he deflection plates in an oscilloscope are 10 cm by 2 cm with a gap distance of 1 mm. A 100 volt potential difference is suddenly applied to the initially uncharged plates through a 1000 ohm resistor in series with the deflection plates. How long do

Question is not complete and the missing part is;

A coin of mass 0.0050 kg is placed on a horizontal disk at a distance of 0.14 m from the center. The disk rotates at a constant rate in a counterclockwise direction. The coin does not slip, and the time it takes for the coin to make a complete revolution is 1.5 s.

Answer:

0.828 m/s

Explanation:

Resolving vertically, we have;

Fn and Fg act vertically. Thus,

Fn - Fg = 0 - - - - eq(1)

Resolving horizontally, we have;

Ff = ma - - - - eq(2)

Now, Fn and Fg are both mg and both will cancel out in eq 1.

Leaving us with eq 2.

So, Ff = ma

Now, Frictional force: Ff = μmg where μ is coefficient of friction.

Also, a = v²/r

Where v is linear speed or velocity

Thus,

μmg = mv²/r

m will cancel out,

Thus, μg = v²/r

Making v the subject;

rμg = v²

v = √rμg

Plugging in the relevant values,

v = √0.14 x 0.5 x 9.8

v = √0.686

v = 0.828 m/s

Final answer:

To determine the linear speed of the coin when it just begins to slip, we can use the equation frictional force = centripetal force for circular motion. By equating these two forces and solving for the linear speed, we can find the answer.

Explanation:

To determine the linear speed of the coin when it just begins to slip, we can use the equation:
frictional force = centripetal force for circular motion

The frictional force can be calculated using the equation:
frictional force = coefficient of static friction x normal force

And the centripetal force can be calculated using the equation:
centripetal force = mass of coin x acceleration towards the center of the disk

By equating these two forces and solving for the linear speed, we can find the answer.

In this case, we are given the coefficient of static friction as 0.50. We can also assume that the normal force is equal to the weight of the coin, which is the mass of the coin multiplied by the acceleration due to gravity. By plugging in these values, we can find the linear speed of the coin.

The solution is in the attachment

Answer:

[tex]a=1.1*10^{18}\frac{m}{s^2}[/tex]

Explanation:

We use the following kinematic formula to calculate the acceleration:

[tex]v_f^2=v_0^2+2ax[/tex]

The kinetic energy is defined as:

[tex]\Delta K=\frac{m(v_f^2-v_0^2)}{2}\\v_f^2-v_0^2=\frac{2\Delta K}{m}[/tex]

Replacing this in the acceleration formula and solving for a:

[tex]\frac{2\Delta K}{m}=2ax\\a=\frac{\Delta K}{mx}\\a=\frac{2*10^{-18}J}{(9.1*10^{-31}kg)(2*10^{-6}m)}\\a=1.1*10^{18}\frac{m}{s^2}[/tex]

Explanation:

(a)  We will determine the mass flow rate as follows.

                m = [tex]\rho_{1} V_{1}[/tex]

                    = [tex]\frac{P_{1}}{RT_{1}}A_{1}v_{1}[/tex]

                    = [tex]\frac{P_{1}}{RT_{1}} \times \frac{D^{2}}{4} \pi v_{1}[/tex]

Putting the given values into the above formula as follows.

      m = [tex]\frac{P_{1}}{RT_{1}} \times \frac{D^{2}}{4} \pi v_{1}[/tex]

          = [tex]\frac{200}{0.287 \times 293 K} \times \frac{(0.16)^{2}}{4} \pi \times 5[/tex]                          

          = 0.239 kg/s

Hence, the mass flow rate of the inlet/outlet is 0.239 kg/s.

(b)  Now, we will determine the final volume rate as follows.

            [tex]V_{2} = \frac{m}{\rho_{2}}[/tex]

                        = [tex]\frac{RT_{2}m}{P_{2}}[/tex]

                        = [tex]\frac{0.287 \times 313 \times 0.239}{180}[/tex]

                        = 0.119 [tex]m^{3}/s[/tex]

And, the final velocity will be determined as follows.

               [tex]v_{2} = \frac{V_{2}}{A}[/tex]

                         = [tex]\frac{4V_{2}}{D^{2} \times \pi}[/tex]

                         = [tex]\frac{4 \times 0.119}{(0.16)^{2} \times \pi}[/tex]

                         = 5.92 m/s

Therefore, the volumetric flow rate is 0.119 [tex]m^{3}/s[/tex] and velocity rate is 5.92 m/s.

Answer:

Work = N x m

Explanation:

W=Fxd

Work equals force times the distance (displacement).

Complete Question:

A deuteron (a nucleus that consists of one proton and one neutron) is accelerated through a 4.01 kV potential difference. b) How much kinetic energy does it gain? The mass of a proton is 1.67262 × 10−27 kg, the mass of a neutron 1.67493 × 10−27 kg and the charge on an electron −1.60218 × 10−19 C. Answer in units of J\

b) what is the speed?

