If R = 20 Ω, what is the equivalent resistance between points A and B in the figure?​

Answer:A  200

Explanation:

Vp=1.41*Vrms

Vp=169.7 v

Z=Vp/Ip

Z=169.7/.8484

Z=200.03 ohm

Explanation:

EE = ½ kx²

EE = ½ (800 N/m) (6 m)²

EE = 14,400 J

Answer:

14,400 J

Explanation:

Its the answer

Answer:

Conventions used in SI to indicate units are as follows:

Answer:

The approximate  spring constant is  [tex]k = 55533.33 \ N/m[/tex]

Explanation:

From the question we are told that

   The  mass of the person is  [tex]m = 68 \ kg[/tex]

     The  dip of the car is  [tex]x = 1.2 \ cm = 0.012 \ m[/tex]

Generally according to hooks law  

        [tex]F = k * x[/tex]

here the force F is the weight of the person which is mathematically represented as

         [tex]F = m * g[/tex]

=>    [tex]m * g = k * x[/tex]

=>     [tex]k = \frac{m * g }{x }[/tex]

=>    [tex]k = \frac{68 * 9.8}{ 0.012}[/tex]

=>   [tex]k = 55533.33 \ N/m[/tex]

Answer:

El sistema de posicionamiento global (conocido mundialmente como GPS) podría utilizarse para proponer recorridos alternativos para llegar a un lugar específico, como un parque, a través de la creación de un recorrido guiado por una aplicación móvil con diferentes rutas de acceso al lugar.

Así, por ejemplo, se crearían diferentes rutas de acceso desde un punto A hasta un punto B, teniendo en cuenta factores como: rapidez, congestión vehicular, pago o no de peajes, posibilidad de acceso a pie y determinados factores extra que influyan en la forma de llegar al lugar. Todo ello plasmado en un mapa interactivo en el cual se señalen las rutas disponibles mediante el marcado del mapa en cuestión.

Answer:

The  apparent depth is  [tex]D' = 0.00376 \ m[/tex]

Explanation:

From the question we are told that

     The  depth of the water is  [tex]D = 5.0 \ mm = 5.0 *10^{-3} \ m[/tex]

      The  refractive index of water is  [tex]n = 1.33[/tex]

Generally the apparent depth of the coin is mathematically represented as

          [tex]D' = D * [\frac{ n_a}{n} ][/tex]

Here  [tex]n_a[/tex]  is the refractive index of  air the value is  [tex]n_a = 1[/tex]

So

        [tex]D' = 5.0 *10^{-3} * [\frac{1}{1.33} ][/tex]

        [tex]D' = 0.00376 \ m[/tex]

The apparent depth will be 0.00376 m.

The index of refraction of a substance also known as the refraction index is a dimensionless quantity that specifies how quickly light passes through it in optics.

d is the depth of the water =5.0 mm =5.0 ×10⁻³

n is the refractive index of water =1.33

[tex]\rm n_a[/tex] is the refractive index of wire=1

The apparent depth of the coin is given as;

[tex]\rm D'=D \times \frac{n_a}{n} \\\\ \rm D'=5.0 \times 10^{-3} \times \frac{1}{1.33} \\\\ \rm D'=0.00376 \ m[/tex]

Hence the apparent depth will be 0.00376 m.

To learn more about the index of refraction refer to the link;

https://brainly.com/question/23750645

Answer: 56°

Explanation:

Brewster's angle refers to the angle at the point where light of a certain polarization passes through a transparent dielectric surface and is transmitted perfectly such that no reflection is made.

The formula is;

[tex]= Tan^{-1} (\frac{n_{2} }{n_{1}} )[/tex]

[tex]= Tan^{-1} (\frac{1.5 }{1} )[/tex]

= 56.30993247

= 56°

Answer:

m = 22,877 kg / s

Explanation:

Let's solve this exercise in parts, first look for the amount of heat generated by the plant and then the amount of water to dissipate this heat

The plant generates a power of 300 MW at a rate of 40%, let's use a direct ratio rule to find the heat. If the power is 400 MW it corresponds to 40%, what heat (Q) corresponds to the other 60%

           Q = 300 60% / 40%

           Q = 450 MW

having the amount of heat generated we can use the calorimeter equation,

           Q = m [tex]c_{e}[/tex] [tex](T_{f} - T_{o})[/tex]

            m = Q / c_{e} (T_{f} - T_{o})

let's use the maximum temperature change allowed

           (T_{f} - T_{o}) = 5

the specific heat of sea water is 3934 J / kg ºC, note that it is less than that of pure water, due to the salts dissolved in sea water

power and energy are related

              W = Q / t

               Q = W t

let's calculate

             m = 450 10⁶ / (3934 5)

             m = 22,877 kg / s

Answer: the same as Lily's

Explanation:

Angular velocity has to do with the speed at which an object will be able to rotate. We are informed that Bob and Lily are riding on a merry-go-round.

