For a demonstration, a professor uses a razor blade to cut a thin slit in a piece of aluminum foil. When she shines a laser pointer (λ=680nm) through the slit onto a screen 5.5 m away, a diffraction pattern appears. The bright band in the center of the pattern is 8.0 cm wide. What is the width of the slit?

Answer:

The force that would be applied on the rope just to start moving the wagon is 122 N

Explanation:

Frictional force opposes motion between two surfaces in contact. It is the force that must be applied before a body starts to move. Static friction  opposes the motion of two bodies that are in contact but are not moving. The magnitude of static friction to overcome for the body to move  can be calculated using equation 1.

F = μ x mg .............................. 1

where F is the frictional force;

          μ is the coefficient of friction ( μs, in this case, static friction);

          m  is mass of the object and;

          g is the acceleration due to gravity( a constant equal to 9.81 m/[tex]s^{2}[/tex])

from the equation we are provide with;

       μs  = 0.25

       m = 50 kg

       g =  9.81 m/[tex]s^{2}[/tex]

      F =?

Using equation 1

F = 0.25 x 50 kg x  9.81 m/[tex]s^{2}[/tex]

F = 122.63 N  

Therefore a force of 122 N must be applied to the rope just to start the wagon.

Explanation:

Below is an attachment containing the solution.

Answer:

b) 8.

Explanation:

Below is an attachment containing the solution.

Answer:

The frequency 400 hz is not possible .

Explanation:

Given that,

Frequency = 200 hz

Length = l

Suppose, The given frequencies are,

600 Hz, 1000 Hz, 1400 Hz, 1800 Hz and 400 hz.

The possible resonance frequencies are

We need to calculate the fundamental frequency

Using formula of fundamental frequency for pipe

[tex]F=\dfrac{nv}{4L}[/tex]

Where, n = odd number

Put the value of frequency

[tex]200= \dfrac{nv}{4L}[/tex]

We need to calculate the first over tone

Using formula of fundamental frequency

n = 3,

[tex]F=\dfrac{nv}{4L}[/tex]

Put the value into the formula

[tex]F_{2}=3\times\dfrac{v}{4l}[/tex]

[tex]F_{2}=3\times200[/tex]

[tex]F_{2}=600\ Hz[/tex]

We need to calculate the second over tone

Using formula of fundamental frequency

n = 5,

[tex]F=\dfrac{nv}{4L}[/tex]

Put the value into the formula

[tex]F_{3}=5\times200[/tex]

[tex]F_{3}=1000\ Hz[/tex]

We need to calculate the third over tone

Using formula of fundamental frequency

n = 7,

[tex]F=\dfrac{nv}{4L}[/tex]

Put the value into the formula

[tex]F_{4}=7\times200[/tex]

[tex]F_{4}=1400\ Hz[/tex]

We need to calculate the fourth over tone

Using formula of fundamental frequency

n = 9,

[tex]F=\dfrac{nv}{4L}[/tex]

Put the value into the formula

[tex]F_{5}=9\times200[/tex]

[tex]F_{5}=1800\ Hz[/tex]

Hence, The frequency 400 hz is not possible .

Final answer:

In a pipe that is closed at one end and open at the other, only odd multiples of the fundamental frequency will resonate. Frequencies that are even multiples of the fundamental 200 Hz, such as 400 Hz, 600 Hz, 800 Hz, will not resonate.

Explanation:

The lowest-pitch tone to resonate in a pipe of length l that is closed at one end and open at the other end is 200 Hz. This pipe supports harmonic frequencies following the sequence fn = n(v/4L), where n = 1, 3, 5..., v is the speed of sound, and L is the length of the pipe. Therefore, frequencies that will not resonate in the same pipe are those that are even multiples of the fundamental frequency, such as 400 Hz, 600 Hz, 800 Hz, and so on.

Answer:

The lowest possible frequency of sound is 971.4 Hz.

