Two boxes connected by a light horizontal rope are on a horizontal surface. The coefficient of kinetic fric-tion between each box and the surface is ilk = 0.30. Box B has mass 5.00 kg, and box A has mass m. A force F with magnitude 40.0 N and direction 53.1° above the horizontal is applied to the 5.00-kg box, and both boxes move to the right with a = 1.50 m/s2. A) What is the tension T in the rope that connects the boxes? B) What is m?

Answer:

Explanation:

Given

apparent frequency is 1 % lower than the original frequency

Speed of sound [tex]v=343\ m/s[/tex]

suppose f is the original frequency  

apparent frequency [tex]f'=0.99f[/tex]

using Doppler's formula

[tex]f'=f\cdot \frac{v-v_o}{v+v_s} [/tex]

where  

[tex]f'[/tex]=apparent frequency  

[tex]v[/tex]=velocity of sound in the given media

[tex]v_s[/tex]=velocity of source

[tex]v_o[/tex]=velocity of observer  

[tex]0.99f=f(\dfrac{343-v_o}{v+0})[/tex]

[tex]0.99\times 343=343-v_o[/tex]

[tex]v_o=343\times 0.1[/tex]

[tex]v_o=34.3\ m/s[/tex]

Answer:

the probability that at least 125 of the 500 components will have failure times larger than 20 is 0.7939

Explanation:

see the attached file

Geocentric Theory

In astronomy, the geocentric model is a superseded description of the Universe with Earth at the center. Under the geocentric model, the Sun, Moon, stars, and planets all orbited Earth

Heliocentric Theory

Heliocentrism is the astronomical model in which the Earth and planets revolve around the Sun at the center of the Solar System. Historically, heliocentrism was opposed to geocentrism, which placed the Earth at the center

G-Theory is the earth is the center of the universe.

H-Theory is the sun is the center of the universe.

Explanation:

Work = Force x Displacement

Force = Weight of student

Weight = Mass x Acceleration due to gravity

Mass, m = 80 kg

Acceleration due to gravity, g = 9.81 m/s²

Weight = 80 x 9.81 = 784.8 N

Displacement = 8 m

Work = 784.8 x 8 = 6278.4 J

The amount of work done by the student to elevate his body to this height is 6278.4 J

Power is the ratio of work to time taken

         [tex]P=\frac{W}{t}\\\\P=\frac{6278.4}{12}\\\\P=523.2W[/tex]

Power is 523.2 Watts

The power consumed by the student is 523.2 watts.

[tex]Work = Force \times Displacement[/tex]

Force = Weight of student  

[tex]Weight = Mass \times Acceleration\ due\ to\ gravity[/tex]

Mass, m = 80 kg

Acceleration due to gravity, g = 9.81 m/s²

[tex]Weight = 80 \times 9.81 = 784.8 N[/tex]  

Displacement = 8 m

[tex]Work = 784.8 \times 8 = 6278.4 J[/tex]

So,

Power consumed should be

[tex]= 6278.4 \div 12[/tex]

= 523.2

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Answer:

The central atom in a lewis structure tends to be the least electronegative atom.

Explanation:

Lewis structures show each atom and its position in the structure of the molecule using its chemical symbol.

In Lewis structure, the central atom tends to be the least electronegative atom. Usually the central atom will be the one that has the most unpaired valence electrons or  least electronegative.

Therefore, the best phrase that completes the sentence is "least electronegative".

Answer:

The central atom is also usually the least electronegative.

Explanation:

The central atom is usually the least electronegative element in the molecule or ion; hydrogen and the halogens are usually terminal.

The least electronegative elements in the center of lewis structures because an atom in the central position shares more of its electrons than does a terminal atom. Atoms with higher electronegative are generally more reluctant to share its electrons.

a)

[tex]F_{E_x}=1.19\cdot 10^{-3}N[/tex] (+x axis)

[tex]F_{B_x}=0[/tex]

[tex]F_{B_y}=0[/tex]

b)

[tex]F_{E_x}=1.19\cdot 10^{-3} N[/tex] (+x axis)

[tex]F_{B_x}=0[/tex]

[tex]F_{B_y}=3.21\cdot 10^{-3}N[/tex] (+z axis)

c)

[tex]F_{E_x}=1.19\cdot 10^{-3} N[/tex] (+x axis)

[tex]F_{B_x}=3.21\cdot 10^{-3} N[/tex] (+y axis)

[tex]F_{B_y}=3.21\cdot 10^{-3}N[/tex] (-x axis)

Explanation:

a)

The electric force exerted on a charged particle is given by

[tex]F=qE[/tex]

where

q is the charge

E is the electric field

For a positive charge, the direction of the force is the same as the electric field.

