A straight wire carries a current of 10 A at an angle of 30° with respect to the direction of a uniform 0.30-T magnetic field. Find the magnitude of the magnetic force on a 0.50-m length of the wire.

The time required to raise the temperature of the ice from -17.2 °C to 0 °C is approximately 6.22 minutes.

To calculate the time required for the temperature of the ice to rise from -17.2 °C to 0 °C, we can use the equation Q=mcΔT, where Q is the heat supplied, m is the mass of the ice, c is the specific heat of ice, and ΔT is the change in temperature.

The heat supplied (Q) during this process is used to raise the temperature of the ice without causing a phase change. Given that Q=820J/min and c=2100J/kg⋅K, we rearrange the equation to solve for time (t): t=Q/mcΔT

Substituting the given values (m=0.570kg, ΔT=0°C−(−17.2°C)=17.2°C), we find t≈820J/min/ (0.570kg)(2100J/kg⋅K)(17.2°C) ≈6.22min.

This calculation assumes that all the heat supplied is used to raise the temperature of the ice without considering the heat required for a phase change. To account for the heat of fusion during the phase change from ice to water at 0 °C, an additional calculation would be needed. However, the question specifies raising the temperature to 0 °C, so the heat of fusion is not considered in this particular scenario.

Understanding the principles of heat transfer, specific heat, and phase changes is essential in solving thermodynamics problems. These principles are widely applicable in various scientific and engineering fields, providing insights into the behavior of materials under different conditions.

Answer:

Nuclear engineers require a degree in physics in order to have the job.

Explanation:

Answer:

[tex]7.5x10^{14}Hz[/tex], [tex]9.37x10^{14}Hz[/tex].

Explanation:

The ultraviolet light belongs to the electromagnetic spectrum.

The electromagnetic spectrum is the distribution of radiation due to the different frequencies at which it radiates and its different intensities. That radiation is formed by electromagnetic waves, which are transverse waves formed by an electric field and a magnetic field perpendicular to it.

Any radiation from the electromagnetic spectrum has a speed of [tex]3x10^{8}m/s[/tex] in vacuum.

Therefore, in order to know the frequency the following equation can be used:

[tex]c = \nu \cdot \lambda[/tex]  (1)

[tex]\nu = \frac{c}{\lambda}[/tex]  (2)

Where [tex]\nu[/tex] is frequency, c is the speed of light and [tex]\lambda[/tex] is the wavelength.

Notice that it is necessary to express the wavelength in units of meters before it can be used in equation 2.  

[tex]\lambda = 320nm . \frac{1m}{1x10^{9}nm}[/tex] ⇒ [tex]3.2x10^{-7}m[/tex]

[tex]\nu = \frac{3x10^{8}m/s}{3.2x10^{-7}m}[/tex]    

[tex]\nu = 9.37x10^{14}s^{-1}[/tex]    

But [tex]1Hz = s^{-1}[/tex]

[tex]\nu = 9.37x10^{14}Hz[/tex]    

[tex]\lambda = 400nm . \frac{1m}{1x10^{9}nm}[/tex] ⇒ [tex]4x10^{-7}m[/tex]

[tex]\nu = \frac{3x10^{8}m/s}{4x10^{-7}m}[/tex]    

[tex]\nu = 7.5x10^{14}Hz[/tex]    

Hence, the frequency range of UVA is [tex]7.5x10^{14}Hz[/tex]  to [tex]9.37x10^{14}Hz[/tex].

Answer:

D from high to low areas

Answer:

F. See if the slope of Velocity versus Time for both directions are equal but opposite.

Explanation: Velocity is the rate of change of position. The slope of a distance or position-time graph is velocity. Acceleration is the rate of change of velocity. The slope of a velocity-time graph is acceleration.

A velocity-time graph shows changes in the velocity of a moving object over time. The slope of a velocity-time graph represents an acceleration of the moving object.

