The first step to merging is entering the ramp and _____.
A. honking to indicate your location
B. matching your speed
C. signaling your intent
D. telling your passengers where you're going

Answer:

The highest rms voltage will be 8.485 V

Explanation:

For alternating electric current, rms (root means square) is equal to the value of the direct current that would produce the same average power dissipation in a resistive load

If the peak or maximum voltage should not exceed 12 V, then from the relationship

[tex]V_{rms} = \frac{V_{p} }{\sqrt{2} }[/tex]

where [tex]V_{rms}[/tex] is the rms voltage

[tex]V_{p}[/tex] is the peak or maximum voltage

substituting values into the equation, we'll have

[tex]V_{rms} = \frac{12}{\sqrt{2} }[/tex] = 8.485 V

Answer:

The voltage across the capacitor will remain constant

The capacitance of the capacitor will increase

The electric field between the plates will remain constant

The charge on the plates will increase

The energy stored in the capacitor will increase

Explanation:

First of all, if a capacitor is connected to a voltage source, the voltage or potential difference across the capacitor will remain constant. The electric field across the capacitor will stay constant since the voltage is constant, because the electric field is proportional to the voltage applied. Inserting a dielectric material into the capacitor increases the charge on the capacitor.

The charge on the capacitor is equal to

Q = CV

Since the voltage is constant, and the charge increases, the capacitance will also increase.

The energy in a capacitor is given as

E = [tex]\frac{1}{2}CV^{2}[/tex]

since the capacitance has increased, the energy stored will also increase.

Answer:

a) The magnitude of the change in the ball's momentum is 1.1 kilogram-meters per second, b) The change in the magnitude of the ball's momentum is -0.055 kilogram-meters per second, c) D. The magnitude of the change in the ball's momentum.

Explanation:

a) This phenomenon can be modelled by means of the Principle of Momentum Conservation and the Impact Theorem, whose vectorial form is:

[tex]\vec p_{o} + Imp = \vec p_{f}[/tex]

Where:

[tex]\vec p_{o}[/tex], [tex]\vec p_{f}[/tex] - Initial and final momentums, measured in kilogram-meters per second.

[tex]Imp[/tex] - Impact due to collision, measured in kilogram-meters per second.

The impact experimented by the ball due to collision is:

[tex]Imp = \vec p_{f} - \vec p_{o}[/tex]

By using the definition of momentum, the expression is therefore expanded:

[tex]Imp = m \cdot (\vec v_{f}-\vec v_{o})[/tex]

Where:

[tex]m[/tex] - Mass of the ball, measured in kilograms.

[tex]\vec v_{o}[/tex], [tex]\vec v_{f}[/tex] - Initial and final velocities, measured in meters per second.

If [tex]m = 0.275\,kg[/tex], [tex]\vec v_{o} = -2.10\,j\,\left [\frac{m}{s} \right][/tex] and [tex]\vec v_{f} = 1.90\,j\,\left [\frac{m}{s} \right][/tex], the vectorial change of the linear momentum is:

[tex]Imp = (0.275\,kg)\cdot \left[1.90\,j+2.10\,j\right]\,\left[\frac{m}{s} \right][/tex]

[tex]Imp = 1.1\,j\,\left[\frac{kg\cdot m}{s} \right][/tex]

The magnitude of the change in the ball's momentum is 1.1 kilogram-meters per second.

b) The magnitudes of initial and final momentums of the ball are, respectively:

[tex]p_{o} = (0.275\,kg)\cdot \left(2.10\,\frac{m}{s} \right)[/tex]

[tex]p_{o} = 0.578\,\frac{kg\cdot m}{s}[/tex]

[tex]p_{f} = (0.275\,kg)\cdot \left(1.90\,\frac{m}{s} \right)[/tex]

[tex]p_{o} = 0.523\,\frac{kg\cdot m}{s}[/tex]

The change in the magnitude of the ball's momentum is:

[tex]\Delta p = p_{f}-p_{o}[/tex]

[tex]\Delta p = 0.523\,\frac{kg\cdot m}{s} - 0.578\,\frac{kg\cdot m}{s}[/tex]

[tex]\Delta p = -0.055\,\frac{kg\cdot m}{s}[/tex]

The change in the magnitude of the ball's momentum is -0.055 kilogram-meters per second.

c) The quantity calculated in part a) is more related to the net force acting on the ball during its collision with the floor, since impact is the product of net force, a vector, and time, a scalar, and net force is the product of the ball's mass and net acceleration, which creates a change on velocity.