In this problem:

[tex]q=+4.9\mu C=+4.9\cdot 10^{-6}C[/tex] is the charge

[tex]E_x=+242 N/C[/tex] is the electric field, along the x-direction

So the electric force (along the x-direction) is:

[tex]F_{E_x}=(4.9\cdot 10^{-6})(242)=1.19\cdot 10^{-3} N[/tex]

towards positive x-direction.

The magnetic force instead is given by

[tex]F=qvB sin \theta[/tex]

where

q is the charge

v is the velocity of the charge

B is the magnetic field

[tex]\theta[/tex] is the angle between the directions of v and B

Here the charge is stationary: this means [tex]v=0[/tex], therefore the magnetic force due to each component of the magnetic field is zero.

b)

In this case, the particle is moving along the +x axis.

The magnitude of the electric force does not depend on the speed: therefore, the electric force on the particle here is the same as in part a,

[tex]F_{E_x}=1.19\cdot 10^{-3} N[/tex] (towards positive x-direction)

Concerning the magnetic force, we have to analyze the two different fields:

- [tex]B_x[/tex]: this field is parallel to the velocity of the particle, which is moving along the +x axis. Therefore, [tex]\theta=0^{\circ}[/tex], so the force due to this field is zero.

[tex]- B_y[/tex]: this field is perpendicular to the velocity of the particle, which is moving along the +x axis. Therefore, [tex]\theta=90^{\circ}[/tex]. Therefore, [tex]\theta=90^{\circ}[/tex], so the force due to this field is:

[tex]F_{B_y}=qvB_y[/tex]

where:

[tex]q=+4.9\cdot 10^{-6}C[/tex] is the charge

[tex]v=345 m/s[/tex] is the velocity

[tex]B_y = +1.9 T[/tex] is the magnetic field

Substituting,

[tex]F_{B_y}=(4.9\cdot 10^{-6})(345)(1.9)=3.21\cdot 10^{-3} N[/tex]

And the direction of this force can be found using the right-hand rule:

- Index finger: direction of the velocity (+x axis)

- Middle finger: direction of the magnetic field (+y axis)

- Thumb: direction of the force (+z axis)

c)

As in part b), the electric force has not change, since it does not depend on the veocity of the particle:

[tex]F_{E_x}=1.19\cdot 10^{-3}N[/tex] (+x axis)

For the field [tex]B_x[/tex], the velocity (+z axis) is now perpendicular to the magnetic field (+x axis), so the force is

[tex]F_{B_x}=qvB_x[/tex]

And by substituting,

[tex]F_{B_x}=(4.9\cdot 10^{-6})(345)(1.9)=3.21\cdot 10^{-3} N[/tex]

And by using the right-hand rule:

- Index finger: velocity (+z axis)

- Middle finger: magnetic field (+x axis)

- Thumb: force (+y axis)

For the field [tex]B_y[/tex], the velocity (+z axis) is also perpendicular to the magnetic field (+y axis), so the force is

[tex]F_{B_y}=qvB_y[/tex]

And by substituting,

[tex]F_{B_y}=(4.9\cdot 10^{-6})(345)(1.9)=3.21\cdot 10^{-3} N[/tex]

And by using the right-hand rule:

- Index finger: velocity (+z axis)

- Middle finger: magnetic field (+y axis)

- Thumb: force (-y axis)

The force on a charged particle in an electric field is given by Felectric = qE, and in a magnetic field while moving, it is Fmagnetic = qv × B, with the right-hand rule determining the direction of Fmagnetic. A stationary particle only experiences Felectric. When moving, it may experience both forces, depending on the motion's relationship to the field's direction.

The force on a charged particle in an electric field is given by Felectric = qE, where q is the charge and E is the electric field. If the particle is stationary, only the electric force acts on it. When moving in a magnetic field, a magnetic force Fmagnetic = qv × B also acts on it, where v is the velocity and B is the magnetic field. The direction of this force is perpendicular to both v and B as per the right-hand rule.

For a particle with charge q = +4.9 μC:

Answer:

False. Field is non-zero

Explanation:

If you were moving along with the electrons, they would appear stationary to you. You would measure a current of zero. However, the fixed positive charges in the wire seem to move backwards relative to you, creating the equivalent current as if you weren't moving. You would measure the same field, but the field would be caused by the 'backward' motion of positive particles.