So, the value of the slope at a particular time represents the acceleration of the object at that instant.

Answer:

The total length is 0.65m.

Explanation:

The total length [tex]x_{tot}[/tex] of the spring is equal to its length [tex]x_0[/tex] right now (0.50 m ) plus the amount [tex]x[/tex] by which it is compressed due to weight of the man:

[tex]x_{tot} = x_0+x[/tex]

The spring compression is given by Hooke's law:

[tex]F = -kx[/tex]

which in our case gives

[tex]-Mg = -kx[/tex]

solving for [tex]x[/tex] we get:

[tex]x= \dfrac{Mg}{k }[/tex]

putting in [tex]M = 150kg, g= 10m/s^2[/tex] and [tex]k = 10,000N/m[/tex] we get:

[tex]x= \dfrac{(150kg)(10m/s^2)}{10,000N/m^2}[/tex]

[tex]x = 0.15m[/tex]

Hence, the total length of the spring is

[tex]x_{tot} = 0.50m+0.15m[/tex]

[tex]\boxed{x_{tot} = 0.65m.}[/tex]

Explanation:

Mass of the diskshaped grindstone, m = 1.1 kg

Radius of disk, r = 0.09 m

Angular velocity, [tex]\omega=73.3\ rad/s[/tex]

Time, t = 42.4 s

We need to find the frictional torque exerted on the grindstone. Torque in the rotational kinematics is given by :

[tex]\tau=I\alpha[/tex]

I is moment of inertia of disk, [tex]I=\dfrac{mr^2}{2}[/tex]

[tex]\tau=\dfrac{mr^2\alpha }{2}\\\\\tau=\dfrac{1.1\times (0.09)^2\times 73.3 }{2\times 42.4}\\\\\tau=7.7\times 10^{-3}\ N-m[/tex]

So, the frictional torque exerted on the grindstone is  [tex]7.7\times 10^{-3}\ N-m[/tex].

Answer:

Explanation:

velocity of light in a medium of refractive index V = V₀ / μ

V₀  is velocity of light in air and μ is refractive index of light.

time to travel in tube with air =  length of tube / velocity of light

8.72 ns = L / V₀  L is length of tube .

time to travel in tube with jelly =  length of tube / velocity of light

8.72+ 1.82  = L / V  L is length of tube .

10.54 ns = L / V

dividing the equations

10.54 / 8.72 = V₀  / V

10.54 / 8.72 =   μ

1.21 =  μ

refractive index of jelly = 1.21 .

Answer:

The magnitude of the acceleration  is [tex]a = 0.33 m/s^2[/tex]

The direction is [tex]- \r k[/tex] i.e the negative direction of the z-axis

Explanation:

 From  the question we are that

       The mass of the particle [tex]m = 1.8*10^{-3} kg[/tex]

         The charge on the particle is [tex]q = 1.22*10^{-8}C[/tex]

         The velocity is [tex]\= v = (3.0*10^4 m/s ) j[/tex]

        The the magnetic field is  [tex]\= B = (1.63T)\r i + (0.980T) \r j[/tex]

The charge experienced  a force which is mathematically represented as

                    [tex]F = q (\= v * \= B)[/tex]

    Substituting value

         [tex]F = 1.22*10^{-8} (( 3*10^4 ) \r j \ \ X \ \ ( 1.63 \r i + 0.980 \r j )T)[/tex]

            [tex]= 1.22 *10^{-8} ((3*10^4 * 1.63)(\r j \ \ X \ \ \r i) + (3*10^4 * 0.980) (\r j \ \ X \ \ \r j))[/tex]

            [tex]= (-5.966*10^4 N) \r k[/tex]

Note :

           [tex]i \ \ X \ \ j = k \\\\j \ \ X \ \ k = i\\\\k \ \ X \ \ i = j\\\\j \ \ X \ \ i = -k \\\\k \ \ X \ \ j = -i\\\\i \ \ X \ \ k = - j\\[/tex]

Now force is also mathematically represented as

        [tex]F = ma[/tex]

Making a the subject

      [tex]a = \frac{F}{m}[/tex]

   Substituting values

     [tex]a =\frac{(-5.966*10^4) \r k}{1.81*10^{-3}}[/tex]

        [tex]= (-0.33m/s^2)\r k[/tex]

        [tex]= 0.33m/s^2 * (- \r k)[/tex]

Answer:

The answer is A, friend.

