet, where diffraction is minimum.

Learn more about diffraction here:

https://brainly.com/question/11176463

#SPJ5

Answer:

Do u have a picture of the graph?

Explanation:

I can solve it with refraction

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

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:

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:

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:

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:

21,920 Pascals

Explanation:

P = ρgh

P = (1096 kg/m³) (10 m/s²) (2 m)

P = 21,920 Pa

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: 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:

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 = 1.15 10³ g

Explanation:

For this exercise we will use the relations of the centripetal acceleration

     a = v² / r

where is the linear speed of the rotor and r is the radius of the rotor

let's use the relationships between the angular and linear variables

          v = w r

let's replace

          a = w² r

let's reduce the angular velocity to the SI system

        w = 550 rev / min (2pi rad / 1 rev) (1 min / 60 s)

         w = 57.6 rad / s

let's calculate

       a = 57.6²  3.4

       a = 1.13 10⁴ m / s²

To calculate this value in relation to g, let's find the related

       a / g = 1.13 10⁴ / 9.8

       a = 1.15 10³ g

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:

Seismic attenuation describes the energy loss experienced by seismic waves as they propagate. It is controlled by the temperature, composition, melt content, and volatile content of the rocks through which the waves travel.

Explanation:

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:

9,375

Explanation:

Data provided

The mass of the car m = 1500 Kg.

The diameter of the circular track D = 200 m.

For the computation of magnitude and direction of the net force on the car first we need to find out the radius of the circular path which is shown below:-

The radius of the circular path is

[tex]R = \frac{D}{2}[/tex]

[tex]= \frac{200}{2}[/tex]

= 100 m

after the radius of the circular path we can find the magnitude of the centripetal force with the help of below formula

[tex]Force F = \frac{mv^2}{R}[/tex]

[tex]= \frac{1500\times (25)^2}{100}[/tex]

= 9,375

Therefore for computing the magnitude of the centripetal force we simply applied the above formula.

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:

t = 0.31s

Explanation:

In order to calculate the time that the object takes to travel from the point with its maximum speed to the point with the maximum acceleration, you first use the following formulas, for the maximum speed and the maximum acceleration:

[tex]v_{max}=\omega A\\\\a_{max}=\omega^2A[/tex]

A: amplitude

v_max = 1.38m/s

a_max = 6.83m/s^2

w: angular frequency

From the previous equations you can obtain the angular frequency w.

You divide vmax and amax, and solve for w:

[tex]\frac{v_{max}}{a_{max}}=\frac{\omega A}{\omega^2 A}=\frac{1}{\omega}\\\\\omega=\frac{a_{max}}{v_{max}}=\frac{6.83m/s^2}{1.38m/s^2}=4.94\frac{rad}{s}[/tex]

Next, you take into account that the maximum speed is obtained when the object passes trough the equilibrium point, and the maximum acceleration for the maximum elongation, that is, the amplitude. In such a trajectory the time is T/4 being T the period.

You calculate the period  by using the information about the angular frequency:

[tex]T=\frac{2\pi}{\omega}=\frac{2\pi}{4.94rad/s}=1.26s[/tex]

Then the required time is:

[tex]t=\frac{T}{4}=\frac{1.26s}{4}=0.31s[/tex]

Answer:

  d sin tea = m λ

Explanation:

When we have a two-slit system, the optical path difference determines whether the intensity reaching an observation screen is maximum or zero.

To find this difference in optical path, we assume that the screen is much farther than the gap is, we draw a perpendicular from ray 1 to the second ray

       OP = d sin θ

now to have constructive interference and see a bright line this leg must be an integer number of wavelengths, ose

            d sin tea = m λ

where

d             is the distance between the two slits

θ     complexion the angle sea the point hold it between the two slits

λ    the wavelength of the coherent light used

m   an integer, which counts the number of lines of interference

Units in the SI system

d, lam in meters

θ degrees

m an integer

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:

Force of Rope = 122.5 N

Force of Rope = 480.2N

Explanation:

given data

length = 3.00 m

mass = 25.0 kg

clown mass = 79.0 kg

angle = 30°

solution

we get here Force of Rope on with and without Clown that is

case (1) Without Clown

pivot would be on the concrete pillar so Force of Rope will be

Force of Rope × 3m = (25kg)×(9.8ms²)×(1.5m)

solve it and we get

Force of Rope = 122.5 N

and

case (2) With Clown

so here pivot is still on concrete pillar and clown is standing on the board middle  and above the centre of mass so Force of Rope will be

Force of Rope × 3m = (25kg+73kg)×(9.8ms²)×(1.5m)

solve it and we get

Force of Rope = 480.2N