Since we are further told that Bob rides on a horse near the outer edge of the circular platform, and Lily rides on a horse near the center of the circular platform and that he merry-go-round is rotating at a constant angular speed.

Based on the above analysis, Bob's angular speed will be thesame as that of Lily.

Answer:

The minimum velocity of the ball during its flight is 22.98 m/s.

Explanation:

The velocity of the ball v = 30 m/s

The angle it makes with the horizontal ∅ = 40°

The minimum velocity of the ball during flight will be the horizontal axis component of the velocity, as acceleration is zero on this axis.

[tex]V_{x}[/tex] = v cos ∅

[tex]V_{x}[/tex]  = 30 cos 40°

[tex]V_{x}[/tex] = 30 x 0.766 = 22.98 m/s

Complete Question

A voltaic cell utilizes the following reaction and operates at 298 K:

3Ce4+(aq)+Cr(s)→3Ce3+(aq)+Cr3+(aq).

What is the emf of this cell under standard conditions? Express your answer using three significant figures.

Answer:

The value is [tex]E^o_{cell} = 2.35 V[/tex]

Explanation:

From the question we are told that

   The ionic equation is  

               [tex]3 Ce^{4 +} _{(aq)} + Cr _{(s)} \to 3 Ce^{3+} _{(aq)} + Cr^{3r} _{(aq)}[/tex]

Now under standard conditions the reduction  half reaction  is

      [tex]Ce^{4+} + e \to Ce^{3+} ; \ \ E^o_r = 1.61 V[/tex]

And the oxidation half reaction is

      [tex]Cr^{3+} + 3e^{-} \to Cr ; \ \ \ E^o_o = - 0.74 V[/tex]

The emf of this cell under standard conditions  is mathematically represented as

     [tex]E^o_{cell} = E^o _r - E^o _o[/tex]

substituting values

     [tex]E^o_{cell} = 1.61 - (- 0.74)[/tex]

    [tex]E^o_{cell} = 2.35 V[/tex]

Answer:

B. If the container is cooled, the gas particles will lose kinetic energy and temperature will decrease.

C. If the gas particles move more quickly, they will collide more frequently with the walls of the container and pressure will increase.

E. If the gas particles move more quickly, they will collide with the walls of the container more often and with more force, and pressure will increase.

#FreeMelvin

Explanation:

the missing figure in the Question has been put in the attachment.

Then from the figure we can observe that

the center of the sphere is positive, therefore, negative charge will be  induced at A.

As B is grounded there will not be any charge on B

Hence the answer is A is negative and B is charge less.

Answer:

a) The value of g at such location is:

[tex]g=9.8005171\,\frac{m}{s^2}[/tex]

b) the period of the pendulum with the length is 0.970 m is:

[tex]T=1.9767 sec[/tex]

Explanation:

Recall the relationship between the period (T) of a pendulum and its length (L) when it swings under  an acceleration of gravity g:

[tex]L=\frac{g}{4\,\pi^2} \,T^2[/tex]

a) Then, given that we know the period (2.0 seconds), and the pendulum's length (L=0.993 m), we can determine g at that location:

[tex]g=\frac{4\,\pi^2\,L}{T^2}\\g=\frac{4\,\pi^2\,0.993}{(2)^2}\\g=\pi^2\,(0.993)\,\frac{m}{s^2} \\g=9.8005171\,\frac{m}{s^2}[/tex]

b) for this value of g, when the pendulum is shortened to 0.970 m, the period becomes:

Answer:

v_{f} = 74 m/s, F = 230 N

Explanation:

We can work on this exercise using the relationship between momentum and moment

        I = ∫ F dt = Δp

bold indicates vectors

we can write this equations in its components

X axis

       Fₓ t = m ( -v_{xo})

Y axis  

        t = m (v_{yf} - v_{yo})

in this case with the ball it travels horizontally v_{yo} = 0

Let's use trigonometry to write the final velocities and the force

        sin 30 = v_{yf} / vf

        cos 30 = v_{xf} / vf

        v_{yf} = vf sin 30

        v_{xf} = vf cos 30

         sin40 = F_{y} / F

         F_{y} = F sin 40

         cos 40 = Fₓ / F

         Fₓ = F cos 40

let's substitute

      F cos 40 t = m ( cos 30 - vₓ₀)