Explanation:

Given that,

Distance between  loudspeakers = 2.00 m

Height = 5.50 m

Sound speed = 340 m/s

We need to calculate the distance

Using Pythagorean theorem

[tex]AC^2=AB^2+BC^2[/tex]

[tex]AC^2=2.00^2+5.50^2[/tex]

[tex]AC=\sqrt{(2.00^2+5.50^2)}[/tex]

[tex]AC=5.85\ m[/tex]

We need to calculate the path difference

Using formula of path difference

[tex]\Delta x=AC-BC[/tex]

Put the value into the formula

[tex]\Delta x=5.85-5.50[/tex]

[tex]\Delta x=0.35\ m[/tex]

We need to calculate the lowest possible frequency of sound

Using formula of frequency

[tex]f=\dfrac{nv}{\Delta x}[/tex]

Put the value into the formula

[tex]f=\dfrac{1\times340}{0.35}[/tex]

[tex]f=971.4\ Hz[/tex]

Hence, The lowest possible frequency of sound is 971.4 Hz.

Answer:

t = 0.1 s

Explanation:

Given:

- The distance crushed (Stopping-Distance) s = 1.20 m

- The mass of the car m = 1,550 kg

-  The initial velocity vi = 24.0 m/s

Find:

- How long does the collision last t?

Solution:

- The stopping distance s is the average velocity v times time t as follows:

                                s = t*( vf + vi ) / 2

Where,

             vf = 0 m/s ( Stopped )

                                s = t*( vi ) / 2

                                t = 2*s / vi

                                t = 2*1.20 / 24

                                t = 0.1 s

Answer:

t=0.1seconds.

Explanation:

The mass of the car m = 1,550 kg  

We know the stopping distance, 1.20m,

we know the final velocity, 0m/s (its stopped),

the starting velocity, 24m/s.

d=t(v2+v1)/2

Rearange to solve for t, and remove the v2^2 as its zero

t=2d/v1  

t=2(1.20m)/(24m/s)

t=0.1seconds.

Answer:

They have a frequency of 2 hertz

Explanation:

Frequency is the number of occurrences of a repeating event per unit of time and Wavelength is the distance from one crest to another, or from one trough to another.

In the question, it is stated that the leaf does two full up and down bobs, this means that it completes 2 full cycles in one second. Therefore, its frequency is 2/s

where, s⁻¹ is hertz

so, They have a frequency of 2 hertz

Answer:

0.27J

Explanation:

[tex]C_eq= C_1 + C_2\\= 2+4\\= 6UF\\U = (1/2) CV^2\\= (1/2)(6 - 6)(300 * 300)\\= 0.27J[/tex]

Answer:

he failed thousands of times

Explanation:

There is no known number for his failings. Edison may have failed in many of his experiments and in his schooling, but he had something better working in his favor. He had great determination and persistence.

He failed thousands of times in an attempt to develop an electric light, the great Edison simply viewed each unsuccessful experiment as the elimination of a solution that wouldn’t work, thereby moving him that much closer to a successful solution.

The diameter of one leg of the stool is missing and it's 2cm.

Answer:

(ΔL/L) = 0.00729%

Explanation:

If the Weight of the man is W, the weight will be distributed equally on the 3 legs and so the reactions for each leg will be W/3 or F/3.

Now, Youngs modulus(Y) of douglas fir wood is about 1.3 x 10^(10) N/m^2. Gotten from youngs modulus of common materials.

Now, weight of man is 70kg.

Now diameter of one leg is 2cm.so radius of one leg = 2/2 = 1cm = 1 x 10^(-2)m

Area for one leg is; π( 1 x 10^(-2)m)^2 = 3.14 x 10^(-4)m

Now as stated earlier, the force on one leg is; F/3.

Now F = mg = 70 x 9.81 = 686.7N

So, force on one leg = 686.7/3 = 228. 9N

Now we know youngs modulus(Y) = Stress/Strain.

Stress = F/A while Strain = ΔL/L

Therefore Y = (F/A) / (ΔL/L)

And therefore, (ΔL/L) = F/(AY)

So (ΔL/L) = 228.9/(3.14 x 10^(-4))x(1.3 x 10^(10)) = 7. 29 x 10^(-5)

When expressed in percentage, it becomes 0.00729%

An accurate calculation for the percent decrease in the stool legs' length under a 70 kg man's weight requires knowledge of the materials properties and dimensions of the stool. General concepts of stress, strain, and deformation in materials are discussed, including their relationship to applied forces and material properties as described by Hooke's Law.

