Old cannons were built on wheeled carts, both to facilitate moving the cannon and to allow the cannon to recoil when fired. When a 150 kg cannon and cart recoils at 1.5 m/s, at what velocity would a 10 kg cannonball leave the cannon?

Answer:

Explanation:

total mass, M = 15.8 kg

initial velocity, u = 0 m/s

m1 = 4.5 kg

m2 = 5.4 kg

v1 = 27.5 m/s at an angle 18°

v2 = 21.4 m/s at an angle 23°

As there is no external force acts on the system, so the moemntum of the system is conserved.

Momentum before collision = momentum after collision

as the momentum before collision is zero, so the momentum after collision is also zero.

The given question is incomplete. The complete question is as follows.

The block has a weight of 75 lb and rests on the floor for which [tex]\mu k[/tex] = 0.4. The motor draws in the cable at a constant rate of 6 ft/sft/s. Neglect the mass of the cable and pulleys.

Determine the output of the motor at the instant [tex]\theta = 30^{o}[/tex].

Explanation:

We will consider that equilibrium condition in vertical direction is as follows.

           [tex]\sum F_{y} = 0[/tex]

         N - W = 0

           N = W

or,      N = 75 lb

Again, equilibrium condition in the vertical direction is  as follows.

        [tex]\sum F_{x} = 0[/tex]

       [tex]T_{2} - F_{k}[/tex] = 0

         [tex]T_{2} = \mu_{k} N[/tex]

                  = [tex]0.4 \times 75 lb[/tex]

                  = 30 lb

Now, the equilibrium equation in the horizontal direction is as follows.

         [tex]\sum F_{x} = 0[/tex]

       [tex]T Cos (30^{o}) + T Cos (30^{o}) = T_{2}[/tex]

          [tex]2T Cos (30^{o}) = T_{2}[/tex]

    or,             T = [tex]\frac{T_{2}}{2 Cos (30^{o})}[/tex]

                        = [tex]\frac{30}{2 Cos (30^{o})}[/tex]

                        = [tex]\frac{30}{1.732}[/tex]

                        = 17.32 lb

Now, we will calculate the output power of the motor as follows.

             P = Tv

                = [tex]17.32 lb \times 6[/tex]

                = [tex]103.92 \times \frac{1}{550} \times \frac{hp}{ft/s}[/tex]

                = 0.189 hp

or,             = 0.2 hp

Thus, we can conclude that output of the given motor is 0.2 hp.

Answer:

The out put power is 0.188 hp.

Explanation:

Given that,

Weight = 75 lb

Coefficient of friction = 0.4

Rate = 6 ft/s

Suppose, Determine the output of the motor at the instant θ = 30°.

For block,

We need to calculate the force in vertical direction

Using balance equilibrium equation in vertical

[tex]\sum{F_{y}}=0[/tex]

[tex]N-W=0[/tex]

[tex]N=W[/tex]

Put the value into the formula

[tex]N=75\ lb[/tex]

Using balance equilibrium equation in horizontal

[tex]\sum{F_{x}}=0[/tex]

[tex]T_{2}-f_{k}=0[/tex]

[tex]T_{2}=\mu_{k}N[/tex]

Put the value into the formula

[tex]T_{2}=0.4\times75[/tex]

[tex]T_{2}=30\ lb[/tex]

For pulley,

We need to calculate the force

Using balance equilibrium equation in horizontal

[tex]\sum{F_{x}}=0[/tex]

[tex]T\cos\theta+T\cos\theta=T_{2}[/tex]

[tex]2T\cos30=T_{2}[/tex]

[tex]T=\dfrac{T_{2}}{2\cos30}[/tex]

Put the value into the formula

[tex]T=\dfrac{30}{2\times\cos30}[/tex]

[tex]T=17.32\ lb[/tex]

We need to calculate the out put power

Using formula of power

[tex]P=Tv[/tex]

Put the value into the formula

[tex]P=17.32\times6[/tex]

[tex]P=103.92\ lb.ft/s[/tex]

[tex]P=0.188\ hp[/tex]

Hence, The out put power is 0.188 hp.

Mass of the block is 0.557 kg

Solution:

The quantity of heat (q)  liberated during any process is equal to the product of the mass of the block (m), specific heat (C) and the change in temperature (ΔT). Specific heat of lead is 130 J/kg °C.