      F sin 40 t = m (v_{f} sin 30-0)

we have two equations and two unknowns, so the system can be solved

        F cos 40 0.1 = 0.4 (v_{f} cos 30 - 20)

        F sin 40 0.1 = 0.4 v_{f} sin 30

we clear fen the second equation and subtitles in the first

         F = 4 sin30 /sin40     v_{f}

         F = 3.111 v_{f}

        (3,111 v_{f}) cos 40 = 4 v_{f} cos 30 - 80

        v_{f} (3,111 cos 40 -4 cos30) = - 80

        v_{f} (- 1.0812) = - 80

        v_{f} = 73.99

        v_{f} = 74 m/s

now we can calculate the force

          F = 3.111 73.99

          F = 230 N

Answer:

Kidney

Explanation:

One of the main function of the kidney is to maintain the homeostasis of sodium and potassium ions in the blood and body.

Aldosterone is a key steroid hormone that balances sodium and potassium ions in the blood and body fluid. Potassium and sodium ions generate electric impulse in the body which helps to perform different activities such as muscles flexing.

Kidney function for reabsorption and secretion, in which reabsorption of Na is done nd balances the sodium and potassium in the blood and body.

Answer:

tmin= lambda/2

Explanation:

See attached file pls

Answer:

a

  [tex]Q = 5.34 *10^{19} \ J[/tex]

b

   [tex]T = 0.445 * 365 = 162. 413 \ days[/tex]

Explanation:

From the question we are told that

     The  area of  Manhattan is  [tex]a_k = 87.46 *10^{6} \ m^2[/tex]

      The area of the ice is [tex]a_i = 4* 87.46 *10^{6 } = 3.498 *10^{8}\ m^2[/tex]

        The  thickness is  [tex]t = 500 \ m \\[/tex]

Generally the volume of the ice is mathematically represented is

         [tex]V = a_i * t[/tex]

substituting value

         [tex]V = 500 * 3.498*10^{8}[/tex]

         [tex]V = 1.75 *10^{11}\ m^3[/tex]

Generally the mass of the ice is

       [tex]m_i = \rho_i * V[/tex]

Here [tex]\rho_i[/tex] is the density of ice the value is  [tex]\rho _i = 916.7 \ kg/m^3[/tex]

=>   [tex]m_i = 916.7 * 1.75*10^{11}[/tex]

=>    [tex]m_i = 1.60 *10^{14} \ kg[/tex]

Generally the energy needed for the ice to melt is mathematically represented as

        [tex]Q = m _i * H_f[/tex]

Where [tex]H_f[/tex] is the latent heat of fusion of ice and the value is  [tex]H_f = 3.33*10^{5} \ J/kg[/tex]

=>    [tex]Q = 1.60 *10^{14} * 3.33*10^{5}[/tex]

=>    [tex]Q = 5.34 *10^{19} \ J[/tex]

Considering part b

  We are told that the annual energy consumption is  [tex]G = 1.2*10^{20 } \ J / year[/tex]

So  the time taken to melt the ice is

      [tex]T = \frac{ 5.34 *10^{19}}{ 1.2 *10^{20}}[/tex]

        [tex]T = 0.445 \ years[/tex]

converting to days

      [tex]T = 0.445 * 365 = 162. 413 \ days[/tex]

volume of balloon

= 4/3 T R3

= 4/3 x 3.14 x 6.953

= 1405.47 m3

uplift force

= volume of balloon x density of air x 9.8

= = 1405.47 x 1.29 x 9.8

= 1813.05 x 9.8 N

weight of helium gas

= volume of balloon x density of helium x

9.8

= 1405.47 x .179 x 9.8

= 251.58 x 9.8 N

Weight of other mass = 930 x 9.8 N Total weight acting downwards

= 251.58 x 9.8 +930 x 9.8

= 1181.58 x 9.8 N

If W be extra weight the uplift can balance

1181.58 × 9.8 + W × 9.8 = 1813.05 * 9.8

1181.58+W=1813.05

W= 631.47 kg

Answer:

Explanation:

Given data

voltage supplied Vs= 1.5 volts

resistance R1= 1000 ohms

resistance R2= 500 ohms

The total resistance is

Rt= 1000+ 500

Rt= 1500 ohms

The current I is given as

[tex]I= \frac{Vs}{Rt} \\\\ I= \frac{1.5}{1500} = 0.001mA[/tex]

Voltage across R1

[tex]VR1= Vs(\frac{R1}{R1+R2} )=1.5(\frac{1000}{1000+500} )= 1.5(\frac{1000}{1500} )\\ \\\ VR1= 1v[/tex]

Voltage across R2

[tex]VR2= Vs(\frac{R2}{R1+R2} )=1.5(\frac{500}{1000+500} )= 1.5(\frac{500}{1500} ) \\\ VR2=0.5v[/tex]

In series connection the current is the same for all components while the voltage divides across all components,the voltages consumed by each individual resistance is equal to the source voltage.