In a nutshell, the right choice is option D.

Answer:

    I₂ = 0.25 I₀

Explanation:

To know the light transmitted by a filter we must use the law of Malus

          I = I₀ cos² θ

In this case, the intensity of the light that passes through the first polarizer is I₀, it reaches the second polarized, which is at 45⁰, therefore the intensity I1 comes out of it.

        I₁ = I₀ cos² 45

        I₁ = I₀ 0.5

this is the light that reaches the third polarizer, which is at 45⁰ with respect to the second, from this comes the intensity I₂

       I₂ = I₁ cos² 45

       I₂ = (I₀ 0.5) 0.5

       I₂ = 0.25 I₀

this is the intensity of the light transmitted by the set of polarizers

Q: An appliance with a 20 Ω resistor has a power rating of 15.0 W. Find the maximum current which can flow safely through the appliance g

Answer:

0.866 A

Explanation:

From the question,

P = I²R............................. Equation 1

Where P = power, I = maximum current, R = Resistance.

Make I the subject of the equation

I = √(P/R).................... Equation 2

Given: P = 15 W, R = 20 Ω

Substitute these values into equation 2

I = √(15/20)

I = √(0.75)

I = 0.866 A

Hence the maximum current that can flow safely through the appliance = 0.866 A

The two types of electromagnetic waves that have higher frequencies than the waves that make up ultraviolet light are gamma rays and X-rays.

Electromagnetic waves are components of the electromagnetic spectrum, which is made up of the following:

Each electromagnetic wave have a specific frequency and wavelength.

However, the two types of electromagnetic waves that have higher frequencies than the waves that make up ultraviolet light are gamma rays and X-rays.

Learn more about electromagnetic waves at: https://brainly.com/question/8553652

Answer:

gamma rays and X-rays

Explanation:

d on edge I got 100%

Answer:

A) 1.4167 × 10^(-11) F

B) r_a = 0.031 m

C) E = 3.181 × 10⁴ N/C

Explanation:

We are given;

Charge;Q = 3.40 nC = 3.4 × 10^(-9) C

Potential difference;V = 240 V

Inner radius of outer sphere;r_b = 4.1 cm = 0.041 m

A) The formula for capacitance is given by;

C = Q/V

C = (3.4 × 10^(-9))/240

C = 1.4167 × 10^(-11) F

B) To find the radius of the inner sphere,we will make use of the formula for capacitance of spherical coordinates.

C = (4πε_o)/(1/r_a - 1/r_b)

Rearranging, we have;

(1/r_a - 1/r_b) = (4πε_o)/C

ε_o is a constant with a value of 8.85 × 10^(−12) C²/N.m

Plugging in the relevant values, we have;

(1/r_a - 1/0.041) = (4π × 8.85 × 10^(−12) )/(1.4167 × 10^(-11))

(1/r_a) - 24.3902 = 7.8501

1/r_a = 7.8501 + 24.3902

1/r_a = 32.2403

r_a = 1/32.2403

r_a = 0.031 m

C) Formula for Electric field just outside the surface of the inner sphere is given by;

E = kQ/r_a²

Where k is a constant value of 8.99 × 10^(9) Nm²/C²

Thus;

E = (8.99 × 10^(9) × 3.4 × 10^(-9))/0.031²

E = 3.181 × 10⁴ N/C

Answer:

The angle between the blue beam and the red beam in the acrylic block is  

 [tex]\theta _d =0.19 ^o[/tex]

Explanation:

From the question we are told that

     The  refractive index of the transparent acrylic plastic for blue light is  [tex]n_F = 1.497[/tex]

     The  wavelength of the blue light is [tex]F = 486.1 nm = 486.1 *10^{-9} \ m[/tex]

    The  refractive index of the transparent acrylic plastic for red light is  [tex]n_C = 1.488[/tex]