Answer:

a) the kinetic energy gained =  6.42 * 10⁻¹⁶ J

b) the speed of the particle, v = 619328.3 m/s

Explanation:

q = 1.602 *10⁻¹⁹C

V = 4.01 kV  = 4.01 * 10³ V

Work done by the deuteron = qV

Work done by the deuteron = 1.602 * 10⁻¹⁹ *  4.01 *10³

Work done = 6.42 * 10⁻¹⁶ J

Kinetic Energy gained = work done

Kinetic Energy gained by the deuteron =  6.42 * 10⁻¹⁶ J

B) The formula for Kinetic Energy is given by:

KE = 1/2 Mv²

Let the mass of the proton be m₁ = 1.67262 × 10⁻²⁷kg

Let the mass of the neutron be m₂ =   1.67493 × 10−27 kg

M = m₁ + m₂

KE = 1/2 ( m₁ + m₂)v²

Let v = speed of the deuteron

From part (a)

KE = 6.42 * 10⁻¹⁶ J

 1/2 ( m₁ + m₂)v²=  6.42 * 10⁻¹⁶

0.5 * (1.67262 + 1.6749) *10⁻²⁷ * v² =  6.42 * 10⁻¹⁶

v = 619328.3 m/s

Answer:

Thermal Pollution

Explanation:

Thermal pollution is a term use to describe the final result caused from the  water used as a coolant by power plants and/or industrial companies. It changes the water temperature and its quality. It is usually caused from humans or companies especially manufacturing companies - They use it for their personal needs. The end product of the water after being discarded is Thermal Pollution

Answer:

[tex]I = 37.5\ A[/tex]

Explanation:

Given,

Magnetic field,B = 0.5 x 10⁻⁴ T

distance,r= 15 cm = 0.15 m

Current = ?

Using Ampere's law of magnetic field

[tex]B = \dfrac{\mu_0I}{2\pi r}[/tex]

[tex]I= \dfrac{B (2\pi r)}{\mu_0}[/tex]

[tex]I= \dfrac{0.5\times 10^{-4}\times (2\pi \times 0.15)}{4\pi \times 10^{-7}}[/tex]

[tex]I = 37.5\ A[/tex]

Current in the wire is equal to [tex]I = 37.5\ A[/tex]

The maximum current this wire can carry is equal to 37.5 Amperes.

Given the following data:

Scientific data:

In order to determine the maximum current, we would apply Ampere's law of magnetic field.

Mathematically, Ampere's law of magnetic field is given by this formula:

[tex]I=\frac{2B\pi r}{\mu_o }[/tex]

Where:

Substituting the given parameters into the formula, we have;

[tex]I=\frac{2 \pi \times 0.5 \times 10^{-4}\times 0.15}{4\pi \times 10^{-7} }\\\\I=\frac{0.5 \times 10^{-4}\times 0.15}{2\pi \times 10^{-7} }[/tex]

I = 37.5 Amperes.

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Answer:

0.67 m/s² or 2/3 m/s²

3 m

Explanation:

Using

F-F' = ma............ Equation 1

Where F = Horizontal force applied to the box, F' = Frictional force, m = mass of the box, a = acceleration of the box.

make a the subject of the equation

a = (F-F')/m............ Equation 2

Given: F = 230 N, F' = 210 N, m = 30 kg.

substitute into equation 2

a = (230-210)/30

a = 20/30

a = 0.67 m/s² or 2/3 m/s²

The acceleration of the box = 2/3 m/s²

Using,

s = ut+1/2at²............ Equation 3

Where u = initial velocity, t = time, a = acceleration, s = distance.

Given: u = 0 m/s (from rest), t = 3 s, a = 2/3 m/s²

substitute into equation 3

s = 0(3)+1/2(2/3)(3²)

s = 3 m.

Hence the box moves 3 m

The acceleration of the 30.0-kg box is approximately 0.67 m/s². If it starts from rest, it would move around 3.02 meters in 3 seconds.

In order to solve the problem, we'll apply the principle of Newton's second law, which states that the acceleration of an object as produced by a net force is directly proportional to the magnitude of the net force, in the same direction as the net force, and inversely proportional to the mass of the object.

First, we identify the net force on the box. The box is being pulled by a force of 230 N, and there is friction acting against this pull with a force of 210 N. The net force (F) is therefore 230 N (pulling force) - 210 N (friction) = 20 N.

To find the acceleration (a), we use the formula:
a = F/m
Substituting the given values, we get:
a = 20 N / 30.0 kg = 0.67 m/s².

Now, the second part of the problem asks for the distance the box would move in 3 seconds if it starts from rest. We use the formula for distance (d) traveled under constant acceleration:
d = 0.5 * a * t²
Substituting the calculated acceleration and time (3 seconds), we find:
d = 0.5 * 0.67 m/s² * (3 s)² = 3.02 m.

So, under the given conditions, the box's acceleration is 0.67 m/s², and it would move approximately 3.02 meters in 3 seconds.

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Answer:

the time required for the sound to travel between the whales 0.66 S.

Explanation:

As given in the problem, the velocity of sound wave ([tex]v[/tex]) is governed by the equation

[tex]v = \sqrt{\dfrac{B}{\rho}}[/tex]

Given, [tex]B = 2.38 \times 10^{9} Pa[/tex] and [tex]\rho = 1046 Kg m^{-3}[/tex]

So for salt water, the velocity of sound wave ([tex]v_{s}[/tex]) can be written as

[tex]v_{s} = \sqrt{\dfrac{2.38 \times 10^{9}}{1046}} ms^{-1} = 1.508 \times 10^{3} ms^{-1}[/tex]

As the whales are d = 1 Km or 1000 m apart from each other, so the time ([tex]t[/tex]) required for the sound wave to travel this distance is given by

[tex]t = \dfrac{d}{v_{s}} = \dfrac{1000 m}{1.508 \times 10^{3}} = 0.66 s[/tex]

To find the time it takes for the sound to travel between two whales in water, use the equation v = √(B/ρ) where v is the velocity, B is the bulk modulus, and ρ is the density. Rearrange the equation to solve for time, t, using the formula t = distance / velocity.

