To practice Problem-Solving Strategy 15.2 Standing Waves.In lab, your instructor generates a standing wave using a thin string of length L = 2.25 m fixed at both ends. You are told that the standing wave is produced by the superposition of traveling and reflected waves, where the incident traveling waves propagate in the +x direction with an amplitude A = 2.45 mm and a speed vx = 14.5 m/s . The first antinode of the standing wave is a distance of x = 37.5 cm from the left end of the string, while a light bead is placed a distance of 18.8 cm to the right of the first antinode. What is the maximum transverse speed vy of the bead? Make sure to use consistent distance units in your calculations.To check your equation for the standing wave's transverse velocity, find the maximum transverse velocity at x = 75.0 cm .​

Answer:

u = 10.02m/s

Explanation:

a = f/m

a = 20/12 = 1.67m/s²

U =2aS

u = 2 x 1.67 x 3

U = 10.02m/s

Answer:

The Total current drawn is 20.83 Ampere.

Explanation:

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Answer:

See detailed solution below

Explanation:

a) From C= εoεrA/d

Where;

C= capacitance of the capacitor

εo= permittivity of free space

εr= relative permittivity

A= cross sectional area

d= distance between the plates

Since the relative permittivity of air=1 and permittivity of free space = 8.85 × 10^−12 Fm−1

Then;

C= 8.85 × 10^−12 Fm−1 × 0.2m^2/0.009 m

C= 196.67 × 10^-12 F or 1.967 ×10^-10 F

b) Q= CV = 1.967 ×10^-10 F × 140 V = 2.75 × 10^-8 C

c) E= V/d = 140 V/0.009m = 15.56 Vm-1

d) W= 1/2 CV^2 = 1/2 × 1.967 ×10^-10 F × (140)^2 =1.93×10^-6J

Part II

When the distance is now 0.014 m

a) C= 8.85 × 10^−12 Fm−1 × 0.2m^2/0.014 m = 1.26×10^-10 F

b) W= 1/2 Q^2/C = 1/2 × ( 2.75 × 10^-8 C)^2 / 1.26×10^-10= 3×10^-6 J

Note that the voltage changes when the distance is changed but the charge remains the same

Answer:

T= 2p√m/k

Explanation:

This is because the period of oscillation of the mass of spring system is directly proportional to the square root of the mass and it is inversely proportional to the square root of the spring constant.

The period of a mass on a spring is given by the equation

T=2π√m/k.

Where T is the period,

M is mass

K is spring constant.

An increase in mass in a spring increases the period of oscillation and decrease in mass decrease period of oscillation.

When there is the relationship between the Period (T) caused by the oscillation of the mass should be considered as the T= 2p√m/k.

The mass of the spring system with respect to period of oscillation should be directly proportional to the square root of the mass and it is inversely proportional to the square root of the spring constant.

So the following equation should be considered

T=2π√m/k.

Here,

T is the period,

M is mass

K is spring constant.

An increase in mass in a spring rises the period of oscillation and reduce in mass decrease period of oscillation.

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Answer:

HSBC keen vs kg get it yyyyyuuy

Explanation:

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nnnbvvvvvggfxrugdfutdfjhyfggigftffghhjjhhjyhrdffddfvvvvvvvvvvvbbbbbbbbbvvcxccghhyhhhjjjhjnnnnnnnnnnnnnbhbfgjgfhhccccccvvjjfdbngxvncnccbnxcvbchvxxghfdgvvhhihbvhbbhhvxcgbbbcxzxvbjhcxvvbnnxvnn

Answer:

1. Grating

2. Interference

3. Diffraction

4. Specular dot

Explanation:

1. Composed of numerous narrowly spaced slits and grooves  ........  Grating

2. Having the same wavelength, frequency, and in-phase Interaction of waves where they meet in space .......   Interference

3. The bending of waves near a boundary or as a wave passes through an opening ...... Diffraction

4. The zeroth order direct reflection fringe  ......  Specular dot

Answer:

A.  Z = 185.87Ω

B.  I  =  0.16A

C.  V = 1mV

D.  VL = 68.8V

E.  Ф = 30.59°

Explanation:

A. The impedance of a RL circuit is given by the following formula:

[tex]Z=\sqrt{R^2+\omega^2L^2}[/tex]       (1)