We have to rearrange the equation to find the mass of the block of lead as,

[tex]\begin{array}{l}\boldsymbol{q}=\boldsymbol{m} \times \boldsymbol{C} \times \boldsymbol{\Delta} \mathbf{T}\\ \\\boldsymbol{m}=\frac{\mathbf{q}}{\boldsymbol{c} \times \mathbf{\Delta} \mathbf{T}}\\ \\\mathbf{m}=\frac{427 \mathbf{J}}{13 \mathbf{0}_{\mathbf{4} *}^{\prime} \mathbf{r} \times \mathbf{s} \mathbf{9} \mathbf{0}^{*} \mathbf{c}}=\mathbf{0} . \mathbf{5} \mathbf{5} \mathbf{7} \mathbf{k} \mathbf{g}\end{array}[/tex]

Thus the mass (m) = 0.577 kg

Answer:

the correct answer is 0.57

Explanation:

Answer:

Concentration in mg/m^3 = 3845.39mg/m^3

in PPM =  3.84539 ppm

Explanation:

From the dimensions, we compute the volume (V)

V = 10 x 40 x 30 = 12000ft^3

Volatilize volume = 1/4 *12000 = 3000ft^3 at NTP

Volume of pyridine in the room = ( 12000 - 3000) = 9000ft^3

converting to meters = 9000 * (0.3048)^3m^3

= 254. 85m^3

Computing concentration(C) =  mass/s.g/volume

S.g converted to mg per meter cube = 0.98x1000= 0.98 * 10^3g/m^3 = 0.98 * 10^6mg/m^3

Hence concentration =  0.98 * 10^6/254.85 = 3845.39 mg/m^3

Parts per million denotes low concentration of a solution

1ppm = 1mg per liter , and 1 liter = 0.001m^3

Hence, this  =  3845.39/1000 = 3.84539 PPM

Answer:

Explanation:

capacitance of parallel plate capacitor

c = ε A / d , d is distance between plates , A is surface area , ε is constant

As d becomes two times , Capacitance c = 1/ 2 times ie c / 2

potential V = Q / C

Q is constant , potential

v = Q / c /2

= 2 . Q / C

= 2 V

So potential difference becomes 2 times.

NEW P D = 398 X 2

= 796 V.

Answer:

1)   v = 2.20 10⁵ m / s , 2)  r = 4.86 10⁻³ m,    r = 4.92 10⁻³ m

Explanation:

1) A speed selector is a section where the magnetic and electrical forces have opposite directions, so

           [tex]F_{m} = F_{e}[/tex]

           q v B = q E

           v = E / B

           V = E s

           E = V / s

            v = V / s B

            v = 148 / (0.0016 0.42)

            v = 2.20 10⁵ m / s

2) when the isotopes enter the spectrometer, we can use Newton's second law

             F = m a

Acceleration is centripetal

             a = v² / r

             q v B = m v² / r

              r = m v / qB

The mass of uranium 235 is

           m = 3.903 10⁻²⁵ kg

The radius of this isotope is

             r = 3,903 10⁻²⁵ 2.20 10⁵ / (92  1.6 10⁻¹⁹  1.2)

             r = 4.86 10⁻³ m

The mass of the uranium isotope 238 is

              m = 238 a = 238 1.66 10-27 = 395.08 10-27 kg

The radius is

             r = 395.08 10⁻²⁷ 2.20 10⁵ / (92 1.6 10⁻¹⁹  1.2)

             r = 4.92 10⁻³ m

Final answer:

The speed of the ions entering the mass spectrometer is 352.38 m/s. The difference in the diameters of the orbits of the 238U and 235U ions can be calculated using the formula for the radius of a charged particle's orbit in a magnetic field.

Explanation:

To find the speed of the ions entering the mass spectrometer, we can use the equation for the force on a charged particle in a magnetic field: F = qvB, where F is the force, q is the charge, v is the velocity, and B is the magnetic field. The force is equal to the electric force, qE, so we can set these two equations equal to each other and solve for v: qvB = qE. Since q is the charge of the ion and E is the electric field strength, we can rearrange this equation to solve for v, giving us v = E/B.

Plugging in the values given in the question, we find that the speed of the ions entering the mass spectrometer is 148 V / 0.42 T = 352.38 m/s.

To find the difference in the diameters of the orbits of the 238U and 235U ions, we can use the formula for the radius of a charged particle's orbit in a magnetic field: r = mv / (qB), where m is the mass of the ion, v is its velocity, q is its charge, and B is the magnetic field. Since the ions are singly ionized, their charges are +e, where e is the charge of an electron. Plugging in the values given in the question, we can calculate the radius of the orbits of the 238U and 235U ions, and then subtract these values to find the difference in their diameters.