The critical area density for a mirror in space where radiation pressure cancels out gravitational attraction can be computed using the principles of radiation pressure and gravitational force. By equating the momentum transfer from reflected sunlight to the gravitational pull of the Sun, one can find the area density at which these forces balance.

The question posed is about finding the critical area density for a mirror located in space at a distance R from the Sun, where the radiation pressure from sunlight would balance the gravitational attraction exerted by the Sun. The solution to this problem involves using the principle of momentum transfer from sunlight and equating it to the gravitational force to derive the density at which the forces are balanced.

The radiation pressure P exerted by sunlight can be calculated using the formula P = 2I/c, where I is the intensity of sunlight and c is the speed of light. Given that sunlight above Earth's atmosphere has an intensity of 1.30 kW/m², the radiation pressure P would be twice this value divided by the speed of light, due to the reflection phenomenon (momentum is doubled as the direction is reversed).

On the other hand, the gravitational force acting on an object is given by the formula F = GMm/R², where G is the gravitational constant, M is the mass of the Sun, m is the mass of the object (or spacecraft with the mirror), and R is the distance from the Sun. The mass m can be represented in terms of the area density ρA, where ρ is the area density and A is the area of the mirror.

Setting the radiation pressure equal to the gravitational force and solving for ρ will yield the critical value of area density at which the two forces cancel out.

The critical value of area density for the mirror at which the radiation pressure exactly cancels out the gravitational attraction from the sun is:

[tex]\[ \sigma = \frac{P_{sun}}{4\pi G M_{sun} R^2 c} \][/tex]

To find the critical value of area density (σ) for the mirror at which the radiation pressure exactly cancels out the gravitational attraction from the sun, we need to set the radiation pressure equal to the gravitational force per unit area.

Let's denote the radiation pressure as [tex]\( P_{rad} \)[/tex] and the gravitational force per unit area as [tex]\( P_{grav} \)[/tex]

The radiation pressure [tex]\( P_{rad} \)[/tex] can be calculated using the formula:

[tex]\[ P_{rad} = \frac{I}{c} \][/tex]

where [tex]\( I \)[/tex] is the intensity of the sunlight at the mirror's position and [tex]\( c \)[/tex] is the speed of light.

The intensity [tex]\( I \)[/tex] at a distance [tex]\( R \)[/tex] from the sun can be found using the inverse square law:

[tex]\[ I = \frac{P_{sun}}{4\pi R^2} \][/tex]

where [tex]\( P_{sun} \)[/tex] is the total power output of the sun.

The gravitational force per unit area [tex]\( P_{grav} \)[/tex] is given by:

[tex]\[ P_{grav} = \frac{F_{grav}}{A} = \frac{G M_{sun} m_{mirror} / R^2}{A} \][/tex]

where [tex]\( G \)[/tex] is the gravitational constant, [tex]\( M_{sun} \)[/tex] is the mass of the sun, [tex]\( m_{mirror} \)[/tex] is the mass of the mirror, and [tex]\( A \)[/tex] is the area of the mirror.