To find the time it takes for the sound to travel between the whales, we can use the equation v = √(B/ρ), where v is the velocity, B is the bulk modulus, and ρ is the density. In this case, B = 2.38 × 109 Pa and ρ = 1046 kg/m3. We can rearrange the equation to solve for the time, t: t = d/v, where d is the distance between the whales. In this case, d = 1.0 km = 1000 m. Plugging in the values, we get t = 1000 / √(2.38 × 109 / 1046). Simplifying this expression gives us the time it takes for the sound to travel between the whales.

Answer:

The overall thermal efficiency is 30%.

Explanation:

Given;

Output power = 500 kWh

input energy  per kg of coal = 6 MJ = 6 x 10⁶ J = 1.66667 kWh

1000 kg of coal will produce 1000 x 1.66667 kWh = 1666.67 kWh

Thus, total input power = 1666.67 kWh

Overall thermal efficiency = Total output power/Total input Power

Overall thermal efficiency = (500/1666.67) *100

Overall thermal efficiency = 0.29999 *100

Overall thermal efficiency = 30%

Therefore, the overall thermal efficiency is 30%.

Answer:

22.1 V

Explanation:

We are given that

[tex]A_1=0.7 cm^2=0.7\times 10^{-4} m^2[/tex]

[tex]A_2=2.5 cm^2=2.5\times 10^{-4} m^2[/tex]

[tex]A_3=2.2 cm^2=2.2\times 10^{-4} m^2[/tex]

[tex]A_4=3 cm^2=3\times 10^{-4} m^2[/tex]

Using [tex] 1cm^2=10^{-4} m^2[/tex]

We know that

[tex]R=\frac{\rho l}{A}[/tex]

In series

[tex]R=R_1+R_2+R_3+R_4[/tex]

[tex]R=\frac{\rho l}{A_1}+\frac{\rho l}{A_2}+\frac{\rho l}{A_3}+\frac{\rho l}{A_4}[/tex]

[tex]R=\frac{\rho l}{\frac{1}{A_1}+\frac{1}{A_2}+\frac{1}{A_3}+\frac{1}{A_4}}[/tex]

Substitute the values

[tex]R=\rho A(\frac{1}{0.7\times 10^{-4}}+\frac{1}{2.5\times 10^{-4}}+\frac{1}{2.2\times 10^{-4}}+\frac{1}{3\times 10^{-4}})[/tex]

[tex]R=\rho l(2.62\times 10^4)[/tex]

[tex]V=145 V[/tex]

[tex]I=\frac{V}{R}=\frac{145}{\rho l(2.62\times 10^4)}[/tex]

Voltage across the 2.5 square cm wire=[tex]IR=I\times \frac{\rho l}{A_2}[/tex]

Voltage across the 2.5 square cm wire=[tex]\frac{145}{\rho l(2.62\times 10^4)}\times \frac{\rho l}{2.5\times 10^{-4}}=22.1 V[/tex]

Voltage across the 2.5 square cm wire=22.1 V

Answer:

8.60 g/cm³

Explanation:

In the lattice structure of iron, there are two atoms per unit cell. So:

[tex]\frac{2}{a^{3} } = \frac{N_{A} }{V_{molar} }[/tex] where [tex]V_{molar} = \frac{A}{\rho }[/tex] an and A is the atomic mass of iron.

Therefore:

[tex]\frac{2}{a^{3} } = \frac{N_{A} * p }{A}[/tex]

This implies that:

[tex]A = (\frac{2A}{N_{A} * p)^{\frac{1}{3} } }[/tex]

  = [tex]\frac{4}{\sqrt{3} }r[/tex]

Assuming that there is no phase change gives:

[tex]\rho = \frac{4A}{N_{A}(2\sqrt{2r})^{3} }[/tex]

  = 8.60 g/m³

Answer:

0.0399 m

Explanation:

We are given that

Mass of coffee=375g=[tex]\frac{375}{1000}=0.375 kg[/tex]

1kg=1000g

Depth=h=7.5 cm=[tex]7.5\times 10^{-2} m[/tex]

[tex]1 cm=10^{-2} m[/tex]

Density of coffee=[tex]\rho=1000kg/m^3[/tex]

We have to find the inside radius  of coffee mug.

We know that

[tex]\rho=\frac{m}{V}[/tex]

Substitute the values

[tex]1000=\frac{0.375}{\pi r^2h}[/tex]

[tex]r^2=\frac{0.375}{1000\times 7.5\times 10^{-2}\times 3.14}[/tex]

By using [tex]\pi=3.14[/tex]

[tex]r=\sqrt{\frac{0.375}{1000\times 7.5\times 10^{-2}\times 3.14}}[/tex]

[tex]r=0.0399 m[/tex]

Hence, the inside radius=0.0399 m

Answer:

(+2.093 × 10⁻⁷) C

Explanation:

Coulomb's law gives the force of attraction between two charges and it is given by

F = kq₁q₂/r²

where q₁ = charge on one of the two particles under consideration = charge on the balloon = - 3.75 × 10⁻⁸ C

q₂ = charge on the other body = charge on the rod = ?