Answer:

159.1 ton

Explanation:

The solution is shown in the attached file

Answer:

The safe load is 159 ton

Explanation:

The safe load is equal to:

[tex]L=\frac{WHBV^{2/3} }{K}[/tex]

Where:

W = weight of hammer = 2.5 ton

H = drop height = 22 ft

B = number of blows used to drive the last batch = 36/4 = 9 ft³

K = dimensionless constant = 28

V = uncompacted volume = 27 ft³

Replacing values:

[tex]L=\frac{2.5*22*9*(27^{2/3}) }{28} =159ton[/tex]

Answer:

a. 20m/s

b.50N

c. Turkey has a larger mass than the ball. Neglible final acceleration and therefore remains stationery.

Explanation:

a. Given the force as 50N, times as 0.2seconds and the weight of the ball as 0.5 kg, it's final velocity can be calculated as:

[tex]F\bigtriangleup t=m\bigtriangleup v\\\\50N\times 0.2s=0.5kg\times \bigtriangleup v\\\\\bigtriangleup v=2(50N\times0.2)\\\\=20m/s[/tex]

Hence, the velocity of the ball after the kick is 20m/s

b.The force felt by the turkey:

#Applying Newton's 3rd Law of motion, opposite and equal reaction:

-The turkey felt a force of 50N but in the opposite direction to the same force felt by the ball.

c. Using the law of momentum conservation:

-Due to ther external forces exerted on the turkey, it remains stationery.

-The turkey has a larger mass than the ball. It will therefore have a negligible acceleration if any and thus remains stationery.

-Momentum is not conserved due to these external forces.

The velocity of the ball after the kick is 200 m/s. The turkey feels a force of 50N. The turkey doesn't go flying backward because of its greater mass compared to the ball.

The velocity of the ball after the kick can be calculated using the equation:

velocity = force / mass * time

Substituting the values, we get:

velocity = 50N / 0.5kg * 0.2s = 200 m/s

The force that the turkey feels can be determined using Newton's third law, which states that for every action, there is an equal and opposite reaction. So, the force the turkey feels will be the same as the force of the kick, which is 50N.

The turkey doesn't go flying backward because the turkey has a greater mass than the ball. According to Newton's second law, the acceleration of an object is inversely proportional to its mass. Since the turkey has a greater mass, it experiences less acceleration compared to the ball. Conservation of momentum is upheld in this situation, as the momentum of the turkey and ball together remains the same before and after the kick.

Answer:

#Check explanation below for a detailed description.

Explanation:

A plot of absorbance vs temperature is used to determine  [tex]T_m[/tex] of a duplex DNA.

-The duplex DNA denatures.

-The denaturation creates two single strands which enhances Ultraviolet absorption ( increased temperatures increases absorption rates).

-Higher concentrations of [tex]Sodium \ Chloride[/tex] hampers stability of the DNA.

Explanation:

A trapeze is rotating with 1 rotation per second .

Thus its angular velocity ω = 2π n

here n is the number of rotations per second

Thus ω = 2π b because n = 1 in this case

Suppose the moment of inertia of his is = I

Then angular momentum L₁ = I  ω = 2 I π

In the second case , the moment of inertia becomes = 0.4 I

Let his angular velocity is  ω₀

Thus angular momentum L₂ = 0.4 I  ω₀

Because no external torque is applied , therefore angular momentum will remain constant .

Thus L₁ = L₂

Therefore  2 I π = 0.4 I x 2 n₀ π

here n₀ is the number of rotations per second

n₀ = [tex]\frac{1}{0.4}[/tex] = [tex]\frac{5}{2}[/tex] = 2.5

Answer:

|Ax| =1.03 m  (directed towards negative x-axis)

|Ay|= 5.30 (directed towards negative y-axis)

Explanation:

Let A is a vector = 5.4 m

θ = 259°

to Find Ax, Ay

Sol:

according the condition it lies in 3rd quadrant

we know that Horizontal Component Ax = A cos θ

Ax = 5.4 Cos 259°

Ax = - 1.03 m

|Ax| =1.03 m  (directed towards negative x-axis)

Now Ay = A sin θ

Ay = 5.4 Sin 259°

Ay = -5.30

|Ay|= 5.30 (directed towards negative y-axis)

(a) -1.030m

(b) -5.301m

Given a vector F in the xy plane, of magnitude F and in a direction θ counterclockwise from the positive direction of the x-axis;

The x-component ([tex]F_{X}[/tex]) of vector F is given by;

[tex]F_{X}[/tex] = F cos θ    ---------------------(i)

And;

The y-component ([tex]F_{Y}[/tex]) of vector F is given by;