Moving at the same speed as the drift speed of electrons does not result in a zero magnetic field. The movement of electrons in a wire creates a magnetic field, and this field would still be present even if you were moving at the same speed as the electrons.

When moving along a wire at the same speed as the drift speed of the electrons, you will not measure a magnetic field of zero. The drift speed of electrons refers to the average velocity at which the electrons move in a conductor when an electric field is applied. This speed is generally very slow, but it does not mean that there is no magnetic field.

The movement of electrons in a wire creates a magnetic field around the wire, even if the drift speed is small. This is because the electric current generated by the movement of electrons is what produces the magnetic field.

So, even if you were to move at the same speed as the drift speed of the electrons in the wire, you would still measure a non-zero magnetic field because the movement of electrons in the wire is what generates the magnetic field.

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Answer:

The electric field inside the wire will remain the same or constant, while the drift velocity will by a factor of four.

Explanation:

Electron mobility, μ = [tex]\frac{v_d}{E}[/tex]

where

[tex]v_d[/tex] = Drift velocity

E = Electric field

Given that the electric field strength = 1.48 V/m,

Therefore since the electric potential depends on the length of the wire and the attached potential difference, then when the electron mobility is increased 4 times the Electric field E will be the same but the drift velocity will increase four times. That is

4·μ = [tex]\frac{4*v_d}{E}[/tex]

Answer:

Explanation:

Usually, the electron drift velocity in a material is directly proportional to the electric field, which means that the electron mobility is a constant (independent of electric field).

μ × E = Vd

Where,

D = electric field

Vd = drift velocity

μ = electron mobility

μ2 = μ × 4

μ/Vd1 = 4 × μ/Vd2

Vd2 = 4 × Vd1

The electric field is the same but drift velocity increases.

Ion

Explanation:

At the synaptic terminal, voltage-gated ion channels open, thereby stimulating the synaptic vesicles to release the neurotransmitters by exocytosis.

These ion channels are the signaling molecules in neurons. They are the transmembrane proteins that form ion channels. The membrane potential changes the conformation of the channel proteins that regulates their opening and closing. These channels play an important role in neurotransmitter release in presynaptic nerve endings.

For example - Ca²⁺ gated ion channel.

Answer:

(a) I_A=1/12ML²

(b) I_B=1/3ML²

Explanation:

We know that the moment of inertia of a rod of mass M and lenght L about its center is 1/12ML².

(a) If the rod is bent exactly at its center, the distance from every point of the rod to the axis doesn't change. Since the moment of inertia depends on the distance of every mass to this axis, the moment of inertia remains the same. In other words, I_A=1/12ML².

(b) The two ends and the point where the two segments meet form an isorrectangle triangle. So the distance between the ends d can be calculated using the Pythagorean Theorem:

[tex]d=\sqrt{(\frac{1}{2}L) ^{2}+(\frac{1}{2}L) ^{2} } =\sqrt{\frac{1}{2}L^{2} } =\frac{1}{\sqrt{2} } L=\frac{\sqrt{2} }{2} L[/tex]

Next, the point where the two segments meet, the midpoint of the line connecting the two ends of the rod, and an end of the rod form another rectangle triangle, so we can calculate the distance between the two axis x using Pythagorean Theorem again:

[tex]x=\sqrt{(\frac{1}{2}L)^{2}-(\frac{\sqrt{2}}{4}L) ^{2} } =\sqrt{\frac{1}{8} L^{2} } =\frac{1}{2\sqrt{2}} L=\frac{\sqrt{2}}{4} L[/tex]

Finally, using the Parallel Axis Theorem, we calculate I_B:

[tex]I_B=I_A+Mx^{2} \\\\I_B=\frac{1}{12} ML^{2} +\frac{1}{4} ML^{2} =\frac{1}{3} ML^{2}[/tex]

A) Moment of inertia about an axis passing through the point where the two segments meet :   [tex]I_{A} = \frac{1}{12} ML^{2}[/tex]

B) Moment of inertia  passing through the point where the midpoint of the line connects  to its two ends :   [tex]Ix_{} = \frac{1}{3} ML^{2}[/tex]

A) The moment of inertia about an axis passing through the point where the two segments meet is [tex]I_{A} = \frac{1}{12} ML^{2}[/tex]  given that the rod is bent at the center and distance from all the points to the axis remains the same, the moment of inertia about the center will remain the same.