Since [tex]\( m_{mirror} = \sigma A \), where \( \sigma \)[/tex] is the area density of the mirror, we can write:

[tex]\[ P_{grav} = \frac{G M_{sun} \sigma A / R^2}{A} = \frac{G M_{sun} \sigma}{R^2} \][/tex]

Now, we set [tex]\( P_{rad} = P_{grav} \)[/tex] to find the critical value of [tex]\( \sigma \)[/tex]:

[tex]\[ \frac{I}{c} = \frac{G M_{sun} \sigma}{R^2} \][/tex]

Substituting [tex]\( I \)[/tex] from the intensity equation, we get:

[tex]\[ \frac{P_{sun}}{4\pi R^2 c} = \frac{G M_{sun} \sigma}{R^2} \][/tex]

Solving for [tex]\( \sigma \)[/tex], we find:

[tex]\[ \sigma = \frac{P_{sun}}{4\pi R^2 c} \cdot \frac{1}{G M_{sun}} \][/tex]

[tex]\[ \sigma = \frac{P_{sun}}{4\pi G M_{sun} R^2 c} \][/tex]

Complete question:- Suppose that the mirror described in Part

To calculate the wavelength of the light, use the equation d sin θ = mλ. Plugging in the values, we find that the wavelength is 563 nm.

In order to calculate the wavelength of the light, we can use the equation d sin θ = mλ, where d is the slit separation, θ is the angle of the bright fringe, m is the order of the fringe, and λ is the wavelength of the light. In this case, we are given the slit separation (0.0100 mm) and the angle of the third bright line (10.95°). We can rearrange the equation to solve for λ:

λ = d sin θ / m

Plugging in the values, we get:

λ = (0.0100 mm) sin(10.95°) / 3

λ = 5.63 x 10-7 mm or 563 nm

Answer:

Explanation:

The picture attached shows the full explanation and i hope it helps. Thank you

Final answer:

The magnetic force on a supersonic jet with a 0.500-μC static charge, flying over the Earth's south magnetic pole at 660 m/s, is 7.50 × 10-7 N, directed to the north. This effect is considered negligible given the small magnitude of the force.

Explanation:

When an aircraft with a static charge flies through the Earth's magnetic field, it experiences a magnetic force due to the interaction between its charge, its velocity, and the magnetic field. This is a concept from electromagnetism, specifically the Lorentz force. The magnitude of the magnetic force (F) can be calculated using the equation F = qvBsin(θ), where q is the charge on the aircraft, v is the speed of the aircraft, B is the magnetic field strength, and θ is the angle between the velocity of the charge and the direction of the magnetic field.

For a supersonic jet with a 0.500-μC (or 0.500x10-6 C) charge flying due west over the Earth's south magnetic pole, where the magnetic field (B) is 8.00 × 10-5 T and points straight down (θ = 90 degrees), the magnitude of the magnetic force is:

F = (0.500x10-6 C)(660 m/s)(8.00 × 10-5 T)sin(90 degrees) = 7.50 × 10-7 N, perpendicular to both the magnetic field lines and the velocity of the jet (which means it points either north or south). Since the jet is flying due west and the magnetic field is down, the right-hand rule indicates the force will push the jet to the north.

Regarding part (b), since the magnetic force magnitude, 7.50 × 10-7 N, is quite small compared to the typical forces experienced by a supersonic jet, such as thrust, drag, and lift, it can be considered a negligible effect. It's unlikely to have any significant impact on the flight of the aircraft.

Answer:

please read the answer below

Explanation:

We have that both electric field and magnetic field are given by:

[tex]\vec{E}=E_osin(kx-\omega t)\hat{j}\\\\\vec{B}=B_osin(kx-\omega t)\hat{k}[/tex]

I complete with bold words the answers:

_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _

a. In these formulas, it is useful to understand which variables are parameters that specify the nature of the wave. The variables E0 and B0are the magnitudes of the electric and magnetic fields.

2. amplitudes

b. The variable w is called the angular frequency of the wave.

2. angular frequency

c. The variable k is called the wavenumber of the wave.

1. wavenumber

c.

1. E =E0sin(wt)k^

d.

6. x and t

hope this helps!!