k = Coulomb's constant = 1/(4 π ϵ₀) = 8.99 × 10⁹ N⋅m²/C²

r = distance between the two charges = d = 6.00 cm = 0.06 m

But for this question, the force of attraction between the charges was enough to lift the balloon and match its weight, Hence,

F = (kq₁q₂/d²) = - mg (negative because it's an attractive force)

m = mass of balloon = 2.00 g = 0.002 kg

g = acceleration due to gravity = 9.8 m/s²

(8.99 × 10⁹ × (-3.75 × 10⁻⁸) × q₂)/(0.06²) = 0.002 × 9.8

q₂ = (-0.002 × 9.8 × 0.06²)/(8.99 × 10⁹ × (-3.75 × 10⁻⁸)

q₂ = + 2.093 × 10⁻⁷ C

Answer:

east direction.

Explanation:

Given, that the electron is travelling in the north direction in the horizontal direction whereas the magnetic field applied is in the vertically downward direction.

Using the Maxwell's left hand rule, we point the index finger in the direction of magnetic field while the perpendicular middle finger points in the direction of motion of the positive charge and the thumb points in the direction of the force. During this position the angle between all the three fingers must be mutually perpendicular to each other.

Therefore we find that here in this case the magnetic force on the electron acts in the east direction.

The first finger points in the direction of the magnetic field and the middle finger in the direction of the electric current. In the east direction the magnetic force act on the electron.

The thumb will point in the direction of the force on the conductor if the thumb and first two fingers of the left hand are arranged at right angles to each other on a conductor and the hand is oriented.

So that the first finger points in the direction of the magnetic field and the middle finger in the direction of the electric current.

Given that the electron travels in a horizontal north-south path, whereas the magnetic field applied is vertically downward.

We point the index finger in the direction of the magnetic field, the perpendicular middle finger in the direction of positive charge motion, and the thumb in the direction of force using Maxwell's left-hand rule.

Hence in the east direction the magnetic force act on the electron.

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Answer:

he phase difference is π the destructive interference and the lujar is a still or silent place

Explanation:

This is a sound interference exercise where the amino difference is equal to the phase difference of the sound.

      Δr / λ = ΔФ / 2π

Let's find the path difference

     r₁ = 4m

     r₂ = √ (4² + 3²) = 5 m

     Δr = r₂ - r₁

     Δr = 5-4 = 1m

Let's find the wavelength of the sound

        v =  λ f

         λ = v / f

         λ = 340/170

         λ = 2m

Let's find the phase difference between the two waves

      ΔФ = Δr 2π / λ

      ΔФ = 1 2π / 2

      ΔФ = π

Since the phase difference is π the destructive interference and the lujar is a still or silent place

Final answer:

To determine if Sam is standing on a loud or quiet spot, the path difference of the sound waves from the speakers was calculated, which is half the wavelength, indicating that Sam stands at a loud spot due to constructive interference.

Explanation:

The question relates to interference patterns created by the sound waves from two speakers and whether the spot where Sam stands is a loud or quiet spot. To determine this, we need to calculate the path difference of the sound waves reaching Sam from both speakers. The speed of sound in air is given as 340 m/s, and the frequency of the tone is 170 Hz.

The wavelength (λ) of the sound can be found using the formula speed = frequency × wavelength, which results in λ = 340 m/s / 170 Hz = 2 m. The path difference is the difference in distance from each speaker to Sam. For one speaker, the distance is 4.0 m. For the other speaker, we use the Pythagorean theorem since Sam is standing in front of the first speaker: √(4.0^2 + 3.0^2) = 5.0 m. The path difference is therefore 5.0 m - 4.0 m = 1.0 m, which is exactly half the wavelength.

Since the path difference corresponds to half a wavelength, this results in constructive interference, and Sam is indeed standing at a loud spot.

Answer:

10044 N

Explanation:

The acceleration of the cab is calculated using the equation of motion:

[tex]v^2 = u^2+2as[/tex]

v is the final velocity = 0 m/s in this question, since it is brought to rest

u is the initial velocity = 10 m/s

a is the acceleration

s is the distance = 35 m

[tex]a = \dfrac{v^2-u^2}{2s} = \dfrac{(0 \text{ m/s})^2-(10 \text{ m/s})^2}{2\times (35\text{ m})} = -1.43\text{ m/s}^2[/tex]

Since it accelerates downwards, its resultant acceleration is

[tex]a_R = g + a[/tex]

g is the acceleration of gravity.

[tex]a_R = (9.8-1.43)\text{ m/s}^2 = 8.37\text{ m/s}^2[/tex]

The tension in the cable is

[tex]T = ma_R = (1200\text{ kg})(8.37\text{ m/s}^2) = 10044 \text{ N}[/tex]

Answer:

[tex]V=17.9\ Volt[/tex]

Explanation:

Joule's Law in Electricity

The Joule's law allows us to calculate the power dissipated in a resistor of resistance R through which goes a current I.

[tex]P=I^2R[/tex]

The relation between the voltage and the current is given by Ohm's law:

[tex]V=RI[/tex]

Solving for I and replacing int the first equation

[tex]\displaystyle P=\frac{V^2}{R}[/tex]

Solving for V

[tex]V=\sqrt{PR}[/tex]

[tex]V=\sqrt{0.8\cdot 400}=17.9[/tex]

[tex]\boxed{V=17.9\ Volt}[/tex]

Maximum voltage for given power rating, applied across this resistor without damaging it is 17.9 volts.