B) Determine the moment of inertia about an axis passing through the point midpoint of the line which connects the two ends

First step: determine the distance between the ends ( d )

  After applying  Pythagoras theorem

d = [tex]\frac{\sqrt{2} }{2} L[/tex]  

Next step :  determine distance between the two axis ( x )

After applying Pythagoras theorem

x = [tex]\frac{\sqrt{2} }{4} L[/tex]    

Final step : Calculate the value of  Iₓ

applying  Parallel Axis Theorem

Iₓ = Iₐ + Mx²

   = [tex]\frac{1}{12} ML^{2}[/tex]  + [tex]\frac{1}{4} ML^{2}[/tex]

∴  [tex]Ix_{} = \frac{1}{3} ML^{2}[/tex]

Hence we can conclude that Moment of inertia about an axis passing through the point where the two segments meet :   [tex]I_{A} = \frac{1}{12} ML^{2}[/tex],  Moment of inertia  passing through the point where the midpoint of the line connects its two ends :   [tex]Ix_{} = \frac{1}{3} ML^{2}[/tex]

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The first part of the question is missing and it is:

A boat of mass 250 kg is coasting, with its engine in neutral, through the water at speed 3.00 m/s when it starts to rain. The rain is falling vertically, and it accumulates in the boat at the rate of 10.0 kg/hr.

Answer:

a(o) = 18.03 x 10^(-3) m/s^(-2)

Explanation:

First of all, if we make a momentum balance in the direction of motion (the x-direction), we'll discover that the change of the momentum will be equal to the forces acting on the boat.

Hence, there is only the drag force, which acts against the direction of motion as:

d(m·v)/dt = - k·v² (since f_d = - k·v²)

where k = 0.5 Ns²/m²

Now let's simplify the time derivative on the Left, by applying product rule of differentiation and we obtain:

v·dm/dt + m·dv/dt = - k·v²

where dm/dt = 10kg/hr (this is the change of the mass of the boat)

Furthermore, acceleration is the time derivative of time velocity, Thus;

a = dv/dt = - (k·v² + v·dm/dt) / m

Now, for the moment,when the rain starts; since we know all the values on the right hand side, let's solve for the acceleration ;

a(o) = - (ko·vo² + vo·dm/dt) / mo

= -[ (0.5(3)²) + (3x10/3600)/250

(dt = 3600secs because 1hr = 3600 secs)

So a(o) = 18.03 x 10^(-3) m/s^(-2)

Answer:

The frequency of the ripples is 2Hz, and their period is 0.5 seconds.

Explanation:

Since the ripples on the pond are making the leaf oscillate up and down at a rate of two times per second, we can calculate the period T and the frequency f of the ripples on the pon[tex]T=\frac{1}{0.5Hz} =[/tex]d.

The frequency, by definition, is the number of waves per unit of time. In this case, we have two waves per second, so the frequency is 2s⁻¹, or 2Hz.

The period is the inverse of the frequency, so

[tex]T=\frac{1}{2Hz} =0.5s[/tex]

Then, the period is equal to 0.5 seconds.

Answer:

Velocity of sound is 512 m/s

Explanation:

for first resonance condition we know that

[tex]\frac{\lambda}{4} = L[/tex]

[tex]\frac{\lambda}{4} = 0.40[/tex]

[tex]\lambda = 1.6 m[/tex]

now the frequency is given as

[tex]f = 320 Hz[/tex]

now to find the velocity of sound we have

[tex]v = \lambda f[/tex]

[tex]v = 1.6 (320)[/tex]

[tex]v = 512 m/s[/tex]

Explanation:

Below is an attachment containing the solution.

Answer:

Conservation of energy

Explanation:

Gustav Kirchhoff, a German physicist proposed a rule called Kirchhoff's loop rule also known as Kirchhoff's second rule which states that for any closed loop, the potential difference including the voltage supply is zero. That means energy supplied by a voltage source must absorbed for balance by other components in the loop

This fulfils the law of conservation of energy

Answer:

The total distance, side to side, that the top of the building moves during such an oscillation = 31 cm

Explanation:

Let the total side to side motion be 2A. Where A is maximum acceleration.

Now, we know know that equation for maximum acceleration is;

A = α(max) / [(2πf)^(2)]

So 2A = 2[α(max) / [(2πf)^(2)] ]

α(max) = (0.025 x 9.81) while frequency(f) from the question is 0.2Hz.