Answer:

The angular accelerations of the hoops are related by the following equation [tex]\alpha _1 = 2\alpha_2[/tex].

Explanation:

Net force on the hoop is given by;

[tex]F_{net} = ma[/tex]

where;

a is linear acceleration

m is the mass

Net torque on the hoop is given by;

[tex]\tau_{net} =I\alpha[/tex]

where;

I is moment of inertia

α is the angular acceleration

But, τ = Fr

[tex]Fr = I \alpha\\\\\alpha = \frac{Fr}{I} \\\\\alpha = \frac{Fr}{mr^2} \\\\\alpha = \frac{F}{mr} \\\\\alpha = \frac{1}{r} (\frac{F}{m} )\\\\(since\ the \ force\ and \ mass \ are \ the \ same, \frac{F}{m} = constant=k)\\\\ \alpha = \frac{k}{r}\\\\k = \alpha r[/tex]

[tex]\alpha _1 r_1= \alpha_2 r_2[/tex]

let the angular acceleration of the smaller hoop = α₁

let the radius of the smaller hoop = r₁

then, the radius of the larger loop, r₂ = 2r₁

let the angular acceleration of the larger hoop = α₂

[tex]\alpha _1 r_1= \alpha_2 r_2\\\\\alpha_2= \frac{ \alpha _1 r_1}{r_2} \\\\\alpha_2=\frac{\alpha _1 r_1}{2r_1} \\\\\alpha_2= \frac{\alpha _1}{2} \\\\\alpha _1 = 2\alpha_2[/tex]

Therefore, the angular accelerations of the hoops are related by the following equation [tex]\alpha _1 = 2\alpha_2[/tex]

Complete question is;

A rope, under a tension of 153 N and fixed at both ends, oscillates in a second harmonic standing wave pattern. The displacement of the rope is given by

y = (0.15 m) sin[πx/3] sin[12π t].

where x = 0 at one end of the rope, x is in meters, and t is in seconds. What are (a) the length of the rope, (b) the speed of the waves on the rope, and (c)the mass of the rope? (d) If the rope oscillates in a third - harmonic standing wave pattern, what will be the period of oscillation?

Answer:

A) Length of rope = 4 m

B) v = 24 m/s

C) m = 1.0625 kg

D) T = 0.11 s

Explanation:

We are given;

T = 153 N

y = (0.15 m) sin[πx/3] sin[12πt]

Comparing this displacement equation with general waveform equation, we have;

k = 2π/λ = π/2 rad/m

ω = 2πf = 12π rad/s

Since, 2π/λ = π/2

Thus,wavelength; λ = 4 m

Since, 2πf = 12π

Frequency;f = 6 Hz

A) We are told the rope oscillates in a second-harmonic standing wave pattern. So, we will use the equation;

λ = 2L/n

Since second harmonic, n = 2 and λ = L = 4 m

Length of rope = 4 m

B) speed is given by the equation;

v = fλ = 6 × 4

v = 24 m/s

C) To calculate the mass, we will use;

v = √T/μ

Where μ = mass(m)/4

Thus;

v = √(T/(m/4))

Making m the subject;

m = 4T/v²

m = (4 × 153)/24²

m = 1.0625 kg

D) Now, the rope oscillates in a third harmonic.

So n = 3.

Using the formula f = 1/T = nv/2L

T = 2L/nv

T = (2 × 4)/(3 × 24)

T = 0.11 s

Answer:

The distance is  [tex]d = 1.39 *10^{-4} \ m[/tex]

Explanation:

From the question we are told that

   The wavelength is  [tex]\lambda = 593 \ nm = 593 *10^{-9} \ m[/tex]

    The distance of the screen is   x  =  4.80 m

    The  location of the fourth order bright fringe is  y  =  8.20 cm = 0.082 m

    The order of the fringe is  n  =  4

Generally the position of a fringe with respect to the central fringe is mathematically represented as

           [tex]y = \frac{ n * x * \lambda }{d}[/tex]

Where d is the distance between the slits, so making d the subject

          [tex]d = \frac{\lambda * x * n }{ y }[/tex]

substituting values

          [tex]d = \frac{ 593 *10^{-9} * 4.80 * 4 }{ 0.082 }[/tex]

           [tex]d = 1.39 *10^{-4} \ m[/tex]

Answer:

Explanation:

Bohr's atomic model was replaced by the  modern atomic model because of its limitations, which included :

(a) Only applicable for Hydrogen and like atoms ( He+1, Li+2 )

(b) Couldn't explain Zeeman Effect (splitting of spectral lines due external magnetic field ) and Stark Effect (splitting of spectral lines due to external electric field).