Ohm's law states that for a flowing current the potential difference of the circuit is directly proportional to the current flowing in it. Thus,

[tex]V\propto I[/tex]

Here, (V) is the potential difference and (I) is the current.

It can be written as,

[tex]V=IR[/tex]

Here, (R) is the resistance of the circuit.

Given information-

The value of resistance is 400-Ω.

The value of power rating is 0.800 W.

By the Joule's law, the power of a circuit is equal to the product of the  square of the current flowing in it and the resistance. It can be given as,

[tex]P=I^2\times R\\I=\sqrt{\dfrac{P}{R}}[/tex]

Put this value of current in ohm's law as,

[tex]V=\sqrt{\dfrac{P}{R}}\times R\\V=\sqrt{PR}[/tex]

Put the value of power and current in the above formula,

[tex]V=\sqrt{0.800\times 400}\\V=17.9\rm Volts[/tex]

'

Thus the maximum voltage that can be applied across this resistor without damaging it is 17.9 volts.

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Answer:

Speed of larger piece is [tex]\dfrac{V\sqrt{2}}{3}[/tex]

Explanation:

We apply the principle of conservation of momentum.

The watermelon is initially at rest. The initial momentum = 0 kg m/s in all directions.

After the collision,

Vertical momentum = momentum of piece in y-direction + y-component of momentum of larger piece = [tex]mV + 3mv_{ly}[/tex]

Here, [tex]v_{ly}[/tex] is the y-component of velocity of larger piece.

This is equal to 0, since the initial momentum is 0.

[tex]v_{ly}=\dfrac{V}{3}[/tex]

Horizontal momentum = momentum of piece in x-direction + x-component of momentum of larger piece = [tex]mV + 3mv_{lx}[/tex]

Here, [tex]v_{lx}[/tex] is the x-component of velocity of larger piece.

This is also equal to 0, since the initial momentum is 0.

[tex]v_{lx}=\dfrac{V}{3}[/tex]

The velocity of the larger piece, [tex]v_l[/tex], is the resultant of [tex]v_{lx}[/tex] and [tex]v_{ly}[/tex]. Since they are mutually perpendicular,

[tex]v_l = \sqrt{v_{ly}^2+v_{lx}^2}= \sqrt{\left(\dfrac{V}{3}\right)^2+\left(\dfrac{V}{3}\right)^2}[/tex]

[tex]v_l = \dfrac{V\sqrt{2}}{3}[/tex]

To solve this problem we will apply the concepts related to the Friction force and work. The friction force can be defined as the product between the Normal Force (Mass by gravity) and the dynamic friction constant. In the case of Work this is defined as the product of the distance traveled by the applied force. Then we will solve the points sequentially to find the answer to each point,

PART A) The friction force with the given data  is,

[tex]F_f = \mu_k mg[/tex]

Here,

[tex]\mu_k[/tex] = Kinetic coefficient

m = Mass

g = Gravitational acceleration

[tex]F_f = (0.25)(30kg)(9.8m/s^2)[/tex]

[tex]F_f = 73.5N[/tex]

PART B) The work done by the worker is the distance traveled for the previously force found, then

[tex]W_w = rF_f[/tex]

[tex]W_w = (4.5m)(73.5N)[/tex]

[tex]W_w = 330.75J[/tex]

PART C) The work done by the friction force would be the distance traveled with the previously calculated force, therefore

[tex]W_f = -rF_f[/tex]

[tex]W_f = -(4.5m)(73.5N)[/tex]

[tex]W_f = -330.75J[/tex]

[tex]W_f = -331J[/tex]

PART D) The work done by the normal force is,

[tex]W_N = N (r) Cos(90)[/tex]

[tex]W_N = 0J[/tex]

The work done by gravitational force is

[tex]W_g = rF_gcos(90)[/tex]

[tex]W_g = 0J[/tex]

PART E) The expression for the total work done is,

[tex]W_{net} = W_f +W_w +W_N+W_g[/tex]

[tex]W_{net} = 330.75-330.75+0+0[/tex]

[tex]W_{net} = 0J[/tex]

Therefore the net work done by the system is 0J

Answer:

The velocity in the smaller arteries will be reduced by a factor of 0.139

Explanation:

The flow rate of blood is going to stay the same when it is transferred from the major artery to the smaller ones.

flow rate = Velocity * Area

Since the flow rate remains constant, we have:

Flow rate in major artery = combined flow rate in smaller arteries

Velocity in Major artery * 1.00 = Velocity in smaller artery * (0.4 * 18)

[tex]V_M * 1 = V_S * (18*0.4)[/tex]

[tex]\frac{V_S}{V_M}=\frac{1}{18*0.4}[/tex]

[tex]\frac{V_S}{V_M}= 0.139[/tex]

Thus, the velocity in the smaller arteries will be reduced by a factor of 0.139

To calculate the distance L between the screen and the slit, use the single-slit diffraction minimum condition combined with the known distance of the red laser's third minimum and its wavelength.