[tex]F_{Y}[/tex] = F sin θ            -----------------------(ii)

Now to the question;

Let the vector be A

Therefore;

The magnitude of vector A is A = 5.4m

The direction θ of A counterclockwise from the positive direction of the x-axis = 259°

(a) The x-component ([tex]A_{X}[/tex]) of the vector A is therefore given by;

[tex]A_{X}[/tex] = A cos θ       ------------------------(iii)

Substitute the values of θ and A into equation (iii) as follows;

[tex]A_{X}[/tex] = 5.4 cos 259°

[tex]A_{X}[/tex] = 5.4 x (-0.1908)

[tex]A_{X}[/tex] = -1.030

Therefore, the x-component of the vector is -1.030m

(b) The y-component ([tex]A_{Y}[/tex]) of the vector A is therefore given by;

[tex]A_{Y}[/tex] = A sin θ       ------------------------(iv)

Substitute the values of θ and A into equation (iv) as follows;

[tex]A_{Y}[/tex] = 5.4 sin 259°

[tex]A_{Y}[/tex] = 5.4 x (-0.9816)

[tex]A_{Y}[/tex] = -5.301m

Therefore, the y-component of the vector is -5.301m

Option d is correct; relativity formulas are not needed because the Lorentz factor (γ-1) is much less than 1, indicating that the electron's velocity is non-relativistic after being accelerated through a potential difference of 0.200 kV.

We need to determine whether we should use relativity formulas for the electrons accelerated through a potential difference. When electrons gain kinetic energy (K) by acceleration through a potential difference (V), the energy they gain can be expressed as Ve=K=(γ-1)mc2. Here, γ stands for Lorentz factor, m is the mass of electron, and c is the speed of light. Given that the potential difference ('V') is 0.200 kV, and the rest mass energy ('Eo') of an electron is 0.511 MeV, we calculate:

γ-1= (0.200 keV) / (0.511 MeV) ≈ 0.3916 × 10-3, which implies that γ is approximately 1. So, γ-1 << 1, which in turn means that the electron's velocity is much less than the speed of light, and the relativistic effect can be neglected. Thus, we do not need to use special relativity.

The correct answer is d.

Answer:

Series circuit approximation = R1 >> R2 then R2 ≈ 0

Parallel circuit approximation = R1 << R2 then R2 ≈ 0

Explanation:

in a series circuit, if there is a large difference between two resistors then we can omit the resistor which has low resistance because the higher value resistor dominates.

R1 >> R2 then R2 ≈ 0

in a parallel circuit, if there is a large difference between two resistors then we can omit the resistor which has high resistance because the lower value resistor dominates.

R1 << R2 then R2 ≈ 0

Answer:

The pebble reaches a height of 11.716 m above the starting point.

Explanation:

Given:

Mass of pebble (m) = 50.3 g = 0.0503 kg [1 g = 0.001 kg]

Spring constant (k) = 320.0 N/m

Elongation of catapult (x) = 0.190 m

Height of fly (h) = ?

Air resistance is ignored. So, conservation of energy holds true.

The energy stored in the catapult on stretching it is elastic potential energy. This elastic potential energy is transferred to the pebble in the form of gravitational potential energy. Therefore,

Elastic potential energy = Gravitational potential energy

[tex]\frac{1}{2}kx^2=mgh\\\\h=\frac{kx^2}{2mg}[/tex]

Plug in the given values and solve for 'h'. This gives,

[tex]h=\frac{(320.0\ N/m)(0.190\ m)^2}{2(0.0503\ kg)(9.8\ m/s^2)}\\\\h=\frac{11.552\ Nm}{0.986\ N}\\\\h=11.716\ m[/tex]

Therefore, the pebble reaches a height of 11.716 m above the starting point.

Acceleration of a speck is 0.77 m/s²

Explanation:

Given of the solution-

Diameter (We can represent as d) = 32 cm

radius, (We can represent as r )= 32/2 = 16 cm = 0.16 m

Angular acceleration,(We can represent as  α ) = 46 rpm

[tex]\alpha = 46 * \frac{2\pi }{60}[/tex]

Acceleration, a = ?

We know that the formula for the acceleration is

[tex]a = r * \alpha[/tex]

[tex]a = 0.16 * 46 * \frac{2\pi }{60}\\ \\a = 0.16 * 46 * \frac{2 * 3.14}{60}[/tex]

[tex]a = 0.77m/s^2[/tex]

Therefore, acceleration of a speck is 0.77 m/s²

Answer:

When light goes from one material into another material having a HIGHER index of refraction, its speed and wavelength decrease, but its frequency stays the same.