Therefore 2A = 2 [(0.025 x 9.81) / [((2π(0.2)) ^(2)] ] = 2( 0.245 / 1.58) = 0.31m or 31cm

Answer:

towards right,

lean inward to avoid tripping

Explanation:

The frictional force becomes centripetal force while turinig and presses on the wheel to maintain the grip of wheel on the road. Otherwise the bicycle will slip. So the press on the wheel is inward or towards the right when turning a corner to right

The rider has to lean inward ( to the right) to avoid tripping. Leaning inward, changes the direction of normal force on the wheels from vertical to slant. This helps to counter the torque produced by the frictional force

Answer:

The answers to the questions are;

1. The amount of energy required to move a mass m from the center of the Earth to the surface is 0.5·m·g·R

2. The amount of energy required to move the mass from the surface of the planet to a very large distance away m·g·R.

3. The speed of an object of mass m, dropped into this hole, when it reaches the center of the planet is 9682.41783 m/s.

Explanation:

We note that the Work done W by the force F on the mass to move a small distance is given by

F×dr

The sum of such work to move the body to a required location is

W =[tex]\int\limits {F} \, dr[/tex]

F = mg' = m[tex]\frac{gr}{R}[/tex]

We integrate from 0 to R (the center to the Earth surface)

Therefore W = [tex]\int\limits^R_0 {m\frac{gr}{R}} \, dr[/tex]

Which gives W = [tex]\frac{mg}{R} [\frac{r^2}{2} ]^R_0 = \frac{1}{2}mgR[/tex]

2. To find the work done we have to integrate from the surface to infinity as follows W = [tex]\int\limits^{inf}_R {m\frac{gr}{R}} \, dr[/tex] = [tex]\frac{mg}{R} [\frac{r^2}{2} ]^{inf}_R = mgR[/tex]

The energy required to move the object to a large distance is equal to twice the energy reqired to move the object to the surface.

3 We note that the acceleration due to gravity at the surface is g and reduces to zero at the center of the Earth

v² = u² + 2·g·s

Radius of the Earth = 6371 km

From surface to half radius we have

v₁² = 2×9.81×6371/2×1000 = 62499460.04

v₁ = 7905.66 m/s

From the half the radius of the earth to the Earth center =

v₂² = 7905.66² + 2×9.81/2×6371/2×1000 = 93749215.04

v₂ = 9682.41783 m/s

The speed of an object of mass m, dropped into this hole, when it reaches the center of the planet. is 9682.41783 m/s

1. Energy to Move from Center to Surface: Using the uniform-density approximation, the energy required to move a mass m from Earth's center to the surface is [tex]\(-\frac{GMm}{R}\)[/tex]. 2. Energy to Move from Surface to Infinity: Moving the mass from the surface to infinity requires zero energy. 3. Speed at Earth's Center: The speed of an object dropped through a hole from the surface to the Earth's center is [tex]\(\sqrt{\frac{2GM}{R}}\)[/tex].

1. Energy to Move Mass from Center to Surface:

The gravitational potential energy U is given by:

[tex]\[ U = -\frac{GMm}{r} \][/tex]

where:

- G is the gravitational constant,

- M is the mass of the planet,

- m is the mass being moved,

- r is the distance from the center.

For this scenario, r is the radius of the planet R.

[tex]\[ U_{\text{center to surface}} = -\frac{GMm}{R} \][/tex]

2. Energy to Move Mass from Surface to Infinity:

When moving the mass to a very large distance away [tex](\( \infty \))[/tex], the potential energy becomes zero.

[tex]\[ U_{\text{surface to infinity}} = 0 \][/tex]

3. Speed of Object Dropped Through Earth's Center:[tex]\[ U_{\text{surface}} = K_{\text{center}} \]\[ -\frac{GMm}{R} = \frac{1}{2}mv^2 \][/tex]converted into kinetic energy at the center.

Solve for v:

[tex]\[ v = \sqrt{\frac{2GM}{R}} \][/tex]

Given:

- M is the mass of the Earth,

- m is the mass being moved.

These expressions provide the required energy calculations and the speed of the object dropped through the Earth's center.

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Answer:

5.9x10³⁰ kg = 2.97 solar mass.

Explanation:

The mass of Sirius can be calculated using Kepler's Third Law:

[tex]P^{2} = \frac{4 \pi^{2}}{G (M_{1} + M_{2})} \cdot a^{3}[/tex]

where P: is the period, G: is the gravitational constant = 6.67x10⁻¹¹ m³kg⁻¹s⁻², a: is the size of the orbit, M₁: is the mass of the Earth-like planet =  5.97x10²⁴ kg, and M₂: is the mass of Sirius.  