Answer:

11.206 km/s

Explanation: to leave the surface of the earth, you velocity Ve must be;

Escape velocity Ve = (2gRe)^0.5

Where g = acceleration due to gravity 9.81 m/s^2

Re = earth's radius 6400 km = 6.4x10^6

Ve = (2 x 9.81 x 6.4 x 10^6)^0.5

Ve = (125568000)^0.5

Ve = 11205.7 m/s

= 11.206 km/s

Answer:

20 ohm

Explanation:

V = I x R

R = V/ I

= 120/6

R = 20 ohm

Considering the Ohm's law, the resistance of the coffee machine is 20 Ω.

The current (I) is a measure of the speed at which the charge passes a given reference point in a specified direction. Its unit of measure is amps (A).

The driving force (electrical pressure) behind the flow of a current is known as voltage and is measured in volts (V). That is, voltage is a measure of the work required to move a charge from one point to another.

Resistance (R) is the difficulty that a circuit opposes to the flow of a current and it is measured in ohms (Ω).

Ohm's law establishes the relationship between current, voltage, and resistance in an electrical circuit.

This law establishes that the intensity of the current that passes through a circuit is directly proportional to the voltage of the same and inversely proportional to the resistance that it presents:

I= V÷R

Where:

In this case, you know that:

Replacing in the Ohm's Law:

6 A= 120 V÷R

Solving:

6 A× R= 120 V

R= 120 V÷6 A

R= 20 Ω

Finally, the resistance of the coffee machine is 20 Ω.

Learn more about Ohm's law:

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Answer:

A) 0.1955 days

B) 35.18 days

Explanation: Given that the Earth orbited the Sun in 4 months rather than 1 year, but had exactly the same rotation speed. That is the orbital period will be

365.24/12 = P/4

P = 121.75 days

We have three periods, the orbital period Po = 121.75 days,

the sidereal period Ps = 23.93419 hours and

The mean solar day Pd =24 hours.

Since they had exactly the same rotation speed, the number of solar days in a year is Po/Pd.

The number of sidereal days in a year would be Po/Pd + 1 = 121.75 + 1 = 122.75 days. Since there are more sidereal days in a year this means that the sidereal day is a little bit shorter, 121.75/122.75 = 0.99185 times shorter.

0.99185 × 24 = 23.80448 hours.

24 - 23.80448 = 0.1955 days

Therefore, solar day is 0.99185 days longer than a sidereal day

Given that the Mercury’s side real day is 58.6 days and its orbital period is 88 days. What is the length of Mercury’s solar day?

Using the formula

Ld = PoPs/(Po + Ps)

Ld = Length of Mercury's solar day

Po = orbital period

Ps = Mercury's side real day

Ld = (88 × 58.6)/ (88 + 58.6)

Ld = 5156.8/146.6

Ld = 35.18 days

Final answer:

To determine the resultant internal loadings at the cross section at E, we calculate the reaction forces at points B and C. The reaction force at B is 360 lb directed towards the left due to the thrust bearing, while the reaction force at C is 830 lb directed towards the right due to the journal bearing. Combining these forces, the resultant force at E is 1190 lb directed towards the right.

Explanation:

To determine the resultant internal loadings acting on the cross-section at E, we need to consider the forces acting on the shaft at points B and C. Part a requires determining the reaction force at B due to the thrust bearing, part b requires determining the reaction force at C due to the journal bearing, and part c requires combining the forces to find the resultant internal loadings at E.

For part a, since the shaft is supported by a smooth thrust bearing at B, the reaction force at B will be equal in magnitude and opposite in direction to the applied force P1. Therefore, the reaction force at B is 360 lb directed towards the left.

For part b, since the shaft is supported by a journal bearing at C, the reaction force at C will be equal in magnitude to the applied force P2. Therefore, the reaction force at C is 830 lb directed towards the right.

For part c, to find the resultant force at E, we need to combine the forces at B and C. Since the forces are along the same line of action, we can simply add the magnitudes of the forces. The resultant force at E is equal to the sum of the reaction forces at B and C, which is (360 lb) + (830 lb) = 1190 lb directed towards the right.