R: resistance of the circuit = 160-Ω

w: angular frequency = 220 rad/s

L: inductance of the circuit = 0.430H

You replace in the equation (1):

[tex]Z=\sqrt{(160\Omega)^2+(220rad/s)^2(0.430H)^2}=185.87\Omega[/tex]

The impedance of the circuit is 185.87Ω

B. The current amplitude is:

[tex]I=\frac{V}{Z}[/tex]                     (2)

V: voltage amplitude = 30.0V

[tex]I=\frac{30.0V}{185.87\Omega}=0.16A[/tex]

The current amplitude is 0.16A

C. The current I is the same for each component of the circuit. Then, the voltage in the resistor is:

[tex]V=\frac{I}{R}=\frac{0.16A}{160\Omega}=1*10^{-3}V=1mV[/tex]            (3)

D. The voltage across the inductor is:

[tex]V_L=L\frac{dI}{dt}=L\frac{d(Icos(\omega t))}{dt}=-LIsin(\omega t)\\\\V_L=-(0.430H)(160\Omega)sin(220 t)=68.8sin(220t)\\\\V_L_{max}=68.8V[/tex]

E. The phase difference is given by:

[tex]\phi=tan^{-1}(\frac{\omega L}{R})=tan^{-1}(\frac{(220rad/s)(0.430H)}{160\Omega})\\\\\phi=30.59\°[/tex]

Answer:

E = 2750 J at h = 5 m

Explanation:

The gravitational potential energy is given by :

[tex]E=mgh[/tex]

In this case, m is the mass of swimmer is constant at every heights. So,

At h = 1 m, E = 550 J

[tex]550=m\times 10\times 1\\\\m=55\ kg[/tex]

So, at h = 5 m, gravitational potential energy is given by :

[tex]E=55\times 10\times 5\\\\E=2750\ J[/tex]

So, the correct option is (B).

Answer:

Approximately [tex]0.45\; \rm m \cdot s^{-1}[/tex].

Explanation:

The mechanical energy of an object is the sum of its potential energy and kinetic energy. Consider this question from the energy point of view:

Mechanical energy of the block [tex]0.04\; \rm m[/tex] away from the equilibrium position:

While the block moves back to the equilibrium position, it keeps losing (mechanical) energy due to friction:

[tex]\begin{aligned}& \text{Work done by friction} = (-0.3\; \rm N) \times (0.04 \; \rm m) = -0.012\; \rm J\end{aligned}[/tex].

The opposite ([tex]0.012\; \rm N[/tex]) of that value would be the amount of energy lost to friction. Since there's no other form of energy loss, the mechanical energy of the block at the equilibrium position would be [tex]0.032\; \rm N - 0.012\; \rm N = 0.020\; \rm N[/tex].

The elastic potential energy of the block at the equilibrium position is zero. As a result, all that [tex]0.020\; \rm N[/tex] of mechanical energy would all be in the form of the kinetic energy of that block.

Given that the mass of this block is [tex]0.020\; \rm kg[/tex], calculate its speed:

[tex]\begin{aligned}v &= \sqrt{\frac{2\, \mathrm{KE}}{m}} \\ &= \sqrt{\frac{2 \times 0.020\; \rm J}{0.20\; \rm kg}} \approx 0.45\; \rm m\cdot s^{-1}\end{aligned}[/tex].

Answer:

The atomic number

Explanation:

Answer:

The horizontal force acting on m2 is F + 9.8m1

Explanation:

Given;

Block m1 on left of block m2

Make a sketch of this problem;

                         F →→→→→→→→→→→-------m1--------m2

Apply Newton's second law of motion;

F = ma

where;

m is the total mass of the body

a is the acceleration of the body

The horizontal force acting on block m2 is the force applied to block m1 and force due to weight of block m1

F₂ = F + W1

F₂ = F + m1g

F₂ = F + 9.8m1

Therefore, the horizontal force acting on m2 is F + 9.8m1

The force acting on the block of mass m₂ is  [tex]\frac{m_2F}{m_1+m_2}[/tex]

Given that there are two blocks of mass m₁ and m₂.

m₁ is on the left of block m₂. They are in contact with each other.

A force F is applied on m₁ to the right.

