calculate the length of wire.
​

Answer:

2bsnsnsnns181991oiwiw

Answer:

the answer should be b) 3.8 x 10-5 v

9514 1404 393

Answer:

  1.114 kg/m³

Explanation:

The total mass of the air in the balloon and the balloon + cargo will be the mass of the displaced air. If d is the density of the air in the balloon, then we have ...

  2910d +308 = 2910×1.22

Solving for d, we find ...

  2910d = 2919(1.22) -308

  d = 1.22 -308/2910

  d ≈ 1.114 . . . kg/m³

The density of the hot air is about 1.114 kg/m³.

Answer:

Option D.

Explanation:

From the question given above, the following data were obtained:

Force applied (F) = 5 N

Extention (e) = 0.075 m

Spring constant (K) =?

The spring constant for the spring can be obtained as follow:

F = Ke

5 = K × 0.075

Divide both side by 0.075

K = 5 / 0.075

K = 67 N/m

Thus, the spring constant for the spring is 67 N/m

The masses of the gliders provided in the question differ from the masses mentioned in the "Required" section. I'll use the first masses throughout.

Momentum is conserved, so the total momentum of the system is the same before and after the collision:

m₁ v₁ + m₂ v₂ = m₁ v₁' + m₂ v₂'

==>

(0.160 kg) (0.710 m/s) + (0.296 kg) (-2.23 m/s) = (0.160 kg) v₁' + (0.296 kg) v₂'

==>

-0.546 kg•m/s ≈ (0.160 kg) v₁' + (0.296 kg) v₂'

where v₁' and v₂' are the gliders' respective final velocities. Notice that we take rightward to be positive and leftward to be negative.

Kinetic energy is also conserved, so that

1/2 m₁ v₁² + 1/2 m₂ v₂² = 1/2 m₁ (v₁' )² + 1/2 m₂ (v₂' )²

or

m₁ v₁² + m₂ v₂² = m₁ (v₁' )² + m₂ (v₂' )²

==>

(0.160 kg) (0.710 m/s)² + (0.296 kg) (-2.23 m/s)² = (0.160 kg) (v₁' )² + (0.296 kg) (v₂' )²

==>

1.55 kg•m²/s² ≈ (0.160 kg) (v₁' )² + (0.296 kg) (v₂' )²

Solve for v₁' and v₂'. Using a calculator, you would find two solutions, one of which we throw out because it corresponds exactly to the initial velocities. The desired solution is

v₁' ≈ -3.11 m/s

v₂' ≈ -0.167 m/s

and take the absolute values to get the magnitudes.

If you want to instead use the masses from the "Required" section, you would end up with

v₁' ≈ -3.18 m/s

v₂' ≈ -0.236 m/s

Explanation:

Given that,

Maximum potential, V = 4. mV

Distance, d = 0.350 m

Frequency of the wave, f = 100 Hz

(a) The maximum electric field strength created is given by:

[tex]E=\dfrac{V}{d}\\\\E=\dfrac{4.1\times 10^{-3}}{0.350 }\\\\E=0.0117\ V/m[/tex]

(b) The corresponding maximum magnetic field strength in the electromagnetic wave is given by :

[tex]B=\dfrac{E}{c}\\\\B=\dfrac{0.0117}{3\times 10^8}\\\\B=3.9\times 10^{-11}\ T[/tex]

(c) The wavelength of the electromagnetic wave can be calculated as :

[tex]\lambda=\dfrac{c}{f}\\\\\lambda=\dfrac{3\times 10^8}{100}\\\\=3\times 10^6\ m[/tex]

So, the wavelength of the electromagnetic wave is [tex]3\times 10^6\ m[/tex].

Answer:

UY Scuti is slightly larger than VY Canis Majoris

Explanation:

These stars are millions of miles away and cannot be seen by the naked eye.

Beetlejuice is another large star that can be seen by the eye.

Answer:

Blackbody radiator, temperature, ultraviolet, catastrophe, electric current, frequency, spectrum, photons

Explanation:

# a p e x

1 and 2 ) A blackbody radiator is an object that absorbs energy, then emits electromagnetic radiation based on the temperature of the object. This comes directly from the definition in the passages.

3 and 4 ) Ultraviolet catastrophe describes when old physicists assumed as frequency increased the waves would go from visible to ultraviolet because that is what comes next on the spectrum. Instead of this happening, the light became white light and it was an apparent 'catastrophe'

Appropriate words for blank position shown below,

A blackbody radiator is defined as an object that absorbs all electromagnetic radiation that falls on it at all frequencies over all angles of incidence.

Ultraviolet light is a type of electromagnetic radiation that makes black-light posters glow

A movement of positive or negative electric particles produce current.

Frequency is defined as the number of cycles or vibrations undergone during one unit of time by a body in periodic motion.