To determine the percent decrease in the length of the stool legs when a 70 kg man sits on it, we need additional information such as the material properties of the stool legs and any relevant physical dimensions. However, with the given scenario, we can talk generally about stress and strain in materials and how they are calculated in relation to force and deformation. Stress is the force per unit area applied to a material, while strain is the deformation experienced by the material relative to its original length. For an object like a stool leg, the deformation (change in length) due to a person sitting on it could be calculated using Hooke's Law if the deformation remains within the linear elastic range of the material.

Without specific data on the stool's material, cross-sectional area, and elastic modulus, we cannot calculate the exact percent decrease in length of the stool's legs. If such information were provided, we could use the formula σ = F/A, where σ is the stress, F is the force, and A is the cross-sectional area; and the formula ε = δL/L, where ε is the strain, δL is the change in length, and L is the original length. Combined with Hooke's Law (σ = Eε, E being the elastic modulus), we could find the deformation and thus the percent decrease in length.

Answer: Two marbles separated by 300 kilometers

Explanation: Hope i helped have a great day and please mark brainliest i would appreciate it!

Answer:

      [tex]2.42liter[/tex]

Explanation:

The question is incomplete; you need some additional data.

I will assume the missing data from a similar question and keep the final pressure of 703 torr given in your question.

An ideal gas at a pressure of 1.20 atm is contained in a bulb of unknown volume. A stopcock is used to connect this bulb with a previously evacuated bulb that has a volume of 0.720 L as shown here. When the stopcock is opened the gas expands into the empty bulb

If the temperature is held constant during this process and the final pressure is 703 torr , what is the volume of the bulb that was originally filled with gas?

Since the amount of gas and the temperature remain constant, we may use Boyle's law:

             [tex]PV=constant\\\\P_1V_1=P_2V_2[/tex]

Form the data:

        [tex]P_1=1.20atm\\\\P_2=703torr\\\\V_1=unknown\\\\V_2=V_1+0.720liter[/tex]

1. Convert 703 torr to atm:

   [tex]P_2=703torr\times 1atm/760torr=0.925atm[/tex]

2. Sustitute the data in the equation:

       [tex]1.2atm\times V_1=0.925atm\times(V_1+0.720liter)[/tex]

3. Solve:

      [tex]1.2V_1=0.925V_1+ 0.666liter\\\\0.275V_1=0.666liter\\\\V_1=2.4218liter\approx 2.42liter[/tex]

The answer is rounded to 3 significant figures, according to the data.

The maximum potential difference between points A and B is 100 V, and the maximum energy stored in the three-capacitor arrangement is determined by the capacitor with the smallest capacitance.

The maximum potential difference (voltage) between points A and B in a series combination of capacitors is limited by the capacitor with the lowest breakdown voltage. Given C2 = 20.0 μF and C3 = 25.0 μF, and no capacitor can exceed 100 V, the maximum potential difference that can exist between points A and B is 100 V.

The maximum energy that can be stored in the three-capacitor arrangement is determined by the capacitor with the least energy storage capability. In this case, the capacitors are in series, so the total energy stored in the arrangement is the energy stored by the capacitor with the smallest capacitance, which is 20 μF, at the maximum allowed voltage of 100 V.

Answer:

a. [tex]6.032\times10^{-6}C/m^2[/tex]

b.[tex]6.816\times10^5N/C[/tex]

Explanation:

#Apply  surface charge density, electric field, and Gauss law to solve:

a. Surface charge density is defined as charge per area denoted as [tex]\sigma[/tex]

[tex]\sigma=\frac{Q}{4\pi r_{out}^2}[/tex], and the strength of the electric field outside the sphere [tex]E=\frac{\sigma _{new}}{\epsilon _o}[/tex]

Using Gauss Law, total electric flux out of a closed surface is equal to the total charge enclosed divided by the permittivity.

[tex]\phi=\frac{Q_{enclosed}}{\epsilon_o}\\\\\sigma=\frac{Q}{4\pi r_{out}^2}\\\\\sigma=\frac{0.370\times 10^{-6}}{4\pi \times (0.295m)^2}\\\\=3.383\times10^{-7}C/m^2[/tex]  #surface charge outside sphere.