       The  wavelength of the red light is [tex]C = 656.3 nm = 656.3 *10^{-9} \ m[/tex]

    The incidence angle is  [tex]i = 45^o[/tex]

Generally from Snell's law the angle of refraction of the blue light  in the acrylic block  is mathematically represented as

       [tex]r_F = sin ^{-1}[\frac{sin(i) * n_a }{n_F} ][/tex]

Where  [tex]n_a[/tex] is the refractive index of air which have a value of[tex]n_a = 1[/tex]

So

     [tex]r_F = sin ^{-1}[\frac{sin(45) * 1 }{ 1.497} ][/tex]

      [tex]r_F = 28.18^o[/tex]

Generally from Snell's law the angle of refraction of the red light in the acrylic block is mathematically represented as

       [tex]r_C = sin ^{-1}[\frac{sin(i) * n_a }{n_C} ][/tex]

Where  [tex]n_a[/tex] is the refractive index of air which have a value of[tex]n_a = 1[/tex]

So

     [tex]r_C = sin ^{-1}[\frac{sin(45) * 1 }{ 1.488} ][/tex]

      [tex]r_F = 28.37^o[/tex]

The angle between the blue beam and the red beam in the acrylic block

     [tex]\theta _d = r_C - r_F[/tex]

substituting values

       [tex]\theta _d = 28.37 - 28.18[/tex]

       [tex]\theta _d =0.19 ^o[/tex]

Answer:

 m = 3,265 10⁻²⁰  E

Explanation:

For this exercise we can use Newton's second law applied to our system, which consists of a capacitor that creates the uniform electric field and the drop of oil with two extra electrons.

             ∑ F = 0

             [tex]F_{e}[/tex] - W = 0

the electric force is

             F_{e} = q E

as they indicate that the charge is two electrons

             F_{e} = 2e E

The weight is given by the relationship

             W = mg

we substitute in the first equation

               2e E = m g

               m = 2e E / g

let's put the value of the constants

              m = (2 1.6 10⁻¹⁹ / 9.80) E

               m = 3,265 10⁻²⁰  E

 The value of the electric field if it is a theoretical problem must be given and if it is an experiment it can be calculated with measures of the spacing between plates and the applied voltage, so that the system is in equilibrium

Answer:

c. The incident light must have at least as much energy as the electron work function

Explanation:

In photoelectric effect, electrons are emitted from a metal surface when a light ray or photon strikes it. An electron either absorbs one whole photon or it absorbs none. After absorbing a photon, an electron either leaves the surface of metal or dissipate its energy within the metal in such a short time  interval that it has almost no chance to absorb a second photon. An increase in intensity of light source  simply increase the number of photons and thus, the number of electrons, but the energy of electron  remains same. However, increase in frequency of light increases the energy of photons and hence, the

energy of electrons too.

Therefore, the energy of photon decides whether the electron shall be emitted or not. The minimum energy required to eject an electron from the metal surface, i.e. to overcome the  binding force of the nucleus is called ‘Work Function’

Hence, the correct option is:

c. The incident light must have at least as much energy as the electron work function

Answer:

please brainliest!!!

Explanation:

V1/√T1 =V2/√T2

V1 = 331m/s

T1 = 0°C = 273k

V2 = ?

T2 = 35°c = 308k

Answer:

Q = 590,940 J

Explanation:

Given:

Specific heat (c) = 1.75 J/(g⋅°C)

Mass(m) = 2.01 kg = 2,010

Change in temperature (ΔT) = 191 - 23 = 168°C

Find:

Heat required (Q)

Computation:

Q = mcΔT

Q = (2,010)(1.75)(168)

Q = 590,940 J

Q = 590.94 kJ

Answer:

P = cte / V

therefore pressure and volume are inversely proportional

Explanation:

For this exercise we can join the ideal gases equation

        PV = n R T

they indicate that the amount of matter and the temperature are constant, therefore

         PV = cte

        P = cte / V

therefore pressure and volume are inversely proportional

Answer:

Ok, the minimal quantity of charge that we can find is on the electron or in the proton (the magnitude is the same, but the sign is different)

Where the charge of a single proton is:

C = 1.6x10^-19 C

Now, you need to remember that when we are working with charges, we are working with discrete math:

What means that?