Firts, we need to convert the units:

P = 7 months = 1.84x10⁷ s

a = 1 AU = 1.5x10¹¹ m

Now, from equation (1), we can find the mass of Sirius:

[tex] M_{2} = \frac{4 \pi^{2} a^{3}}{P^{2} G} - M_{1} [/tex]

[tex] M_{2} = \frac{4 \pi^{2} (1.5 \cdot 10^{11} m)^{3}}{(1.84\cdot 10^{7} s)^{2} \cdot 6.67 \cdot 10{-11} m^{3} kg ^{-1} s^{-2}} - 5.97 \cdot 10^{24} kg = 5.9 \cdot 10^{30} kg = 2.97 solar mass [/tex]  

Therefore, the mass of Sirius is 5.9x10³⁰ kg = 2.97 solar mass.

I hope it helps you!

Answer: 2.94kg

Explanation: Using Kepler's law of planetary motion which states that, the square of the orbital period of a planet is proportional to the cube of the semi major axis of it's orbit.

Sirius is the name given to the brightest star in the sky

It could be literally interpreted as;

1/mass of planet = (orbital period in years^2 ÷ semi major axis in AU^3)

Orbital period = 7months is equivalent to 7/12 = 0.5833 years

Semi major axis = 1AU

Therefore, mass of Sirius is given by:

1/Mass = (orbital period^2 ÷ semi major axis^3)

1/Mass = (0.5833^2 ÷ 1^3)

1/Mass = 0.3402 ÷ 1

1/Mass = 0.3402

Therefore,

Mass = 1/0.3402 = 2.94kg

Answer:

Check attachment for the body diagram

Explanation:

Centripetal force is a net force that acts on an object to keep it moving along a circular path.

It is given as

F=ma

Where a =v²/r

F=mv²/r

The centripetal force is directed towards the center i.e inward.

Then, analysing the free body diagram, we notice that the horizontal component of the Normal force (FNsinθ) is directed towards the centre. Then, we can say the last option 6 is correct.

Answer:

4.0 cm

Explanation:

For the compression of the spring, the kinetic energy of the mass equals the elastic potential energy of the spring.

So, 1/2mv² = 1/2kx² ⇒ x = (√m/k)v

Since m and k are constant since its the same spring x ∝ v

If our speed is now v₁ = 2v, our compression is x₁

x₁ = (√m/k)v₁ = (√m/k)2v = 2(√m/k)v = 2x

x₁ = 2x

Since x = 2.0 cm, our compression for speed = 2v is

x₁ = 2(2.0) = 4.0 cm

If the same block collides with the spring at a speed of 2v, the compression will be 4.0cm.

Given the data in the question;

Using conservation of energy:

Kinetic energy of the mass = Elastic potential energy of the spring

We have:

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

"v" is directly proportional to "x"

Hence,

[tex]\frac{x_1}{x_2} = \frac{v_1}{v_2}[/tex]

We substitute in our given values

[tex]\frac{2.0cm}{x_2} = \frac{v}{2v}\\\\x_2 = \frac{v(2.0cm*2)}{v} \\\\x_2 = (2.0cm*2)\\\\x_2 = 4.0cm[/tex]

Therefore, if the same block collides with the spring at a speed of 2v, the compression will be 4.0cm.

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Answer:

4.96cm

Explanation:

distance  = 5.20km = 5.2 × 10³ m

Wavelength , λ = 550nm = 5.50 × 10⁻⁷m

distance of moon, L = 384000 km = 3.84 × 10⁸m

formula for resolving power of two objects

d = (1.22 × λ ×L) / D

D = (1.22 × λ ×L) / d

    = (1.22 × 5.50 × 10⁻⁷ ×3.84 × 10⁸) / 5.2 × 10³

D = 4.96cm

Explanation:

Below is an attachment containing the solution.

'In transverse waves, the particles of the medium move perpendicular to the direction of the flow of energy' is true for transverse waves only.

'In longitudinal waves, the particles of the medium move parallel to the direction of the flow of energy' is true for longitudinal waves only.

'Many wave motions in nature are a combination of longitudinal and transverse motion' is true for both longitudinal and transverse waves.

Explanation:

Longitudinal waves are those where the direction of propagation of particles are parallel to the medium' particles. While transverse waves propagate perpendicular to the medium' particles.