[tex]\sigma_{new}=\sigma_{s}-\sigma\\\\\sigma_{new}=6.37\times10^{-6}C/m^2-3.383\times10^{-7}C/m^2\\\\\sigma_{new}=6.032\times10^{-6}C/m^2[/tex]

Hence, the new charge density on the outside of the sphere is [tex]6.032\times10^{-6}C/m^2[/tex]

b. The strength of the electric field just outside the sphere is calculated as:

From a above, we know the new surface charge to be [tex]6.032\times10^{-6}C/m^2[/tex],

[tex]E=\frac{\sigma _{new}}{\epsilon _o}\\\\=\frac{6.032\times10^{-6}C/m^2}{\epsilon _o}\\\\\epsilon _o=8.85\times10^{-12}C^2/N.m^2\\\\E=\frac{6.032\times10^{-6}C/m^2}{8.85\times10^{-12}C^2/N.m^2}\\\\E=6.816\times10^5N/C[/tex]

Hence, the strength of the electric field just outside the sphere is [tex]6.816\times10^5N/C[/tex]

Explanation:

Answer: into the atmosphere

Explanation:

As this energy is released into the atmosphere, bubbles of warm air is formed which is released and it is replaced by cooler air. This process is responsible for many of the weather patterns in our atmosphere, and is known as....

convection

While convection is a form of heat transfer from one place to the other involving the movement of fluids. So in the case narrated above the fluid which serves as a medium is Air.

Answer:

Rutherford bombarded aluminum foil with beam of light known as alpha particles. The mass of this alpha particle is equivalent to helium atom.

Explanation:

Rutherford bombarded aluminum foil with beam of light known as alpha particles. The mass of this alpha particle is equivalent to helium atom.

When this alpha particles were made to strike the aluminum foil, some passed through the foil, some were reflected and speed others changed.

The ones reflected encountered heavier particle known as the nucleus, preventing them from passing through it. The whole observations indicated that atom is not is uniformly charged sphere as proposed by J.J Thomson.

Rutherford proposed new model known as the Planetary model of atom, which described atom as containing a nucleus which is revolved by electron, just like planets revolve round the sun. And this nucleus contains opposite charge to electron which is proton, to balance the motion.

Answer:

37057.9V

Explanation:

Electric potential is defined as the work done in moving a unit positive charge from infinity to a point.

Electric potential (E) = Potential Difference (V)/distance between plates(d)

Given; electric field of 5.71 x 10^5 N/C; distance between plates =6.49cm = 0.0649m

Since E = V/d

V = Ed

V = 5.71×10^5×0.0649

V = 37057.9Volts

The potential difference (in V) between the plates is 37057.9V

Answer:

V = 3.71×10⁴ V

Explanation:

Potential difference: This can be defined as the work done in moving a positive charge from infinity to any point in an electric field.

The S.I unit of potential difference is Volt (V).

The expression for potential difference is

V = E×d.............................. Equation 1

Where V = potential difference between the plates, E = Electric field , d = distance of separation between the plates

Given: E = 5.71×10⁵ N/C, d = 6.49 cm = 0.0649 m.

Substitute into equation 1

V = 5.71×10⁵×0.0649

V = 3.71×10⁴ V

Answer: A = square root (2/L)

Explanation: find the attached file for explanation

Final answer:

The wave function of a particle in a one-dimensional box needs to be normalized. The normalization condition requires that the value of A for the wave function Ψ(x) = Asin(πx/L) is found to be A = √(2/L) after solving the normalization integral.

Explanation:

The wave function Ψ(x) of a particle in a one-dimensional box of width L must be normalized so that the total probability of finding the particle within the box is 1. This normalization condition implies that the integral of the square of the absolute value of the wave function over the interval from 0 to L should equal 1.

The normalization integral for the given wave function Ψ(x) = Asin(πx/L) is:

∫ |Asin(πx/L)|² dx = A² ∫ sin²(πx/L) dx = 1

When you solve the integral from 0 to L, the result is:

A² * L/2 = 1

Therefore, solving for A, we get:

A = √(2/L)

So the value of A in terms of L is √(2/L).