Photons are particles which transmit light.

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Answer:

[tex]\triangle h_c =0.204m[/tex]

Explanation:

Diameter [tex]d=30cm[/tex]

Height [tex]h=90cm[/tex]

Fill height [tex]h_f=60cm[/tex]

Angular speed [tex]N=180rpm[/tex]

Generally the equation for Angular velocity is mathematically given by

[tex]\omega=\frac{2 \pi*N}{60}[/tex]

[tex]\omega=\frac{2 \pi*180}{60}[/tex]

[tex]\omega=18.85rads/s[/tex]

Generally the equation for Liquid surface is mathematically given by

[tex]\mu_s=h*\frac{\omega^2*0.15^2}{4*9.81}[/tex]

[tex]\mu_s=0.396m[/tex]

Therefore the liquid drop at center due to rotation is

[tex]\triangle h_c =h-\mu_s[/tex]

[tex]\triangle h_c =0.60-0.396[/tex]

[tex]\triangle h_c =0.204m[/tex]

We have that the percentage of the energy incident on the cell that is converted to electric energy is

[tex]n=11\%[/tex]

From the question we are told that:

Voltage [tex]V=0.46V[/tex]

Power of light [tex]P=1.0W[/tex]

Wavelength [tex]w=620nm[/tex]

50 \% of the photons give their energy to charge carriers,

Generally, the equation for number of Protons is mathematically given by

[tex]N_p=\frac{P}{E}[/tex]

[tex]N_p=\frac{P \lambda}{hc}[/tex]

[tex]N_p=\frac{1}{(6.62*10^(-34)}*\frac{3*10^8}{(570*10^{-9}))}[/tex]

[tex]N_p=2.87*10^{18}[/tex]

Generally, the equation for Number of electron is mathematically given by

[tex]N_e=50 \% *n_3[/tex]

[tex]N_e=0.5*2.87*10^{18}[/tex]

[tex]N_e=1.43*10^{18}[/tex]

Therefore

Total current

[tex]I= e*N_e[/tex]

Where

e=electron Charge

Therefore

[tex]I=1.43*10^{18}*1.6*10^-{19}[/tex]

[tex]I=0.230A[/tex]

Generally, the equation for Power is mathematically given by

[tex]P=VI[/tex]

[tex]P=0.46*0.230[/tex]

[tex]P=0.1058W[/tex]

Therefore

Efficiency

[tex]n=\frac{0.1058}{1}[/tex]

[tex]n=0.1058[/tex]

[tex]n=11\%[/tex]

In conclusion

The percentage of the energy incident on the cell that is converted to electric energy is

[tex]n=11\%[/tex]

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Answer:

   x = 9.32 cm

Explanation:

For this exercise we have an applied torque and the bar is in equilibrium, which is why we use the endowment equilibrium equation

Suppose the counterclockwise turn is positive, let's set our reference frame at the left end of the bar

          - W l / 2 - W_{child} x + N₂ l = 0

             x = [tex]\frac{-W l/2 + n_2 l}{W_{child}}[/tex]             1)

now let's use the expression for translational equilibrium

         N₁ - W - W_(child) + N₂ = 0

indicate that N₂ = 4 N₁

we substitute

           N₁ - W - W_child + 4 N₁ = 0

           5 N₁ -W - W_{child} = 0

           N₁ = ( W + W_{child}) / 5

we calculate

           N₁ = (450 + 250) / 5

          N₁ = 140 N

we calculate with equation 1

           x = -250 1.50 + 4 140 3) / 140

           x = 9.32 cm

Answer:

1231

Explanation:

Answer:

5572.8 N

Explanation:

Applying,

F  = ma.............. Equation 1

Where F = Force, m = mass of the car, a = acceleration.

We can find a by applying,

v² = u²+2as............. Equation 2

Where v = final velocity, u = initial velocity, a = acceleration,  = distance.

From the question,

Given: v = 0 m/s (come to rest), u = 1.7 m/s, s = 0.210 m

Substitute these value into equation 2

0² = 1.7²+2×0.21×a

a = -1.7²/(2×0.21)

a = -2.89/0.42

a = -6.88 m/s²

Also given: m = 810 kg

Substitute these value into equation 1

F = 810(-6.88)

F = -5572.8 N

Hence the force on the bumber is 5572.8 N

Solution :

Owing to the continuous attraction and repulsion force caused by the magnet or the electromagnet around the core of the motor produces a unidirectional torque whose direction is given by the Lorentz force, [tex]$F=q(\vec v \times \vec B)$[/tex], and thus the torque causes the rotation of the electric motor.