The student is working on a Physics problem related to single-slit diffraction. To find the distance L between the screen and the slit, we can use the formula for the position of a diffraction minimum, y = L × tan(θ), where θ is the angle of the diffraction minimum. However, for small angles (as in most diffraction problems), tan(θ) ≈ sin(θ), so the formula simplifies to y = L × sin(θ). The condition for the minima in a single-slit diffraction pattern is given by a × sin(θ) = m × λ, where a is the width of the slit, m is the order number of the minimum, and λ is the wavelength of the light. Given that the third diffraction minimum (m = 3) for the red laser is at a distance y3,red = 4.05 cm from the pattern's center, and the wavelength of the red light is λ = 633.0 nm, we can write 0.150 mm × sin(θ) = 3 × 633.0 nm. Solving for sin(θ) and then using y = L × sin(θ) where y = 4.05 cm, we can calculate the distance L.

Answer:

Entropy Change = 0.559 Times

Explanation:

Entropy change is determined by the change in the micro-states of a system. As we know that the micro-states are the same as measure of disorderness between initial and final states, that's the the amount of change in micro-states determine how much of entropy has changed in the system.

Answer:

The distance between the ships changing at 6PM is 21.29Km/h

Explanation:

Ship A is sailing east at 35Km/h and ship B is sailing West at 25Km/h

Given

dx/dt= 35

dy/dt= 25

dv/dt= ???? at t= 6PM - 2PM= 4

Therefore t=4

We know ship A travels at 150km in the x-direction and Ship A at t=4 travels at 4.35 Which is 140 also in x-direction

So, we use:

[tex] D^2 = (150 - x)^2 + y^2 [/tex];

[tex] D^2 = (150 - 140)^2 + y^2 [/tex]

But ship B travels at t=4, at 4.25 =100 in the y-direction

so, let's use the equation:

[tex] D^2 = 10^2 + 100^2 [/tex]

[tex] = D= sqrt*(10 + 100) [/tex]

Lets use 2DD' = 2xx' + 2yy'

Differentiating with respect to t we have:

D•d(D)/dt = -(10)•dx/dt + 100•dy/dt

=100.5 d(D)/dt = (-10)•35 + (100)•25

When t=4, we have x=(140-150) =10 and y=100

[tex]= D = sqrt*(10^2 + 100^2) [/tex]

=100.5

= 100.5 dD/dt = 10.35 +100.25

= dD/dt = 21.29km/h

The distance between Ship A and Ship B is changing at approximately 21.4 km/h at 6 P.M.

Here, we use related rates. Let's define the positions of the ships -

Ship A's position at 6 P.M. :

Ship B's position at 6 P.M. :

Let’s denote the distance between the two ships at time t as D, x as the east-west distance (positive east) between Ship A and Ship B, and y as the north-south distance (positive north) from the initial position of Ship B.

Given:

We use the Pythagorean theorem to relate x, y, and D:

Differentiating both sides with respect to t:

Simplifying and solving for [tex]\frac{dD}{dt}[/tex] gives:

Plugging in the values:

Therefore, the distance between the ships is changing at approximately 21.4 km/h at 6 P.M.

Answer:

Incomplete question

Check attachment for the diagram of the problem.

Explanation:

The acceleration of the car A is given as

a=3ft/s²

Car B is rounding a curve of radius

r=440ft

Car B is moving at constant speed of Vb=30mi/hr.

Car A reach a speed of 45mi/hr

Note, 1 mile = 5280ft

And 1 hour= 3600s

Then

Va=45mi/hr=45×5280/3600

Va=66ft/s

Also,

Vb=30mi/hour=30×5280/3600

Vb=44ft/s

Now,

a. Let write the relative velocity of car B, relative to car A

Vb = Va + Vb/a

Then,

Using triangle rule, because vectors cannot be added automatically

Vb/a²= Vb²+Va²-2Va•VbCosθ

From the given graphical question the angle between Va and Vb is 60°.

Vb/a²=44²+66² - 2•44•66Cos60

Vb/a²=1936+ 4356 - 5808Cos60

Vb/a² = 3388

Vb/a = √3388

Vb/a = 58.21 ft/s

The direction is given as

Using Sine Rule

a/SinA = b/SinB = c/SinC

i.e.

Va/SinA = Vb/SinB = (Vb/a)/SinC

66/SinA = 44/SinB = 58.21/Sin60

Then, to get B

44/SinB = 58.21/Sin60

44Sin60/58.21  = SinB

0.6546 = SinB

B=arcsin(0.6546)

B=40.89°

b. The acceleration of Car B due to Car A.

Let write the relative acceleration  of car B, relative to car A.

Let Aa be acceleration of car A

Ab be the acceleration of car B.

Ab = Aa + Ab/a

Given the acceleration of car A

Aa=3ft/s²

Then to get the acceleration of car B, using the tangential acceleration formular

a = v²/r

Ab = Vb²/r

Ab = 44²/440

Ab = 4.4ft/s²

Using cosine rule again as above

Ab/a²= Aa²+Ab² - 2•Aa•Ab•Cosθ

Ab/a²= 3²+4.4²- 2•3•4.4•Cos30

Ab/a²= 9+19.36 - 22.863

Ab/a² = 5.497

Ab/a = √5.497

Ab/a = 2.34ft/s²

To get the direction using Sine rule again, as done above

Using Sine Rule

a/SinA = b/SinB = c/SinC

i.e.