According to Newton's laws of motion:

The equation of motion of the blocks can be written as:

F = (m₁ + m₂)a

here, a is the acceleration.

so, acceleration:

a = F / (m₁ + m₂)

Now, the force acting on the block of mass m₂ is:

f = m₂a

[tex]f = \frac{m_2F}{m_1+m_2}[/tex]

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Answer:

a) ±7.08×10⁻⁷ C/m²

b) 1.77×10⁻⁷ C

Explanation:

For a conductor,

σ = ±Eε₀,

where σ is the charge density,

E is the electric field,

and ε₀ is the permittivity of space.

a)

σ = ±Eε₀

σ = ±(8×10⁴ N/C) (8.85×10⁻¹² F/m)

σ = ±7.08×10⁻⁷ C/m²

b)

σ = q/A

7.08×10⁻⁷ C/m² = q / (0.5 m)²

q = 1.77×10⁻⁷ C

Answer:

Explanation:

Expression for time period of pendulum is given as follows

[tex]T=2\pi\sqrt{\frac{l}{g} }[/tex]

where l is length of pendulum and g is acceleration due to gravity .

Putting the given values for first place

[tex]2.67=2\pi\sqrt{\frac{l}{9.77} }[/tex]

Putting the values for second place

[tex]T=2\pi\sqrt{\frac{l}{9.81} }[/tex]

Dividing these two equation

[tex]\frac{T}{2.67} =\sqrt{\frac{9.77}{9.81} }[/tex]

T = 2.66455 s.

Answer:

(a) force is 1066.56N

Explanation:

(a) MV²/R

Complete Question

For a human body falling through air  in  a spread edge position , the numerical value of the constant D is about [tex]D = 0.2500 kg/m[/tex]

What value of D is required to make vt = 42.7 m/s the terminal velocity of a skydiver of mass 85.0 kg . Express your answer using two significant figures?

Answer:

The value of D is   [tex]D = 0.457 \ kg/m[/tex]

Explanation:

From the question we are told that

     The terminal velocity is  [tex]v_t = 42.7 \ m/s[/tex]

     The mass of the skydiver is  [tex]m = 85.0 \ kg[/tex]

      The numerical value of  D  is  [tex]D = 0.2500 kg/m[/tex]

From the unit of D  in the question we can evaluate D as  

       [tex]D = \frac{m * g }{v^2}[/tex]

substituting values  

        [tex]D = \frac{85 * 9.8 }{(42.7)^2}[/tex]

         [tex]D = 0.457 \ kg/m[/tex]

Answer: A. Strong nuclear

The max effective range of strong nuclear force is about 1.2 femtometers ( which is 1.2*10^(-15) meters). This is well below 1 meter. Strong nuclear forces are the forces that hold together a nucleus. Specifically it holds together the protons that would otherwise repel one another due to similar charge.

Answer:

(a) V = 65.625 Volts

(b) V = 131.25 Volts

(c) V = 131.25 Volts

Explanation:

Recall that:

1) in a metal sphere the charges distribute uniformly around the surface, and the electric field inside the sphere is zero, and the potential is constant equal to:

[tex]V=k\frac{Q}{R}[/tex]

2) the electric potential outside of a charged metal sphere is the same as that of a charge of the same value located at the sphere's center:

[tex]V=k\frac{Q}{r}[/tex]

where k is the Coulomb constant ( [tex]9\,\,10^9\,\,\frac{N\,m^2}{C^2}[/tex] ), Q is the total charge of the sphere, R is the sphere's radius (0.24 m), and r is the distance at which the potential is calculated measured from the sphere's center.

Then, at a distance of:

(a) 48 cm = 0.48 m, the electric potential is:

[tex]V=k\frac{Q}{r}=9\,\,10^9 \,\frac{3.5\,\,10^{-9}}{0.48} =65.625\,\,V[/tex]

(b) 24 cm = 0.24 m, - notice we are exactly at the sphere's surface - the electric potential is:

[tex]V=k\frac{Q}{r}=9\,\,10^9 \,\frac{3.5\,\,10^{-9}}{0.24} =131.25\,\,V[/tex]

(c) 12 cm (notice we are inside the sphere, and therefore the potential is constant and the same as we calculated for the sphere's surface:

[tex]V=k\frac{Q}{R}=9\,\,10^9 \,\frac{3.5\,\,10^{-9}}{0.24} =131.25\,\,V[/tex]

Answer:

c) a difference in electric potential

Explanation:

my insta: priscillamarquezz

Answer:

(1) Percent Difference = 2.47%

(2) Percent Error (9.96 m/s²) = 1.63 %

    Percent Error (9.72 m/s²) = 0.82 %

(3) Percent Error (Mean) = 0.41 %

Explanation:

(1)

Percent Difference = [(9.96 m/s² - 9.72 m/s²)/(9.72 m/s²)]*100 %

Percent Difference = 2.47%

(2)

Percent Error = (|Measured Value - Original Value|/Original Value)*100%

Therefore,

Percent Error (9.96 m/s²) = (|9.96 m/s² - 9.8 m/s²|/9.8 m/s²)*100%

Percent Error (9.96 m/s²) = 1.63 %

Now,

Percent Error (9.72 m/s²) = (|9.72 m/s² - 9.8 m/s²|/9.8 m/s²)*100%

Percent Error (9.72 m/s²) = 0.82 %

(3)

First we need to find the mean of values:

Mean = (9.96 m/s² + 9.72 m/s²)/2

Mean = 9.84 m/s²

Therefore,

Percent Error (Mean) = (|9.84 m/s² - 9.8 m/s²|/9.8 m/s²)*100%

Percent Error (Mean) = 0.41 %

The correct answer is B. No, because she is not working on the atomic structures of a solid

Explanation:

Solid-state physics is a sub-discipline of physics that focuses on studying solids, this includes analyzing solids structures, features, and other phenomena that occur in substances in this state of the matter. This means a solid-state physics will not study or gases.

In this context, the fact the scientist is trying to create an electromagnetic field by using solutions and the flow of these show the scientists is not working with solids but with liquids or gases as solids do not flow. Also, her focus is not solids, and therefore she is not a solid-state physicist. Thus, it can be concluded she is not a solid-state physicists because she is not working on the structures of solids.

Answer:

Explanation:

Mass ( m ) = 2 kg

Speed of the object (v) = 6 metre per second

Kinetic energy =?

Now,

We have,

Kinetic Energy = [tex] \frac{1}{2} \times m \times {v}^{2} [/tex]

Plugging the values,

[tex] = \frac{1}{2} \times 2 \times {(6)}^{2} [/tex]

Reduce the numbers with Greatest Common Factor 2

[tex] = {(6)}^{2} [/tex]

Calculate

[tex] = 36 \: joule[/tex]

Hope this helps...

Good luck on your assignment...

The Kinetic energy of the object will be "36 joules".

The excess energy of moving can be observed as that of the movement of an object, component, as well as the group of components. There would never be a negative (-) amount of kinetic energy.

According to the question,

Mass of object, m = 2 kg

Speed of object, v = 6 m/s

As we know the formula,

→ Kinetic energy (K.E),

= [tex]\frac{1}{2}[/tex] × m × v²

By substituting the values, we get

= [tex]\frac{1}{2}[/tex] × 2 × (6)²

=  [tex]\frac{1}{2}[/tex] × 2 × 36

= 36 joule

Thus the above answer is appropriate.

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Answer:

C. R^2

Explanation:

A cyclotron is a particle accelerator which employs the use of electric and magnetic fields for its functioning. It consists of two D shaped region called dees and the magnetic field present in the dee is responsible for making sure the charges follow the half-circle and then to a gap in between the dees.

R is denoted as the radius of the final orbit then the final particle energy is proportional to the radius of the two dees. This however translates to the energy being proportional to R^2.

Answer:

The wavelength is  [tex]\lambda = 0.01367 m[/tex]

The wave number is  [tex]N = 73.18\ m^{-1}[/tex]

The wave function is  [tex]y= 1.57 *10^{-6} sin 2 \pi ( 73.178 x -25100t)[/tex]

Explanation:

From the question we are told that

    The frequency of the sound wave is  [tex]f = 25,100 Hz[/tex]

    The speed of the wave is  [tex]v = 343 m/s[/tex]

The wavelength of the wave is mathematically evaluated as

           [tex]\lambda = \frac{v}{f}[/tex]

substituting values

         [tex]\lambda = \frac{343}{25100}[/tex]