The correct option is E

Explanation:

Refraction is the bending of light ray as it crosses the boundary between the two media of different densities, thus causing a change in it's direction.

When light goes from one material of lower refractive to another material of higher refractive index, the speed of light reduces, because from law of refraction

aNg= (speed of light in air)/(speed of light in glass)

Snell's Law

So generally,

n1/n2 = v2/v1

Let n1 be incident index

Let n2 be refracted index

v2 is refracted speed

v1 is incident speed

Since given that, n1<n2

Then, the right side of the equation will always be less than 1 i.e n1/n2

Then, v2=v1• ( n1/n2)

Since n1/n2 is less that one this show that the speed v2 will reduce.

Since we know that the speed decrease,

We also know that speed, wavelength and speed is related by

v=fλ

Since, the speed is directly proportional to wavelength this shows that as the speed reduces the wavelength also reduces where frequency is the constant of proportionality

v∝λ. .

then f is constant of proportional

v=fλ

So as speed reduces, the wavelength reduces.

frequency of a light wave does not depend on the medium, while wavelength and speed do.

its speed and wavelength decrease, but its frequency stays the same.

Then the answer is E

Answer:

3.6μF

Explanation:

The charge on the capacitor is defined by the formula

q = CV

because the charge will be conserved

q₁ = C₁V₂

q₂ = C₂V₂ where C₂ V₂ represent the charge on the newly connected capacitor  and the voltage drop across the two capacitor will be the same

q = q₁ + q₂ = C₁V₂ + C₂V₂

CV = CV₂ + C₂V₂

CV - CV₂ = C₂V₂

C ( V - V₂) = C₂V₂

C ( V/ V₂ - V₂ /V₂) = C₂

C₂ = 0.9 ( 10 /2) - 1) = 0.9( 5 - 1) = 3.6μF

Final answer:

The back emf of the motor decreases due to reduced speed caused by dirt in the shaft, which, in turn, causes the current drawn from the wall socket to increase.

Explanation:

When dirt prevents the shaft of an electric motor from turning rapidly, the back emf produced by the motor decreases. Since the back emf is proportional to the motor's angular velocity, any decrease in speed results in a decrease in back emf. Because the back emf opposes the applied voltage from the wall socket, a decrease in back emf implies that the voltage across the motor's coil will increase, causing the motor to draw a larger current from the socket. Therefore, the current that the motor draws from the wall socket will increase. The correct answer to the question is (c) The back emf decreases, and the current drawn from the socket increases.

Answer with Explanation:

We are given that

Number of turns=n=169

Radius of coil=r=2.6 cm=[tex]2.6\times 10^{-2} m[/tex]

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

Area of circular coil=[tex]\pi r^2[/tex]

Where [tex]\pi=3.14[/tex]

Using the formula

Area of circular coil=[tex]3.14\times (2.6\times 10^{-2})^2=21.2\times 10^{-4}m^2[/tex]

a.Magnetic dipole  moment=[tex]\mu=3.4 Am^2[/tex]

[tex]I=\frac{\mu}{nA}[/tex]

Using the formula

[tex]I=\frac{3.4}{169\times 21.2\times 10^{-4}}[/tex]

[tex]I=9.5 A[/tex]

b.Magnetic field,  B=0.03 T

[tex]\tau_{max}=\mu B[/tex]

Substitute the value

[tex]\tau_{max}=3.4\times \times 0.03=0.102 Nm[/tex]

c.[tex]\theta=44.9^{\circ}[/tex]

[tex]\tau=\tau_{max}sin\theta[/tex]

Substitute the values

[tex]\tau=0.102sin44.9=0.07 Nm[/tex]

Final answer:

The skier is moving at 11.1 m/s when she gets to the bottom of the hill. This solution is derived using the principles of energy conservation and work done by friction.

Explanation:

A 62.0 kg skier is moving at 6.90 m/s on a frictionless, horizontal, snow-covered plateau when she encounters a rough patch 4.50 m long. The coefficient of kinetic friction between this patch and her skis is 0.300. After crossing the rough patch and returning to friction-free snow, she skis down an icy, frictionless hill 2.50 m high. To determine how fast the skier is moving when she gets to the bottom of the hill, we analyze the problem using principles of energy conservation and work done by friction.

We start with the equation that equates the final kinetic and potential energies to the initial energies along with the work done by friction. This equation is 0.5 mv² + 0 = 0.5 mu² + μmgl + mgh, where v is the final velocity, μ is the coefficient of friction, g is the acceleration due to gravity (9.8 m/s²), l is the length of the rough patch, and h is the height of the hill.