As wave motions are assumed to be of standing waves which comprises of particles moving parallel as well as perpendicular to the medium, most of the wave motions are composed of longitudinal and transverse motion.

So the option stating the medium' particle moves perpendicular to the direction of the energy flow is true for transverse waves. Similarly, the option stating the medium' particle moves parallel to the direction of flow of energy is true for longitudinal waves only.

And the option stating that wave motions comprises of combination of longitudinal and transverse motion is true for both of them.

Final answer:

The statements A, B, and C are true for both transverse and longitudinal waves.

Explanation:

A. In longitudinal waves, the particles of the medium move parallel to the direction of the flow of energy.
B. Many wave motions in nature are a combination of longitudinal and transverse motion.
C. In transverse waves, the particles of the medium move perpendicular to the direction of the flow of energy.

All of the above statements are true for both transverse and longitudinal waves.

Explanation An is valid for longitudinal waves as it were. In longitudinal waves, for example, sound waves, the particles of the medium vibrate lined up with the bearing of the energy move. This can be pictured as pressure and rarefaction of the particles in the medium.

Only transverse waves are covered by statement C. In cross over waves, for example, light waves, the particles of the medium vibrate opposite to the course of the energy move. The particles can be seen moving side to side or up and down, respectively.

B holds true for both kinds of waves. Many wave movements in nature, for example, water waves and seismic waves, include a mix of longitudinal and cross over movement. Water particles, for instance, move in circular orbits as waves pass by, which is a combination of transverse and longitudinal motion.

The width of the slit can be found using the formula θ = λ / a and the ratio of the intensity at θ=45.0∘ to θ=0 can be calculated using the formula I(θ) = (I0) (sinc² (πa sin(θ) / λ)).

The width of the slit can be calculated using the formula for the angular position of the first minima in single-slit diffraction:

θ = λ / a

where θ is the angular position of the minima, λ is the wavelength of the light, and a is the width of the slit. Rearranging the formula, we can solve for a:

a = λ / θ

Substituting the given values, the width of the slit is:

a = (580 nm) / (90.0∘)

In part (b), the ratio of the intensity at θ=45.0∘ to the intensity at θ=0 can be calculated using the formula for intensity in single-slit diffraction:

I(θ) = (I0) (sinc² (πa sin(θ) / λ))

where I(θ) am the intensity at a certain angle, I0 is the intensity at θ=0, a is the width of the slit, λ is the wavelength, and sinc is the sin function. By substituting the values and calculating the ratio, you can find the answer.

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The width of the slit through which monochromatic light of wavelength 580 nm causes the first diffraction minima at  extpm90.0 extdegree is 580 nm. The intensity ratio at heta=45.0 extdegree to  heta=0 can be determined by using the formula I(heta) = I_0 * [sinc(eta)]^2.

The question pertains to Fraunhofer diffraction through a single slit and involves finding the width of the slit and the ratio of light intensities at different angles.

Part (a): Width of the slit

According to the single slit diffraction formula, the first minima occur at angles heta where d sin(heta) = ext{m} extlambda, where d is the width of the slit, heta is the angle of the diffraction minima, m is the order of the minima, and  extlambda is the wavelength of the light.

Given the first minima at heta = ext{extpm}90.0 ext{extdegree}, we have:

sin(90ext{extdegree}) =  ext{1}
ext{m} = ext{1} (since it's the first minima)
d imes 1 = ext{1} imes ext{580 nm}
d =  ext{580 nm}

Therefore, the width of the slit, d, is 580 nm.

Part (b): Intensity ratio

The intensity at angle heta in a single-slit diffraction pattern is given by:

I(heta) = I_0 imes ext{[sinc(heta)]}^2
where
heta =  rac{ext{extpi} d sin(heta)}{extlambda}
and
ext{sinc}(heta) =  rac{ext{sin}(heta)}{heta}

For heta = 45ext extdegree},
ext{beta} can be calculated and the resulting intensity ratio I(45ext{extdegree})/I(0) can be found accordingly.

When the radius of the circle is reduced while the period remains constant, the centripetal acceleration of the ball increases.

When the radius of the circle is reduced to 0.75R while the period remains T, the centripetal acceleration of the ball increases.

Centripetal acceleration is given by the formula ac = v^2 / r, where v is the tangential velocity and r is the radius of the circle. Since the period T remains constant and the radius is reduced, the velocity of the ball must increase in order to cover the smaller circumference in the same amount of time, resulting in an increase in centripetal acceleration.