The effect of a change in the price of new Sony headphones on the equilibrium price of silicone replacement tips can't be accurately calculated from the provided equations without numerical values. An increase in the price of new headphones would likely lead to a surge in demand for replacement parts, thereby increasing their equilibrium price.

The student is asking about the impact of a change in the price of the new Sony headphones on the equilibrium price of replacement tips. The inverse demand function is given by p = pN - 0.1Q, and the inverse supply function is given by p = 2 + 0.012Q. The equilibrium condition is that the quantity demanded equals the quantity supplied, so we set the demand equation equal to the supply equation and solve for Q and p. Now to find the effect of a change in the price of the new headphones on the equilibrium price of the replacement tips, we need to differentiate the equilibrium price with respect to the price of the new headphones. This essentially involves implicit differentiation of the equilibrium condition.

However, due to the nature of the equations, getting a definitive value for dp/dpN would be difficult without numerical values to plug in. As such, the phrase 'dp/dpN is given by nothing' may have been a typographical error or misunderstanding. An increase in the price of new headphones would likely increase demand for replacement parts, thus raising their equilibrium price. But we cannot compute dp/dpN definitively from the provided equations.

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Given that,

Frequency emitted by the bat, f = 47.6 kHz

The speed off sound in air, v = 413 m/s

We need to find the wavelength detected by the bat. The speed of a wave is given by formula as follows :

[tex]v=f\lambda\\\\\lambda=\dfrac{v}{f}\\\\\lambda=\dfrac{413}{47.6\times 10^3}\\\\\lambda=0.00867\ m[/tex]

or

[tex]\lambda=8.67\ mm[/tex]

So, the bat can detect small objects such as an insect whose size is approximately equal to the wavelength of the sound the bat makes i.e. 8.67 mm.

Answer:

if we double the distance the energy stored will be doubled also

Explanation:

The energy stored in a capacitor is given as

Energy stored =1/2(cv²)

Or

= 1/2(Qv)

Where c = capacitance

Q= charge

But the electric field is expressed as

E= v/d

where v= voltage

d= distance

v=Ed

Substituting into any equation above say

Energy stored =1/2(Qv)

Substituting v=Ev

Energy stored =1/2(QEd)

From the equation above it shows that if we double the distance The energy stored will be doubled also

Answer:

Explanation:

Given that,

Period of Planet A around the star

Pa = Ta

Let the semi major axis of planet A from the star be

a(a) = x

Given that, the distance of Planet B from the star (i.e. it semi major axis) is 4 time that of Planet A

a(b) = 4 × a(a) = 4 × x

a(b) = 4x

Also planet B is 4 times massive as planet A

Using Kepler's law

P² ∝a³

P²/a³ = k

Where

P is the period

a is the semi major axis

Then,

Pa² / a(a)³ = Pb² / a(b)³

Pa denotes Period of Planet A and it is Pa = Ta

a(a) denoted semi major axis of planet A and it is a(a) = x

Pb is period of planet B, which is the required question

a(b) is the semimajor axis of planet B and it is a(b) = 4x

So,

Ta² / x³ = Pb² / (4x)³

Ta² / x³ = Pb² / 64x³

Cross multiply

Ta² × 64x³ = Pb² × x³

Divide both sides by x³

Ta² × 64 = Pb²

Then, Pb = √(Ta² × 64)

Pb = 8Ta

Then, the period of Planet B is eight times the period of Planet A.

The correct answer is C

The period of orbit for planet B is 4 times the period of orbit for planet A.

To determine the period of orbit for planet B, we can use Kepler's Third Law, which states that the square of the period of an orbit is proportional to the cube of its average distance from the center of attraction. Since planet B is four times the distance of planet A and four times as massive, its period will be 4^3/4^2 times that of planet A. Simplifying, this means that TB = 4 * TA, so the correct answer is d. 4TA.