Plugging in the given values,

we have: 0.5 v² = 0.5 x 6.9 x 6.9 + 0.3 x 9.8 x 4.5 + 9.8 x 2.50. S

implifying, we get v² = 123.07, which leads to v = 11.1 m/s.

Thus, the skier's speed at the bottom of the hill is 11.1 m/s.

Answer:

(a)

[tex]f=20Hz\\And \\w=125.7s^{-1}[/tex]

(b)

[tex]K=3.490m^{-1}[/tex]

Explanation:

We know that the speed of any periodic wave is given by:

v=f×λ

The wave number k is given by:

K=2π/λ

Given data

Amplitude A=2.50mm

Wavelength λ=1.80m

Speed v=36 m/s

For Part (a)

For the wave frequency we plug our values for v and λ.So we get:

v=f×λ

[tex]36.0m/s=(1.80m)f\\f=\frac{36.0m/s}{1.80m}\\ f=20Hz\\[/tex]

And the angular speed we plug our value for f so we get:

[tex]w=2\pi f\\w=2\pi (20)\\w=125.7s^{-1}[/tex]

For Part (b)

For wave number we plug the value for λ.So we get

K=2π/λ

[tex]K=\frac{2\pi }{1.80m}\\ K=3.490m^{-1}[/tex]

The answers for this question are

a.) frequency of the wave is 20 Hz.

b.) angular frequency of the wave is [tex]125.71428\ s^{-1}[/tex]

c.) wave number of the wave is [tex]k=3.4920\ m^{-1}[/tex].

Given to us:

Amplitude A= 2.50 mm,

Wavelength λ= 1.80 m,

Velocity V= 36.0 m/sec,

a.) To find out frequency and angular frequency of the wave,we know

[tex]frequency=\dfrac{Velocity}{Wavelength}[/tex]

[tex]f=\dfrac{V}{\lambda}[/tex]

Putting the numerical value,

[tex]f=\dfrac{36}{1.80}\\\\f= 20\ Hz[/tex]

Therefore, the frequency of this wave is 20 Hz.

Also, for angular frequency

[tex]f=2\pi \omega[/tex]

Putting the numerical value,

[tex]\omega=2\pi f \\\\\omega=2\times\pi \times 20\\\\\omega= 125.71428\ s^{-1}[/tex]

Therefore, the angular frequency of this wave is [tex]125.71428\ s^{-1}[/tex].

b.) To find out the wave number of the wave (k),

The wave number for an EM field is equal to 2π divided by the wavelength(λ) in meters.

[tex]k=\dfrac{2\pi}{\lambda}[/tex]

Putting the numerical value,

[tex]k=\dfrac{2\pi}{\lambda}\\\\k=\dfrac{2\pi}{1.80}\\\\k=3.4920\ m^{-1}[/tex]

Therefore, the wave number of the wave is [tex]k=3.4920\ m^{-1}[/tex].

Hence, for this wave

a.) frequency of the wave is 20 Hz.

b.) angular frequency of the wave is [tex]125.71428\ s^{-1}[/tex]

c.) wave number of the wave is [tex]k=3.4920\ m^{-1}[/tex].

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Answer:

The horizontal force that must be applied to the box to cause it to start sliding along the surface is 162N

Explanation:

To start sliding the box on the surface it must overcome its static frictional force  under equilibrium condition

The net force on the box is

F - fs = 0

F = fs = us N = us mg

Force = ( 0.3) x ( 55 kg) x ( 9.8 m/s^2) = 161.7 approximately 162 N

Horizontal force is defined as the forces that equal and opposite in the direction. The horizontal resultant force is always zero.

Given that:

Mass of box = 55 Kg

Coeeficient of Static friction = 0.30

Coefficient of kinetic friction = 0.20

The box if have to slide, then it would have to overcome the static force, such that:

F - Fs = 0

F = Fs

The force will be:

F = [tex]\rm \mu \times mass \times g[/tex]

F = [tex]0.3 \times 55 \times 9.8[/tex]

F = 161.7 Newton

Thus, the force required to slide the box is 161.7 Newton.

To know more about force, refer to the following link:

https://brainly.com/question/4456579

In the scenario described, increasing the height of a mass M in a vertical spring-mass system will result in both gravitational and spring potential energy. Thus, the potential energy in Case 2 is the sum of gravitational potential energy (m*g*D) and the elastic potential energy of the spring (1/2*k*D^2).

The situation described in the question deals with principles of Physics, specifically potential energy and its application to the spring-mass system. Consider both cases with the mass M hanging from a spring with a constant k in a vertical orientation. In the first case, the system is at equilibrium and thus the potential energy is defined as zero.