Explanation:

The area of a circle can be represented by  A = π r²            I

Differentiating both sides w.r.t time

[tex]\frac{dA}{dt}[/tex] = 2π r [tex]\frac{dr}{dt}[/tex]                                    II

Dividing II by I , we have

[tex]\frac{dA}{A}[/tex] = 2 x  [tex]\frac{dr}{r}[/tex]

substituting the values

[tex]\frac{dr}{r}[/tex] = [tex]\frac{5.5}{12}[/tex] = 0.46 mi per unit radius

or dr = 1.4 x 0.46 = 0.64 mi/hr

here 1.4 mi is the radius , when area of circle is 6 mi²

Answer:

(a). The lowest frequency is 64.7 Hz.

(b). The next two resonant frequencies for the bugle are 129.4 Hz and 194.2 hz.

Explanation:

Given that,

Length = 2.65 m

Speed of sound = 343 m

We need to calculate the wavelength

Using formula of wavelength

[tex]\lambda=2l[/tex]

Put the value into the formula

[tex]\lambda=2\times2.65[/tex]

[tex]\lambda=5.3\ m[/tex]

(a). We need to calculate the lowest frequency

Using formula of frequency

[tex]f_{1}=\dfrac{v}{\lambda_{1}}[/tex]

Put the value into the formula

[tex]f_{1}=\dfrac{343}{5.3}[/tex]

[tex]f_{1}=64.7\ Hz[/tex]

(b). We need to calculate the next two resonant frequencies for the bugle

Using formula of resonant frequency

[tex]f_{2}=\dfrac{v}{\lambda_{2}}[/tex]

[tex]f_{2}=\dfrac{v}{l}[/tex]

Put the value into the formula

[tex]f_{2}=\dfrac{343}{2.65}[/tex]

[tex]f_{2}=129.4\ Hz[/tex]

For third frequency,

[tex]f_{3}=\dfrac{v}{\lambda_{3}}[/tex]

[tex]f_{3}=\dfrac{3v}{2l}[/tex]

Put the value into the formula

[tex]f_{3}=\dfrac{3\times343}{2\times2.65}[/tex]

[tex]f_{3}=194.2\ Hz[/tex]

Hence, (a). The lowest frequency is 64.7 Hz.

(b). The next two resonant frequencies for the bugle are 129.4 Hz and 194.2 hz.

The next two resonant frequencies for the bugle are 64.72Hz and 323.6Hz

The lowest frequency is expressed as:

f = v/2l

[tex]f_0=\frac{343}{2(2.65)} f_0=\frac{343}{5.3}\\f_0= 64.72 Hz[/tex]

Since the bugle is an open pipe the next two resonant frequencies are;

[tex]f_2 =3f_0=3(64.72) = 194.15Hz\\\\f_3 = 5f_0 = 5(64.72) =323.6Hz \\[/tex]

Hence the next two resonant frequencies for the bugle are 64.72Hz and 323.6Hz

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Answer:

a)T = 8.63  × 10 ⁴ N, b)T = -3.239 × 10 ⁴ N

Explanation:

Given:

W = 27.8 KN = 27.8 × 10 ³ N,

For upward motion: Fnet is upward, Tension T is upward and weight W is downward so

a) a=21.22 m/s² ( not written clearly the unit. if it is acceleration?) then

Fnet = T - W

⇒ T = F + W = ma + W

T = (W/g)a + W                           (W=mg ⇒m=W/g)

T = (27.8 × 10 ³ N / 9.8 ) 21.22 m/s² + 27.8 × 10 ³ N

T = 86,263.265 N

T = 8.63  × 10 ⁴ N

b) For Declaration in upward direction a = -21.22 m/s²

Fnet = T - W

⇒ T = F + W = ma + W

T = (W/g)a + W                        

T = (27.8 × 10 ³ N / 9.8 ) (-21.22 m/s²) + 27.8 × 10 ³ N

T = -32395.5 N

T = -3.239 × 10 ⁴ N

as Tension can not be negative I hope the value of acceleration and deceleration is correct.

Answer:

[tex]\omega = 300 rad/s[/tex]

Explanation:

Given,

rotational acceleration = 30 rad/s²

initial angular speed = 0 m/s

time, t = 10 s

Final angular speed = ?

Using equation of rotation motion

[tex]\omega = \omega_o + \alpha t[/tex]

[tex]\omega =0+30\times 10[/tex]

[tex]\omega = 300 rad/s[/tex]

Rotational velocity after 10 s = 300 rad/s.