        [tex]\lambda = 0.01367 m[/tex]

The wave number is  mathematically represented as

       [tex]N= \frac{1}{\lambda }[/tex]

substituting values

        [tex]N = \frac{1}{ 0.01367 }[/tex]

        [tex]N = 73.18\ m^{-1}[/tex]

The general form of  wave function is

        [tex]y= A sin (kx -wt)[/tex]

given that the amplitude is   [tex]A = 1.57*10^{-6} \ m[/tex]

  While [tex]w[/tex] which is the angular velocity is  represented as  [tex]w = 2 \pi f[/tex]

  and  k which is the angular wave number is mathematically represented as    [tex]k = \frac{2 \pi }{\lambda }[/tex]

   The wave function becomes  

       [tex]y= A sin 2 \pi (\frac{1}{\lambda} x -ft)[/tex]

substituting values

      [tex]y= 1.57 *10^{-6} sin 2 \pi ( 73.178 x -25100t)[/tex]

Answer:

Explanation:

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Answer:

F = umg where u is coefficient of dynamic friction

Explanation:

F = 0.4 x 80 x 9.81 = 313.92 N

Answer:

Explanation:

We shall apply conservation of mechanical energy law to solve the problem .

loss of height = L ( 1 -  cos 37 ) where L is length of rope

loss of potential energy at the bottom = gain of kinetic energy .

mg L ( 1 - cos 37 ) = 1/2 m v² where v is velocity at the bottom

v² = 2 L g ( 1 - cos 37 )

= 2 x 10 x 9.8 ( 1 - cos 37 )

= 39.46

v = 6.28 m /s

Answer:

The  length is  [tex]D = 5 \ cm[/tex]

Explanation:

From the question we are told  that

     The  length of the  hand is  [tex]l = 10.0 \ cm[/tex]

      The  angle at the hand is  held is  [tex]\theta = 30 ^o[/tex]

Generally resolving the length the length of the hand to it vertical component we obtain that the length of the shadow on the vertical wall is mathematically evaluated as

             [tex]D = l * sin(\theta )[/tex]

substituting values

             [tex]D = 10 * sin (30)[/tex]

             [tex]D = 5 \ cm[/tex]

Answer:

The  corresponding  magnetic field is  

Explanation:

From the question we are told that

    The electric field amplitude is  [tex]E_o = 611\ V/m[/tex]

Generally the  magnetic  field amplitude is  mathematically represented as

              [tex]B_o = \frac{E_o }{c }[/tex]

Where c is the speed of light with a constant value

         [tex]c = 3.0 *0^{8} \ m/s[/tex]

So  

        [tex]B_o = \frac{611 }{3.0*10^{8}}[/tex]

         [tex]B_o = 2.0 4 *10^{-6} \ Vm^{-2} s[/tex]

Since 1  T  is  equivalent to  [tex]V m^{-2} \cdot s[/tex]

         [tex]B_o = 2.0 4 *10^{-6} \ T[/tex]

Explanation:

Mass and energy are closely related. Due to mass–energy equivalence, any object that has mass when stationary (called rest mass) also has an equivalent amount of energy whose form is called rest energy, and any additional energy (of any form) acquired by the object above that rest energy will increase the object's total mass just as it increases its total energy. For example, after heating an object, its increase in energy could be measured as a small increase in mass, with a sensitive enough scale.

Answer:

Second image in your list of possible answers

Explanation:

The second option is what you would expect from modulating a sinusoidal carrier wave of higher frequency after being modulated by a square pulse of lower frequency that allows part of the carrier signal to travel during the time the square signal is constant different from zero, and be absent (flat) during the time the square pulse signal has amplitude zero.

The second line is the best picture of a pulse-modulated carrier.

It would be easy to build a circuit where each pulse ... when it comes along ... just switches the carrier OFF for as long as the pulse lasts.

Answer:

The correct answer will be "[tex]\tau =\frac{L}{R}[/tex]".

Explanation:

The time it would take again for current or electricity flows throughout the circuit including its LR modules can be connected its full steady-state condition is equal to approximately 5[tex]\tau[/tex] as well as five-time constants.

It would be calculated in seconds by:

⇒  [tex]\tau=\frac{L}{R}[/tex]

, where