Answer:

[tex]W=K_f-K_i[/tex]

Explanation:

The work done on a particle by external forces is defined as:

[tex]W=\int\limits^{r_f}_{r_i} {F\cdot dr} \,[/tex]

According to Newton's second law [tex]F=ma[/tex]. Thus:

[tex]W=\int\limits^{r_f}_{r_i}{ma\cdot dr} \,\\[/tex]

Acceleration is defined as the derivative of the speed with respect to time:

[tex]W=m\int\limits^{r_f}_{r_i}{\frac{dv}{dt}\cdot dr} \,\\\\W=m\int\limits^{r_f}_{r_i}{dv \cdot \frac{dr}{dt}} \,[/tex]

Speed is defined as the derivative of the position with respect to time:

[tex]W=m\int\limits^{v_f}_{v_i} v \cdot dv \,[/tex]

Kinetic energy is defined as [tex]K=\frac{mv^2}{2}[/tex]:

[tex]W=m\frac{v_f^2}{2}-m\frac{v_i^2}{2}\\W=K_f-K_i[/tex]

Answer:

a) 44.8J

b) 44.8J

c) 127cm

Explanation:

a) The elastic potential energy of a compressed spring is given by the formula:

[tex]U_e=\frac{1}{2} kx^{2}[/tex]

Where U_e is the elastic potential energy, k is the spring constant and x is the distance the spring is compressed. In this case, we have:

[tex]U_e=\frac{1}{2} (35.0N/cm)(16.0cm)^{2} = 4480Ncm[/tex]

To express the result in Joules, we have to use the fact that 1cm=0.01m. Then:

[tex]U_e=4480Ncm=4480N(0.01m)=44.8J[/tex]

In words, the elastic potential energy of the compressed spring is 44.8J.

b) Using the law of conservation of mechanical energy, we have that:

[tex]E_o=E_f\\\\U_{eo}+U_{go}+K_o=U_{ef}+U_{gf}+K_f[/tex]

Taking t=0 the moment in which the block is released, and t=t_f the point of its maximum height, we have that [tex]U_{g0}=0;K_0=0;U_{ef}=0;K_f=0[/tex] because in t=0 the block has no speed and is in tis lowest point; and in t=t_f the block has stopped and isn't in contact with the spring. So, our equation is reduced to:

[tex]U_{gf}=U_{e0}\\\\U_{gf}=44.8J[/tex]

So, the gravitational potential energy of the block in its highest point is 44.8J.

c) Using the gravitational potential energy formula, we have:

[tex]U_g=mgh\\\\\implies h=\frac{U_g}{mg} \\\\h=\frac{44.8J}{(4.90kg)(9.8m/s^{2}) }=0.9m=90cm[/tex]

Using trigonometry, we can compute the distance between the release point and its highest point:

[tex]d=\frac{h}{sin\theta}=\frac{90cm}{sin45\°}=127cm[/tex]

Answer: FALSE

Explanation: Could you help me with a question?

Answer:

Alfred Wegener

Explanation:

Alfred Wegener was a German scientist who first discovered the idea of continental drift.

The continental drift hypothesis says that the large continents drift from one location to another over the broad ocean water bodies with respect to the fixed poles. This was the first step in understanding the interior of the earth and how the tectonic activities take place on earth.

He contributed many pieces of evidence in order to support this hypothesis, but it was initially not accepted as he was not able to explain the main mechanism for the continental motion.

Answer:

THE ANSWER IS: Water flows more rapidly out of the ground-floor faucet.

Explanation:

Water flows more rapidly out of the ground-floor faucet.

Learn more about municipal water supply here:

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The statement " Close spacing represents greater current densities" is true about the spacing of the streamlines in a wire (option F)

Why is this correct?

In a scenario where a conductor is wider on the left and narrower on the right, the electric field lines and current density are depicted by streamlines. With current flowing from the wider end to the narrower end, the total charge and current remain constant. However, the current density fluctuates, being higher at the narrower end.

Hence, when observing streamlines, closer spacing within the wire signifies a higher current density, specifically in the narrower sections compared to the wider ones.

Complete question:

Which is true about the spacing of the streamlines in a wire?