If the minimum positive is the charge of one proton, then the consecutive charge will be the charge of two protons (there is no somethin in between)

So the consecutive charge will be:

C = 2*1.6x10^-19 C = 3.2x10^-19 C.

So, because we are working in discrete math, we can not have any object that has charge between  1.6x10^-19 C and 3.2x10^-19 C.

Particularly, 2.0x10^-19 C is in that range, so we can conclude that:

No, an object can not carry a charge of 2.0x10^-19 C.

Answer:

50000 turns

Explanation:

Vp / Vs = Np / Ns

240 / 12000 = 1000 / Ns

Ns = 50000 turns

Answer:

2.78 cm³/s

Explanation:

From the question,

Q = v/A'.................... Equation 1

Where Q = Average flow rate, A' = inverse of Area, v = velocity of the car.

Given: v = 100 km/h, A' = 10 km/L

Substitute this value into equation 1

Q = 100/10

Q = 10 L/h.

Now, we convert L/h to cm³/s.

Since,

1 L = 1000 cm³, and

1 h = 3600 s

Therefore,

Q = 10(1000/3600) cm³/s

Q = 2.78 cm³/s

Answer:

B.distance between lines increases

Answer:

A. Lines become narrower

Explanation:

I  got it right on my quiz!

I hope this helps!! :))

Answer:

The approximate displacement of the object is  23  m.

Explanation:

Given that:

v = 4t + 5 (m/s)  for 3< t< 7; n= 4

The approximate displacement of the object can be calculated as follows:

The velocities at the intervals of t are :

3

4

5

6

the velocity at the intervals of t =  7 will be left out due the fact that we are calculating the left endpoint Reimann sum

n = 4 since there are 4 values for t, Then there is no need to divide the velocity values

v(3) = 4(3)+5

v(3) = 12+5

v(3) = 17

v(4)= 4(4)+5

v(4) = 16 + 5

v(4) = 21

v(5)= 4(5)+5

v(5) = 20 + 5

v(5) = 25

v(6) = 4(6)+5

v(6) = 24 + 5

v(6) = 29

Using Left end point;

[tex]= \dfrac{1}{4}(17+21+25+29)[/tex]

= 23 m

Answer:

q = C V    charge on 1 capacitor

q = 1 * 10E-6 * 110 = 1.1 *  10E-4  C per capacitor

N = Q / q = 1 / 1.1 * 10E-4  = 9091 capacitors

9.09 × 10³ capacitors must be connected in parallel.

Given C = 1.00μF = 1 × 10⁻⁶ F

          q = 1.00 C

          V = 110 V

The equivalent capacitance is given by

Ceq = q/V

where q = total charge on all the capacitors

             V = potential difference

For N number of identical capacitors in parallel,

Ceq = NC

Therefore,

NC = q/V

N = q/VC

Putting on the values in the above formula,

N = 1/ (110)(1 × 10⁻⁶)

   = 1 / 110 × 10⁻⁶

   = 9.09 × 10³

Learn more about the capacitors here:

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Answer:

V2 = 21.44cm^3

Explanation:

Given that: the initial volume of the bubble = 1.3 cm^3

Depth = h = 160m

Where P2 is the atmospheric pressure = Patm

P1 is the pressure at depth 'h'

Density of water = ρ = 10^3kg/m^3

Patm = 1.013×10^5 Pa.

Patm = 101300Pa

g = 9.81m/s^2

P1 = P2+ρgh

P1 = Patm +ρgh

P1 = 1.013×10^5+10^3×9.81×160.

P1 = 101300+1569600

P1 = 1670900 Pa

For an ideal gas law

PV =nRT

P1V1/P2V2 = 1

V2 = ( P1/P2)V1

V2 = (P1/Patm)V1

V2 = ( 1670900 /101300 Pa) × 1.3

V2 = 1670900/101300

V2 = 16.494×1.3

V2 = 21.44cm^3

The volume of the bubble can be determined using ideal gas law. The volume of the bubble when it reaches surface is 21.44 [tex]\bold {cm^3}[/tex].