In the second case, the mass M is lifted upward a vertical distance D from the equilibrium position, it now possesses gravitational potential energy in addition to the elastic potential energy of the spring. This total energy is preserved as long as no external force behaves on the system. The gravitational potential energy gained by the mass is equal to m*g*D where m is the mass, g is the acceleration due to gravity and D is the vertical distance lifted.

The potential energy stored in the spring when it is stretched or compressed by a distance 'x' is given by the formula U = 1/2*k*x^2, where 'k' is the spring constant and 'x' is the displacement from the equilibrium position (in this case is D).

So, combining both energies, the total potential energy in Case 2, when the mass has been lifted a vertical distance D is m*g*D + 1/2*k*D^2.

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Answer:

2.2 m/s

Explanation:

solution:

To calculate change in stored energy at desired extension

ΔU = 1/2*k*(δx)^2

     = 1/2*3700*(0.37^2-0.180^2)

     = 201 N.m

use work energy theorem

ΔU = ΔK = 1/2*m*v^2 = 201

              = 2.2 m/s

note:

calculation maybe wrong but method is correct.

Explanation:

Below is an attachment containing the solution.

Answer:

Explanation:

in this case, flux and area vectors are parallel . therefore, flux is just the dot product of electric field and area vector.

(flux=EAcos0 =EA)

flux through the face x face

Physics homework question answer, step 1, image 1

the flux through -x face

Physics homework question answer, step 1, image1

Step 2

the flux through y face,

Physics homework question answer, step 2, image 2

the flux through -y face,

Physics homework question answer, step 2, image 2

the flux through the z faces are zero,

therefore, the net flux is,

Physics homework question answer, step 2, image 3

...

Final answer:

The total electric flux IS ΦE = aL2 + (a + bL)L2 + cL3 net charge is q = ε0 ( aL2 + (a + bL)L2 + cL3 )

Explanation:

The question asks to find the total electric flux ΦE through the surface of a cube with an electric field given by E = (a + bx)x + cy, and to determine the net charge q inside the cube.

Part A: Total Electric Flux through the Cube's Surface

To determine the electric flux through the cube, we need to use Gauss's Law, which relates the electric flux through a closed surface to the charge enclosed by the surface.

The electric flux ΦE through a surface is defined by the integral of the electric field E dotted with the differential area dA over the entire surface. Since the sides of the cube are parallel to the coordinate planes, the electric field components perpendicular to the x, y, and z faces will contribute to the flux.

In this case, only the x-component of the electric field (a + bx)x contributes to the flux through the surfaces perpendicular to the x-axis, which are at x = 0 and x = L.

The y-component cy of the electric field contributes to the flux through the surfaces perpendicular to the y-axis, which are at y = 0 and y = L. There is no z-component to the electric field, so the surfaces perpendicular to the z-axis will have no flux.

The total electric flux through the cube is thus:

ΦE = ∫( (a + bx) x + cy ) · dA

For the two faces perpendicular to the x-axis:

ΦE, x=0 = ∫( a x ) · dA = aL2 (at x = 0)

ΦE, x=L = ∫( (a + bL)x ) · dA = (a + bL)L2 (at x = L)

For the two faces perpendicular to the y-axis:

ΦE, y=0 = 0 (since cy is 0 at y = 0)

ΦE, y=L = ∫( cy ) · dA = cL3 (at y = L)

Summing these up:

ΦE = aL2 + (a + bL)L2 + cL3

Part C: Net Charge Inside the Cube

The net charge q inside the cube can be found using Gauss's Law, which states "the total electric flux out of a closed surface is equal to the charge enclosed divided by the permittivity of free space (epsilon_0)."

q = ΦE ε0

Substituting the expression for ΦE we have:

q = ε0 ( aL2 + (a + bL)L2 + cL3 )

Answer:

Explanation:

The ball from which electrons are transferred will acquire Q charge and the ball receiving electrons will acquire - Q charge . force of attraction between them

= k Q² / d²

9 x 10⁹ x Q² / .12² = 17 X 10⁻³

Q² = .0272 X 10⁻¹²

Q = .1650 X 10⁻⁶ C

No of electrons = .1650 x 10⁻⁶ / 1.6 x 10⁻¹⁹

= .103125 x 10¹³

1.03125 x 10¹²

Answer:

a) The magnitude of the magnetic field = 7.1 mT

b) The direction of the magnetic field is the +z direction.