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Answer:

A) Air resistance acts in a direction opposite the the fall of an object reducing it by doing work against the weight of the object due to gravity.

B) using a ping pong ball, the time of fall will be greatly reduced since it has little weight (its mass x acceleration due to gravity) against the air resistance. The net downward force of the weight and the air resistance will be small.

C) No, I wouldn't expect them to fall at the same time. The steel ball will have more weight compared to the ping pong ball and hence it will have a larger net force downwards.

D) If they are both released from a 6 m height, the steel ball will fall to the ground first since it has a larger net force downwards.

Answer:this is obviously because free electrons can easily move in the metal due to potential difference, and they are inserted in the wire from the negative terminal of the battery and pulled away ( come back) from the wire into the battery at positive terminal.

Electrons in a lightbulb filament flow due to the electric fields created by a connected battery, encountering resistance that dissipates energy as heat and allows the bulb to emit light.

Electrons 'know' to flow in a bulb's filament due to the electric fields created when a battery is connected to a circuit. These fields exert a force on the electrons, causing them to move. The flow of electrons is sustained by the potential difference (emf) provided by the battery, but it's not perpetual because as electrons move through the resistive filament, they encounter resistance analogous to friction, which causes energy to be dissipated as heat. This is the principle by which the bulb emits light. An open circuit, like if a wire is cut, prevents the flow of charge, demonstrating the necessity of a closed path for current to flow.

The resistance to electron movement through the filament is due to interactions with the material's atoms, as described by Newton's second law (a = Ftotal/m). This resistance leads to the conversion of electrical energy into heat, which in the case of an incandescent lightbulb, is necessary for it to glow. Without resistance, electrons accelerated by the emf would continue gaining kinetic energy, which does not occur due to this energy transformation. The steady flow of electrons is the result of these resisting forces balancing out the force exerted by the electric field, leading to a constant flow of current as long as the circuit is complete.

The question is incomplete! Complete question along with answer and step by step explanation is provided below.

Question:

A fisherman notices that his boat is moving up and down periodically without any horizontal motion, owing to waves on the surface of the water. It takes a time of 2.60 s for the boat to travel from its highest point to its lowest, a total distance of 0.630 m. The fisherman sees that the wave crests are spaced a horizontal distance of 5.50 m apart.

How much is the wavelength?

How fast are the waves traveling ?

What is the amplitude A of wave?

Given Information:

time = t = 2.60 s

wavelength = λ  = 5.50 m

distance = d = 0.630 m

Required Information:

a)  wavelength = λ  = ?

b) speed = v = ?

c) Amplitude = A = ?

Answer:

a)  wavelength = 5.50 m

b) speed = 1.056 m/s

c) Amplitude = 0.315 m

Step-by-step explanation:

a)

It is given that wave crests are spaced a horizontal distance of 5.50 m apart that is basically the wavelength so,

λ  = 5.50 m

b)

We know that the speed of the wave is given by

v = λf

where λ is the wavelength and f is the frequency of the wave given by

f = 1/T

Where T is the period of the wave.

Since the it is given that boat takes 2.60 s to travel from its highest point to its lowest that is basically half of the period so one full period is

T = 2*2.60

T = 5.2 s

So the frequency is,

f = 1/5.2

f = 0.192 Hz

Therefore, the speed is

v = λf

v = 5.50*0.192

v = 1.056 m/s

c)

The amplitude of the wave is given by

A = d/2

where d is the distance from the highest point to the lowest, therefore, the amplitude is half of it.

A = 0.630/2

A = 0.315 m

Answer:

d.-10.3m

Explanation:

Note for short sightedness the focal length is negative

Let do be object distance=10m

And di= image distance=-300m

Using lens formula

F=do*di/do-di= 10*300/10-300=-10.3m

Answer:

d. 10.3 m

Explanation:

For a nearsighted person,

1/f = (1/v)+(1/u).................... Equation 1

f = focal, v = image distance, u = object distance.

f = vu/(v+u)................. Equation 2

Given: v = -10 m, u = 300 m

Substitute into equation 2

f = (-10×300)/(-10+300)

f = -3000/290

f = -10.34 m

therefore a concave lens of focal length 10.34 m is  required.