The formula of the pressure of the static fluid

P1 = P2+ρgh

Where,

P1 -  pressure at depth 'h'

P2 -  atmospheric pressure = [tex]\bold {1.013x10^5 }[/tex] =  1670900 Pa

h - Depth = 160m  

ρ - Density of water = [tex]\bold {10^3\ kg/m^3}[/tex]

g- gravitational acceleration = [tex]\bold {9.81\ m/s^2}[/tex]

The initial volume of the bubble = [tex]\bold {1.3\ cm^3}[/tex]  

[tex]\bold {P1 = 1.013x10^5+10^3\times 9.81\times 160}\\\\\bold {P1 = 101300+1569600}\\\\\bold {P1 = 1670900\ Pa}[/tex]  

 For an ideal gas,  

PV =nRT  

[tex]\bold {\dfrac {P_1V_1}{P_2V_2 }= 1}[/tex]  

[tex]\bold {V2 = \dfrac { P_1}{P_2V_1}}[/tex]

So,

[tex]\bold {V2 = \dfrac {1670900 }{101300 }\times 1.3}\\\\\bold {V2 =21.44\ cm^3}[/tex]  

Therefore, the volume of the bubble when it reaches surface is 21.44 [tex]\bold {cm^3}[/tex].

To know more air bubble volume,

https://brainly.com/question/10509397

Answer:

Support at Cy = 1.3 x 10³ k-N

Support at Ay = 200 k-N

Explanation:

given:

fb = 300 k-N/m

fc = 100 k-N/m

D = 300 k-N

L ab = 6 m

L bc = 6 m

L cd = 6 m

To get the reaction A or C.

take summation of moment either A or C.

Support Cy:

∑ M at Ay = 0

      (( x1 * F ) + ( D * Lab ) + ( D * L bc + D * L cd )

Cy = -------------------------------------------------------------------

                                      ( L ab + L bc )

Cy = 1.3 x 10³ k-N

Support Ay:

Since ∑ F = 0,           A + C - F - D = 0

                                   A = F  + D - C

                                  Ay = 200 k-N

Answer:

i was going to but its to late

Explanation:

Answer:

The de Broglie wavelength of electron βe = 2.443422 × 10⁻⁹ m

The de Broglie wavelength of proton βp = 5.70 × 10⁻¹¹ m

Explanation:

Thermal kinetic energy of electron or proton = KE

∴ KE = 3kbT/2

given that; kb = 1.38 x 10⁻²³ J/K , T = 1950 K

so we substitute

KE = ( 3 × 1.38 x 10⁻²³ × 1950 ) / 2

kE = 4.0365 × 10⁻²⁰ (  is the kinetic energy for both electron and proton at temperature T )

Now we know that

mass of electron M'e = 9.109 ×  10⁻³¹

mass of proton M'p = 1.6726 ×  10⁻²⁷

We also know that

KE = p₂ / 2m

from the equation, p = √ (2mKE)

{ p is momentum, m is mass }

de Broglie wavelength = β

so β = h / p = h / √ (2mKE)

h = Planck's constant = 6.626 ×  10⁻³⁴

∴ βe =  h / √ (2m'e × KE)

βe = 6.626 ×  10⁻³⁴ / √ (2 × 9.109 ×  10⁻³¹ × 4.0365 × 10⁻²⁰ )

βe = 6.626 ×  10⁻³⁴ / √  7.3536957 × 10⁻⁵⁰

βe = 6.626 × 10⁻³⁴  / 2.71176984642871 × 10⁻²⁵

βe = 2.443422 × 10⁻⁹ m

βp =  h / √ (2m'p ×KE)

βp = 6.626 ×  10⁻³⁴ / √ (2 × 1.6726 ×  10⁻²⁷ × 4.0365 × 10⁻²⁰ )

βp = 6.626 ×  10⁻³⁴ / √ 1.35028998 × 10⁻⁴⁶

βp =  6.626 ×  10⁻³⁴ / 1.16201978468527 ×  10⁻²³

βp = 5.702140 × 10⁻¹¹ m

Answer:

The speed of the top of the wheel is twice the speed of the car.