Removing half the insulation from the coil makes it to rotate just half a turn. When the half insulation is removed, the coil turns half and the rest of the time the connection terminates. The rest half turn will be provided by the angular momentum. Now after this half turn by the angular momentum, the connections will again be connected and again the torque will work on it to rotate the half turn. This continues and the motor rotates.

Answer:

Explanation:

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Answer:

Displacement current flows in the dielectric material(insulated region)

Explanation:

Firstly a capacitor stores charge when a capacitor is charging (or discharging), current flows in the circuit. Also, there is no charge transfer in the dielectric material in the capacitor which is contradictory to the flow of current. Hence, displacement current is the current in the insulated region due to the changing electric flux.

Answer:

The Anatomy of a Lens

Refraction by Lenses

Image Formation Revisited

Converging Lenses - Ray Diagrams

Converging Lenses - Object-Image Relations

Diverging Lenses - Ray Diagrams

Diverging Lenses - Object-Image Relations

The Mathematics of Lenses

Ray diagrams can be used to determine the image location, size, orientation and type of image formed of objects when placed at a given location in front of a lens. The use of these diagrams was demonstrated earlier in Lesson 5 for both converging and diverging lenses. Ray diagrams provide useful information about object-image relationships, yet fail to provide the information in a quantitative form. While a ray diagram may help one determine the approximate location and size of the image, it will not provide numerical information about image distance and image size. To obtain this type of numerical information, it is necessary to use the Lens Equation and the Magnification Equation. The lens equation expresses the quantitative relationship between the object distance (do), the image distance (di), and the focal length (f)

Answer:

elevators

Theatre system

construction pulley

lifts

Answer:

elevator,cargo lift system

Answer:

[tex]P_b = \frac{\rho_b}{\rho_a} \ P_a[/tex]

Explanation:

The pressure at a depth of a fluid is

       P = ρ g y

where ρ is the density of the fluid, y the depth of the gauge measured from the surface of the fluid.

In this case the pressure for fluid A is

      Pa = ρₐ g y

the pressure for fluid B is

      P_b = ρ_b g y

depth y not changes as the gauge is stationary

if we look for the relationship between these pressures

       [tex]\frac{P_a}{P_b} = \frac{ \rho_a}{\rho_b}[/tex]

        [tex]P_b = \frac{\rho_b}{\rho_a} \ P_a[/tex]

therefore we see that the pressure measured for fluid B is different from the pressure of fluid A

if  ρₐ < ρ_b B the pressure P_b is greater than the initial reading

   ρₐ>  ρ_b the pressure in B decreases with respect to the reading in liquid A

Answer:

 N = 107.94 N

Explanation:

For this exercise we must use Newton's second law.

Let's set a reference system with the x-axis parallel to the ground and the y-axis vertical

X axis

        Fₓ = ma

ej and

       N -F_y - W = 0

let's use trigonometry to decompose the applied force

     cos -35 = Fₓ / F

     sin -35 = F_y / F

     Fₓ = F cos -35

     F_y = F sin -35

     Fₓ = 40.0 cos -35 = 32.766 N

     F_y = 40.0 sin -35 = -22.94 N

we substitute

     N = Fy + W

     N = 22.94 + 85

     N = 107.94 N

When the rock is suspended in the air, the net force on it is

∑ F₁ = T₁ - m₁g = 0

where T₁ is the magnitude of tension in the string and m₁g is the rock's weight. So

T₁ = m₁g = 37.8 N

When immersed in water, the tension reduces to T₂ = 32.0 N. The net force on the rock is then

∑ F₂ = T₂ + B₂ - m₁g = 0

where B₂ is the magnitude of the buoyant force. Then

B₂ = m₁g - T₂ = 37.8 N - 32.0 N = 5.8 N

B₂ is also the weight of the water that was displaced by submerging the rock. Let m₂ be the mass of the displaced water; then

5.8 N = m₂g   ==>   m₂ ≈ 0.592 kg

If one takes the density of water to be 1.00 g/cm³ = 1.00 × 10³ kg/m³, then the volume of water V that was displaced was

1.00 × 10³ kg/m³ = m₂/V   ==>   V ≈ 0.000592 m³ = 592 cm³

and this is also the volume of the rock.

When immersed in the unknown liquid, the tension reduces further to T₃ = 20.2 N, and so the net force on the rock is

∑ F₃ = T₃ + B₃ - m₁g = 0

which means the buoyant force is

B₃ = m₁g - T₃ = 37.8 N - 20.2 N = 17.6 N

The mass m₃ of the liquid displaced is then

17.6 N = m₃g   ==>   m₃ ≈ 1.80 kg

Then the density ρ of the unknown liquid is

ρ = m₃/V ≈ (1.80 kg)/(0.000592 m³) ≈ 3040 kg/m³ = 3.04 g/cm³

Answer:

Option (a) is correct.