The right option is d. 10.3 m

Answer:

The total permutation = nP2

The inversions = nC2

Explanation:

Please look at the solution in the attached Word file

The issue of farsightedness or presbyopia can be corrected by a converging lens with a focal length of 34 cm and power of 2.94 Diopters.

The provided problem can be solved using the lens formula which is 1/f = 1/v - 1/u, where f = focal length, v = image distance, and u = object distance. The image distance (di) for clear vision should match the lens-to-retina distance, which is constant and taken as 2 cm for the given question. The lens that the physicist is recommended to use should create an image with a reading distance of 25 cm and therefore this will be ‘v’. However, the lens is worn 2 cm in front of the eye and hence the actual image distance becomes 27 cm (or 0.27 m). The object is however at a distance of 97 cm (0.97 m) as indicated by the problem. When we plug these values into the lens formula, we get f = 1 / {(1/0.27) - (1/0.97)} which is approximately equal to 0.34 m or 34 cm. The optical power of lens (P) is reciprocal of the focal length in meters. Hence P = 1/f = 1/0.34 = 2.94 D.

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To correct the physicist's presbyopia for clear vision at 25 cm, a lens with a focal length of approximately -30.35 cm (or -0.3035 m) and a power of around -3.29 diopters is needed. The lens formula helped in determining these values.

Correcting Farsightedness

A presbyopic eye has difficulty seeing objects clearly at close range. In this case, the physicist's eye can only see clearly beyond 97 cm. To correct this and allow clear vision at 25 cm, we need to determine the focal length and power of the corrective lens when worn 2 cm in front of the eye.

Using the lens formula:

1/f = 1/v - 1/u

where:

Substituting the values, we get:

1/f = 1/95 - 1/23

1/f = (23 - 95) / (95 * 23)

1/f = -72 / 2185

f = -2185 / 72 ≈ -30.35 cm

Therefore, the focal length of the lens is approximately -30.35 cm (or -0.3035 m).

The power of the lens (P) is given by the formula:

P = 1/f (in meters)

Substituting the focal length:

P = 1 / -0.3035 ≈ -3.29 diopters (D)

Hence, the power of the corrective lens needed is approximately -3.29 D.

Answer:

2.46 W

Explanation:

Thermal conductivity k = 53.4 W/m-K

Radius r = 12 mm = 12x10^-3 m

Lenght = 2.5 m

T1 = 361 °C

T2 = 105 ∘C

Area A = ¶r^2 = 3.142 x (12x10^-3)^2

= 0.00045 m^2

Power P = -AkdT/dx

Where dT = 361 - 105 = 256

dx = lenght of heat travel = 2.5 m

P = -0.00045 X 53.4 X (256/2.5)

= -0.024 X 102.4 = -2.46 W

The negative sign indicates that heat is lost from the metal.

Answer:

Total acceleration will be [tex]160.20rad/sec^2[/tex]

Explanation:

We have given radius of the circle r = 0.685 m

Angular acceleration [tex]\alpha =63.8rad/sec^2[/tex]

Angular speed [tex]\omega =15rad/sec[/tex]

Centripetal acceleration will be

[tex]a_c=\omega ^2r=15^2\times 0.685=154.125rad/sec^2[/tex]

Tangential acceleration will be

[tex]a_t=r\alpha =0.685\times 63.8=43.7rad/sec^2[/tex]

(a) Total acceleration will be equal to

[tex]a=\sqrt{a_t^2+a_c^2}[/tex]

[tex]a=\sqrt{43.7^2+154.125^2}=160.20rad/sec^2[/tex]

So total acceleration will be [tex]160.20rad/sec^2[/tex]