That is: 72  m/s

Explanation:

To find the speed of the top of the wheel, we need to combine to velocities: the tangential velocity of the rotating wheel due to rotational motion [tex](v_t=\omega\,R=\omega\,(0.25\,m)\,)[/tex] - with [tex]\omega[/tex] being the wheel's angular velocity,

plus the velocity due to the translation of the center of mass (v = 36 m/s).

The wheel's angular velocity (in radians per second) can be obtained using the tangential velocity for the pure rotational motion and it equals:[tex]\omega=\frac{v_t}{r} =\frac{36}{0.25} \,s^{-1}[/tex]

Then the addition of these two velocities equals:

[tex]\omega\,R+v=\frac{36}{0.25} (0.25)\,\,\frac{m}{s} +36\,\,\frac{m}{s} =72\,\,\frac{m}{s}[/tex]

Answer:

the frequency of revolution of the second particle is f

Explanation:

centripetal force is balanced by the magnetic force for object under magnetic field is given as

Mv²/r= qvB

But v= omega x r

Omega= 2pi x f

f= qB/2pi x M

So since frequency does not depend on the velocity.therefore the frequency of revolution of the second particle remains the same and its equal to f

Answer:

0.46N

Explanation:

See attached file

Answer:

B. over the symbol.

Explanation:

vectors are represented with a symbol carrying an arrow head with also indicates direction

Answer:

(A)

Explanation:

We know , electric potential energy between two charge particles of charges "q" and "Q" respectively is given by kqQ/r where r is the distance between them.

Since the two charged particles are moving apart, the distance between them (r) increases and thus electrical potential energy decreases.

Answer:

4.31 N

Explanation:

Given:

Δy = -55 m

v₀ = 0 m/s

v = -29 m/s

Find: a

v² = v₀² + 2aΔy

(-29 m/s)² = (0 m/s)² + 2a (-55 m)

a = -7.65 m/s²

Sum of forces in the y direction:

∑F = ma

R − mg = ma

R = m (g + a)

R = (2.0 kg) (9.8 m/s² − 7.65 m/s²)

R = 4.31 N

Answer:

Explanation:

The relation between activity and number of radioactive atom in the sample is as follows

dN / dt = λ N where λ is disintegration constant and N is number of radioactive atoms

For the beginning period

dN₀ / dt = λ N₀

58.2 = λ N₀

similarly

41 = λ N

dividing

58.2 / 41 = N₀ / N

N = N₀ x .70446

formula of radioactive decay

[tex]N=N_0e^{-\lambda t }[/tex]

[tex].70446 =e^{-\lambda t }[/tex]

- λ t = ln .70446 =   - .35

t = .35 / λ

λ = .693 / half life

= .693 / 5715

= .00012126

t = .35 / .00012126

= 2886.36

= 2900 years ( rounding it in two significant figures )

Answer:

The  the intensity at an 11° angle to the axis in terms of the intensity of the central maximum is  

   [tex]I_c = \frac{I}{I_o} =8.48 *10^{-8}[/tex]

Explanation:

From the question we are told that

   The  width of the slit is  [tex]D = 0.3 \ mm = 0.3 *10^{-3} \ m[/tex]

    The  wavelength is  [tex]\lambda = 254 \ nm = 254 *10^{-9} \ m[/tex]

     The angle is  [tex]\theta = 11^o[/tex]

The intensity of at [tex]11^o[/tex] to the axis in terms of the intensity of the central maximum. is mathematically represented as

        [tex]I_c = \frac{I}{I_o} = [ \frac{sin \beta }{\beta }] ^2[/tex]

Where [tex]\beta[/tex] is mathematically represented as

        [tex]\beta = \frac{D sin (\theta ) * \pi}{\lambda }[/tex]

substituting values

      [tex]\beta = \frac{0.3 *10^{-3} sin (11 ) * 3.142}{254 *10^{-9} }[/tex]

     [tex]\beta = 708.1 \ rad[/tex]

So

  [tex]I_c = \frac{I}{I_o} = [ \frac{sin (708.1) }{(708.1)}] ^2[/tex]

   [tex]I_c = \frac{I}{I_o} =8.48 *10^{-8}[/tex]