(c) Inconsistent with De-Broglie's Dual nature of matter and Heisenberg Uncertainty principal, etc.

Answer:

The length of the tube is 85 cm

Explanation:

Given;

speed of sound, v = 340 m/s

first harmonic of open-closed tube is given by;

N----->A , L= λ/₄

λ₁ = 4L

v = Fλ

F = v / λ

F₁ = v/4L

Second harmonic of open-closed tube is given by;

L = N-----N + N-----A, L = (³/₄)λ

[tex]\lambda = \frac{4L}{3}\\\\ F= \frac{v}{\lambda}\\\\F_2 = \frac{3v}{4L}[/tex]

Third harmonic of open-closed tube is given by;

L = N------N + N-----N + N-----A, L = (⁵/₄)λ

[tex]\lambda = \frac{4L}{5}\\\\ F= \frac{v}{\lambda}\\\\F_3 = \frac{5v}{4L}[/tex]

The difference between second harmonic and first harmonic;

[tex]F_2 -F_1 = \frac{3v}{4L} - \frac{v}{4L}\\\\F_2 -F_1 = \frac{2v}{4L} \\\\F_2 -F_1 =\frac{v}{2L}[/tex]

The difference between third harmonic and second harmonic;

[tex]F_3 -F_2 = \frac{5v}{4L} - \frac{3v}{4L}\\\\F_3 -F_2 = \frac{2v}{4L} \\\\F_3 -F_2 =\frac{v}{2L}[/tex]

Thus, the difference between successive harmonic of open-closed tube is

v / 2L.

[tex]700H_z- 500H_z= \frac{v}{2L} \\\\200 = \frac{v}{2L}\\\\L = \frac{v}{2*200} \\\\L = \frac{340}{2*200}\\\\L = 0.85 \ m\\\\L = 85 \ cm[/tex]

Therefore, the length of the tube is 85 cm

Answer:

13.33 rad/s

Explanation:

Applying,

v = ωr......................... Equation 1

Where v = linear speed, ω = angular speed and r = radius.

Note that,

r = d/2................. Equation 2

Where d = diameter of the wheel.

Substitute equation 2 into equation 1

v = ωd/2............... Equation 3

make ω the subject of the equation

ω = 2v/d................ Equation 4

Given: v = 4 m/s, d = 60 cm = 0.6 m

Substitute these values into equation 4

ω = 2(4)/0.6

ω = 13.33 rad/s

Answer:

Explanation:

If the river moves towards the south at 3m/s and the both moves towards the east at 4.0m/s, the speed of the boat relative to me will be the resulting displacement of both velocities of the river and that of the boat. This can be gotten using pythagoras theorem.

Let Vr be the relative speed. According to the theorem;

[tex]V_r^2 = V_s^2 + V_e^2\\\\V_r^2 = 3.0^2 + 4.0^2\\\\V_r^2 = 9+16\\\\V_r^2 = \sqrt{25}\\ \\V_r = 5.0m/s[/tex]

Hence this relative speed is 5.0 m/s

Answer:I don’t know

Explanation:

Answer:

His angular velocity will increase.

Explanation:

According to the conservation of rotational momentum, the initial angular momentum of a system must be equal to the final angular momentum of the system.

The angular momentum of a system = [tex]I[/tex]'ω'

where

[tex]I[/tex]' is the initial rotational inertia

ω' is the initial angular velocity

the rotational inertia = [tex]mr'^{2}[/tex]

where m is the mass of the system

and r' is the initial radius of rotation

Note that the professor does not change his position about the axis of rotation, so we are working relative to the dumbbells.

we can see that with the mass of the dumbbells remaining constant, if we reduce the radius of rotation of the dumbbells to r, the rotational inertia will reduce to [tex]I[/tex].

From

[tex]I[/tex]'ω' = [tex]I[/tex]ω

since [tex]I[/tex] is now reduced, ω will be greater than ω'

therefore, the angular velocity increases.