A. Wide spacing represents faster random-motion velocities.

B. Wide spacing represents greater electric field vectors.

C. Wide spacing represents greater current densities.

D. Close spacing represents faster random-motion velocities.

E. Close spacing represents greater electric field vectors.

F. Close spacing represents greater current densities.

Answer:

Power coming from solar radiations is 6.94 * 10^14 times higher that the power consumption of all humans.

Explanation:

Intensity of sunlight = I = 1380 w/m^2

Area of earth  = A = 4*pi*r^2 = 4*pi*(6.37*10^6)^2 = 5.09*10^14 m^2

he intensity is defined as the total power spread over the area of earth (Area of Sphere with radius equal to distance between earth and sun) and given by the following formula:

                        Intenity of sunlight = Power/Area of earth

                                                 I = P/A

                                                 P = IA

                                                 P = (1380)(5.09*10^14)

                                                 P =  7.036*10^17 W

if we take ratio:

                                                 7.036*10^17/1013 = 6.94 * 10^14

Hence, power coming from solar radiations is 6.94 * 10^14 times higher that the power consumption of all humans.

Answer:

128 N

Explanation:

Using

F = Gm'm/r²....................... Equation 1

Where F = Force, G = Universal constant, m = mass of the human, m' = mass of the moon, r = radius of the moon

Given: m = 75 kg, m' = 7.4×10²² kg, r = 1.7×10⁶ m

Constant: G = 6.67×10⁻¹¹ Nm²/kg²

Substitute into equation 1

F =  (6.67×10⁻¹¹ )(75)(7.4×10²²)/(1.7×10⁶)²

F = (3.7×10¹⁴)/(2.89×10¹²)

F = 1.28×10²

F = 128 N

Explanation:

Below is an attachment containing the solution.

Explanation:

    F = k×[tex]\frac{q1q2}{r2}[/tex]

where, K is coulombs constant, which is equal to -                                  9 x10^9 [tex]Nm^2/C^2.[/tex]

Answer:

[tex]\frac{D_{jet}}{D_{prop}}=2.865[/tex]

Explanation:

Given data

Speed of jet Vjet=1190 km/h

Speed of prop driven Vprop=595 km/h

Height of jet 7.5 km

Height of prop driven transport 3.8 km

Density of Air at height 10 km p7.8=0.53 kg/m³

Density of air at height 3.8 km p3.8=0.74 kg/m³

The drag force is given by:

[tex]D=\frac{1}{2}CpAv^2\\[/tex]

The ratio between the drag force on the jet to the drag force  on prop-driven transport is then given by:

[tex]\frac{D_{jet}}{D_{prop}}=\frac{(1/2)Cp_{7.5}Av_{jet}^2}{1/2)Cp_{3.8}Av_{prop}^2} \\\frac{D_{jet}}{D_{prop}}=\frac{p_{7.5}v_{jet}^2}{p_{3.8}v_{prop}}\\\frac{D_{jet}}{D_{prop}}=\frac{(0.53)(1190)^2}{(0.74)(595)^2}\\ \frac{D_{jet}}{D_{prop}}=2.865[/tex]

A

Explanation:

fred ball hit the pin the time for all just put a hit in his right knee and do not need to be on bed at the carbon footprint in his first year of practice at all but his teammates were also on bed and you were a good friend to be with you cute guys that he is the most likely and do you have the ability for you I don't want him on you and

The time it takes for Wilma's ball to hit the pins is equal to that for Barney's ball to hit. The time it takes for Barney's ball to hit the pins is greater than that for Fred's ball to hit. The time it takes for Betty's ball to hit the pins is greater than that for Fred's ball to hit.

The time it takes for Wilma's ball to hit the pins is equal to that for Barney's ball to hit.

The time it takes for Barney's ball to hit the pins is greater than that for Fred's ball to hit.

The time it takes for Betty's ball to hit the pins is greater than that for Fred's ball to hit.

The time it takes for Betty's ball to hit the pins is equal to that for Barney's ball to hit.

The time it takes for Fred's ball to hit the pins is equal to that for Wilma's ball to hit.

The time it takes for Wilma's ball to hit the pins is greater than that for Betty's ball to hit.

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