Aa/SinA = Ab/SinB = (Ab/a)/SinC

3/SinA = 4.4/SinB = 2.34/Sin30

Then, to get B

4.4/SinB = 2.34/Sin30

4.4Sin30/2.34 = SinB

0.9402 = SinB

B=arcsin(0.9402)

B=70.1°

Since B is obtuse, the other solution for Sine is given as

B= nπ - θ.   , when n=1

B=180-70.1

B=109.92°

To determine the velocity and acceleration which car B appears to have to an observer in car A, we need to consider the relative motion between the two cars. The velocity of car B as observed by the observer in car A is approximately 29955/176 ft/sec. The acceleration of car B as observed by the observer in car A is approximately 1/23966164627200 mi^2/s^2.

To determine the velocity and acceleration which car B appears to have to an observer in car A, we need to consider the relative motion between the two cars. Car B is rounding a curve at a constant speed, so its velocity remains constant. However, the observer in car A will perceive car B as having a different velocity and acceleration. The velocity of car B to the observer in car A will depend on the relative motion between the two cars, while the acceleration of car B to the observer in car A will depend on the change in direction of car B's motion.

Let's calculate the velocity and acceleration of car B as observed by an observer in car A:

Velocity: Since car B is rounding a curve with a radius of 440 ft and a constant speed of 30 mi/hr, we can use the formula v = rω to find the angular velocity ω. The angular velocity ω is equal to the speed divided by the radius, so ω = (30 mi/hr) / (440 ft) = (30 mi/hr) / (5280 ft/mi) / (440 ft) = 1/1760 rad/sec. The observer in car A will perceive car B's velocity as the vector sum of its actual velocity in the curve (tangent to the curve) and the observer's velocity in the direction of the curve (opposite to the centripetal force). Since car A has reached a speed of 45 mi/hr, its velocity can be converted to ft/sec as (45 mi/hr) / (5280 ft/mi) = 15/176 ft/sec. Therefore, the velocity of car B as observed by the observer in car A will be (30 mi/hr) + (15/176 ft/sec) = (660/22 ft/sec) + (15/176 ft/sec) = (660/22 + 15/176) ft/sec = (29955/176) ft/sec.

Acceleration: Since car B is rounding a curve at a constant speed, its acceleration is directed towards the center of the curve and has a magnitude of v^2 / r, where v is the velocity and r is the radius. Substituting the values, we get the acceleration as (30 mi/hr)^2 / (440 ft) = ((30 mi/hr)^2) / ((5280 ft/mi) / (440 ft)) = (900 mi^2/hr^2) / (5280 ft/mi) * (440 ft) = (900 mi^2 * ft^2) / (5280 hr^2) * (440) ft = (900 * 5280 * 440) ft^2 / hr^2 = (2119680000/5280) ft^2 / hr^2 = (400800 ft^2/hr^2) = (400800 ft^2/hr^2) * (1/3600 hr^2/s^2) * (1 mi^2 / (5280 ft)^2) = (400800 / 3600) * (1/5280)^2 mi^2/s^2 = (111/9900) * (1/5280)^2 mi^2/s^2 = (11/990) * (1/5280)^2 mi^2/s^2 = (11/990) * (1/5280)^2 mi^2/s^2 = (1/266611200) mi^2/s^2 = (1/266611200) * (5280 ft/mi)^2 = (1/266611200) * 5280^2 ft^2/s^2 = (1/266611200) * 13939200 ft^2/s^2 = (1/266611200) * 13939200 ft^2/s^2 = (1/19) ft^2/s^2 = (1/19) * (1/5280)^2 mi^2/s^2 = (1/19) * (1/13939200) mi^2/s^2 = (1/19) * (1/13939200) mi^2/s^2 = (1/26268580800) mi^2/s^2 = (1/26268580800) * (5280 ft/mi)^2 = (1/26268580800) * 5280^2 ft^2/s^2 = (1/26268580800) * 13939200 ft^2/s^2 = (1/26268580800) * 13939200 ft^2/s^2 = (1/237896) ft^2/s^2 = (1/237896) * (1/5280)^2 mi^2/s^2 = (1/237896) * (1/13939200) mi^2/s^2 = (1/237896) * (1/13939200) mi^2/s^2 = (1/23966164627200) mi^2/s^2.

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Answer:

r = 0.1 Ω

Explanation:

We will use Ohm's Law in this question: V = IR, where I is the current and R is the resistance.

When the switch is open, the voltmeter reads 1.52 V, and there is no current in the circuit. We can deduce that the internal resistor in the battery causes 0.15 V to dissipate into heat, since the voltmeter reads 1.37 V when the switch is closed.

[tex]0.15 = (1.5)r\\r = 0.1~\Omega[/tex]

Answer:

Explanation:

Intensity of unpolarised light = I₀

intensity after passing through first polariser =  I₀ / 2

Angle between first and second polariser is  φ so

intensity after passing through second polariser

= (I₀ / 2) cos²φ

Now angle between second and third polariser

= 90 - φ

intensity after passing though third polariser

= (I₀ / 2) cos²φ cos²( 90 - φ)

= (I₀ / 2) cos²φ sin²φ

= (I₀ / 8) 4cos²φ sin²φ

= (I₀ / 8) sin²2φ

The intensity after the third polarizer will be:

I₃ = (I₀/8)*sin^2(2φ)

For non-polarized light that passes through any polarizer, we say that the intensity is reduced to its half.

Original intensity = I₀

After the first polarizer, the intensity will be:

I₁ = I₀/2.

Now, when it passes through a polarizer such that the difference in angles with the polarization is x, the new intensity will be:

I₂ = I₁*cos^2(x).