Explanation:

The lines in the emission spectrum of hydrogen is due to the transfer of electrons form higher energy levels to the lower energy levels.

When the electrons transfer from one level of energy that is higher level of energy to the other means to the lower level of energy then they emit some photons which having the frequency or the wavelength in the visible region.

Answer:

The rms voltage of new generator is 3.4 V.

Explanation:

f = 3200 Hz

rms voltage, V = 2 V

rms current, i = 28 mA

Now

f' = 4700 Hz

rms current, i' = 70 mA

let the new rms voltage is V'.

[tex]i = \frac{V}{Xc} = V \times 2\pi fC....(1)\\\\i' = V' \times 2 \pi f' C..... (2)\\\\\frac{i}{i'} =\frac{V f}{V' f'}\\\\\frac{28}{70}=\frac{2\times 3200}{V'\times 4700}\\\\V' = 3.4 V[/tex]

Answer:

2Al + 2H2O + 2NaOH ⟶ 3H2 + 2NaAlO2

Chất rắn màu xám bạc của nhôm (Al) tan dần trong dung dịch, sủi bọt khí là hidro (H2).

Explanation:

a =  - 8.34 m/sec²       ( deceleration or negative)  

Equations for UAM  ( uniformly accelerated motion) are:

vf = v₀ ±  a*t           and      s  =  s₀  + v₀*t + (1/2)*a*t²

In our case, the motion is with deceleration, then

vf = v₀ - a*t       and    s = s₀  +  v₀*t  - (1/2)*a*t²  

working on these equatios we get:

vf = v₀ - a*t     (1)           s  -  s₀   =  v₀*t  - (1/2)*a*t²      (2)

v₀ - vf = a*t

t  =  (v₀ - vf)/a

By substitution of (1) in equation (2)

s  -  s₀   =  v₀ * (v₀ - vf)/a  -  (1/2) * a* [(v₀ - vf)/a]²

s  -  s₀   =  (v₀² - v₀*vf)/a  -   (1/2) * a* (1/a²)* (v₀ - vf)²

s  -  s₀   =  1/a * ( v₀² - v₀*vf ) - 1/a* (1/2)  * (v₀ - vf)²

s  -  s₀   =  1/a* [  ( v₀² - v₀*vf ) - (1/2) * (v₀ - vf)²]

a * (s  -  s₀ )  =   v₀² - v₀*vf - v₀²/2 - vf²/2 + v₀*vf

a * (s  -  s₀ )  =  (1/2) * v₀² - (1/2)*vf²

a * (s  -  s₀ )  =  (1/2) * ( v₀² - vf²)

We find an expression to calculate the minimum deceleration to stop the car in time to avoid crashing

s₀ = 50 meters            s  =  0        v₀ =  104 Km/h    vf = 0

1 Km  = 1000 m    and   1 h = 3600 sec

v₀ = 104 Km/h    =  28.88 m/sec

a  =  (1/2) [ (28.88)² - 0 ] / 0 - 50

a =  - 8.34 m/sec²       ( deceleration or negative)  

Answer:

in which topic you need help

Answer:

A change in pressure at one point in an incompressible fluid is felt at every other point in the fluid.

Explanation:

Pascal's principle states that ''pressure applied to an enclosed fluid will be transmitted without a change in magnitude to every point of the fluid and to the walls of the container.''(Science direct).

The implication of this law is; that a change in pressure at one point in an incompressible fluid is felt at every other point in the fluid. Hence the correct answer chosen above.

The Pascal's principle is applied in hydraulic jacks and automobile brakes.

Answer:

  E = 1.25 MV / m

Explanation:

For this exercise let's use Newton's second law

          F = m a

where the force is electric

          F = q E

we substitute

          q E = m a

          E = m a / q

indicate there are 500,000 excess electrons

          q = 500000 e

          q = 500000 1.6 10⁻¹⁹

          q = 8 10⁻¹⁴ C

the mass is m = 10⁻¹¹ kg and the acceleration a = 10⁴ m / s²

let's calculate

          E = 10⁻¹¹ 10⁴ / 8 10⁻¹⁴

          E = 0.125 10⁷ V / m = 1.25 10⁶ V / m

          E = 1.25 MV / m

Answer:

No ejection of photo electron takes place.

Explanation:

When a photon of suitable energy falls on cathode, then the photoelectrons is emitted from the cathode. This phenomenon is called photo electric effect.

The minimum energy required to just  eject an electron is called work function.

The photo electric equation is

E = W + KE

where, E is the incident energy, W is the work function and KE is the kinetic energy.

W = h f

where. h is the Plank's constant and f is the threshold frequency.

Now, when the violet light is falling, no electrons is ejected. When the red light is falling, whose frequency is less than the violet light, then again no photo electron is ejected from the metal surface.