The angle between the second and the first polarizer is φ, then we have:

I₂ = (I₀/2)*cos^2(φ).

Now we also know that the first and the last polarizer are crossed (so there is an angle of 90°). Then if we define θ as the angle between the second and the third polarizer, we will have that:

φ + θ = 90°

then:

θ = 90° - φ

The intensity after the third polarizer will be:

I₃ = (I₀/2)*cos^2(φ)*cos^2(90° - φ)

And we know that:

cos(90° - φ) = sin(φ)

Then we can rewrite:

I₃ = (I₀/2)*cos^2(φ)*sin^2(φ)

But we want a single trigonometric function, then we use the relation:

cos^2(φ)*sin^2(φ) = sin^2(2φ)/4

And replacing that, we get:

I₃ = (I₀/2)*sin^2(2φ)/4 = (I₀/8)*sin^2(2φ)

If you want to learn more about light, you can read:

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The speed of the ball after it has traveled 4.20 m downward is approximately 24.04 m/s.

To calculate the speed of the ball after it has traveled 4.20 m downward, we need to use the principles of free fall and the equations of motion. Since the ball is dropped from a height of 55.0 m, we can calculate the initial velocity using the equation v_i = sqrt(2 * g * h), where v_i is the initial velocity, g is the acceleration due to gravity (9.8 m/s^2), and h is the height (55.0 m). Plugging in these values, we find that the initial velocity is approximately 34.02 m/s.

Next, we can calculate the final velocity using the equation v_f = sqrt(v_i^2 + 2 * g * d), where v_f is the final velocity, v_i is the initial velocity, g is the acceleration due to gravity, and d is the distance traveled downward (4.20 m). Plugging in the values, we get v_f = sqrt((34.02 m/s)^2 + 2 * (9.8 m/s^2) * (4.20 m)) = approximately 24.04 m/s.

Therefore, the speed of the ball after it has traveled 4.20 m downward is approximately 24.04 m/s.

Answer:

[tex]v(1.5)=0.7648\ m/s[/tex]

Explanation:

Dynamics

When a particle of mass m is subject to a net force F, it moves at an acceleration given by

[tex]\displaystyle a=\frac{F}{m}[/tex]

The particle has a mass of m=2.37 Kg and the force is horizontal with a variable magnitude given by

[tex]F=2cos1.1t[/tex]

The variable acceleration is calculated by:

[tex]\displaystyle a=\frac{F}{m}=\frac{2cos1.1t}{2.37}[/tex]

[tex]a=0.8439cos1.1t[/tex]

The instant velocity is the integral of the acceleration:

[tex]\displaystyle v(t)=\int_{t_o}^{t_1}a.dt[/tex]

[tex]\displaystyle v(t)=\int_{0}^{1.5}0.8439cos1.1t.dt[/tex]

Integrating

[tex]\displaystyle v(1.5)=0.7672sin1.1t \left |_0^{1.5}[/tex]

[tex]\displaystyle v(1.5)=0.7672(sin1.1\cdot 1.5-sin1.1\cdot 0)[/tex]

[tex]\boxed{v(1.5)=0.7648\ m/s}[/tex]

Final answer:

To find the x-component of velocity at t= 1.5 seconds for a particle under a sinusoidally varying force, we derive the equation of motion, integrate the acceleration, and calculate the velocity, resulting in [tex]v_x[/tex] = 0.567 m/s.

Explanation:

The question pertains to finding the x-component of velocity (vx) of a particle at t= 1.5 seconds, given the mass and the sinusoidally varying force. Since the force applied on the particle varies as Fx = F0cos(ωt), where F0 = 2 N and ω = 1.1 rad/s, we can find the acceleration and then integrate it with respect to time to find velocity. The acceleration ax is given by Fx/m = (2cos(1.1t))/2.37. To find the change in velocity, we integrate ax with respect to time, giving us vx = ∫ ax dt = ∫ (2cos(1.1t))/2.37 dt. Evaluating this integral from 0 to 1.5 s, we use the definite integral which simplifies to vx(t) = (2/2.37)(sin(1.1(1.5))-sin(1.1(0)))/1.1. The calculation yields vx = 0.567 m/s, indicating the particle's velocity in the x-direction at 1.5 seconds.

Answer:

Explanation:

Force needed to apply start the box is greater than the force needed to keep it moving because static friction is greater than the kinetic friction .

A  threshold force is needed to move the box and when box started to move kinetic friction comes into play.      

Friction force is directly related to the weight of the box as the friction force is

coefficient of friction time Normal reaction .

And Normal reaction is equal to the weight of box if no force is applied.

[tex]f_r=\mu N[/tex]

[tex]N=mg[/tex]

The force to start moving a box is greater than to keep it moving due to static and kinetic friction. The force of friction is directly proportional to the weight of the box; the heavier the box, the greater the friction.

In physics, the force required to start moving an object is often greater than the force required to keep it moving. This is due to a concept known as friction, particularly, static friction and kinetic friction. Static friction is the force that resists the initiation of sliding motion, and it's usually greater than kinetic friction, which is the force that opposes motion of an object when the object is in motion.

Now, relating friction to the weight of the box, the weight of an object is equal to its mass times the acceleration due to gravity, and it's directly proportional to the force of friction. In essence, the heavier the box is, the more the static and kinetic friction that you have to overcome to move and keep it moving.

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