Explanation:

The force, F on a current carrying wire of current I, and length, L, that passes through a magnetic field B at an angle θ to the flow of current is given by

F = (B)(I)(L) sin θ

F/L = (B)(I) sin θ

For this question,

(F/L) = 0.113 N/m

B = ?

I = 16.0 A

θ = 90°

0.113 = B × 16 × sin 90°

B = 0.113/16 = 0.0071 T = 7.1 mT

b) The direction of the magnetic field will be found using the right hand rule.

The right hand rule uses the first three fingers on the right hand (the thumb, the pointing finger and the middle finger) and it predicts correctly that for current carrying wires, the thumb is in the direction the wire is pushed (direction of the force; -y direction), the pointing finger is in the direction the current is flowing (+x direction), and the middle finger is in the direction of the magnetic field (hence, +z direction).

Answer:

weight of the bullet is 0.0500 lb

velocity as it travels from block A to block B is 900 ft/s

Explanation:

given data

horizontal velocity = 1500 ft/s

mass block A = 6-lb

mass block B = 4.95 lb

blocks A velocity = 5 ft/s

blocks B velocity = 9 ft/s

solution

we apply here law of conservation of momentum that is

m × v(o) + m1 × (0) + m2 × (0) = m × v(2) + m1 × v1 + m2 × v2     ................1

put here value and we get m

m = [tex]\frac{m1 \times v1 +m2 \times v2}{v(o) - v2}[/tex]   ..............2

m = [tex]\frac{6 \times 5 + 4.95 \times 9}{1500 - 9}[/tex]  

m = 0.0500 lb

and

when here bullet is pass through the A block then moment is conserve that is

m × v(o) + m × v1 + m1 × v1   ............3

v1 = [tex]\frac{m\times v(o)- m1\times v1}{m}[/tex]    

v1 = [tex]\frac{0.0500\times 1500 - 61\times 5}{0.05}[/tex]  

v1 = 900 ft/s

a) -10,800 N/C

b) [tex]4.79\cdot 10^{10}m/s^2[/tex]

c) [tex]4.88\cdot 10^9[/tex] times g

Explanation:

a)

The magnitude of the electric field produced by a single-point charge is given by

[tex]E=\frac{kQ}{r^2}[/tex]

where

k is the Coulomb's constant

Q is the charge producing the  field

r is the distance at which the field is calculated

In this problem:

[tex]Q=-3.0\cdot 10^{-5}C[/tex] is the c harge producing the field

[tex]r=5.0 m[/tex] is the distance at which we want to calculate the field

Substituting,

[tex]E=\frac{(9.0\cdot 10^9)(-3.0\cdot 10^{-5})}{(5.0)^2}=-10,800 N/C[/tex]

where the negative sign indicates that the direction of the field is towards the charge producing the field (for a negative charge, the electric field is inward, towards the charge)

b)

The force experienced by a charged particle in an electric field is given by

[tex]F=qE[/tex]

where

q is the magnitude of the charge

E is the electric field

Moreover, the force on an object can be written as:

[tex]F=ma[/tex]

where

m is the mass

a is the acceleration

Combining the two equations,

[tex]ma=qE\\a=\frac{qE}{m}[/tex]

In this problem:

[tex]q=1.6\cdot 10^{-19}C[/tex] is the charge of the proton

[tex]E=0.50 kN/C=500 N/C[/tex] is the strength of the electric field

[tex]m=1.67\cdot 10^{-27} kg[/tex] is the mass of the proton

Substituting, we find the acceleration of the proton:

[tex]a=\frac{(1.6\cdot 10^{-19})(500)}{(1.67\cdot 10^{-27})}=4.79\cdot 10^{10}m/s^2[/tex]

c)

The acceleration due to gravity is the acceleration at which every object near the Earth's surface falls down, in absence of air resistance, and it is given by

[tex]g=9.81 m/s^2[/tex]

On the other hand, the acceleration of the proton in this problem is:

[tex]a=4.79\cdot 10^{10} m/s^2[/tex]

To find how many times greater is this acceleration than that due to gravity, we can divide the acceleration of the proton by the value of g. Doing so, we find:

[tex]\frac{a}{g}=\frac{4.79\cdot 10^{10}}{9.81}=4.88\cdot 10^9[/tex]

So, it is [tex]4.88\cdot 10^9[/tex] times greater than g.

Final answer:

The electric field at the given point is -1.08×10⁳ N/C, which points towards the origin. The acceleration of a proton in a 0.50 kN/C electric field is 4.79×10ⁱ³ m/s². This acceleration is approximately 4.88×10ⁱ¹ times greater than the acceleration due to gravity.

Explanation:

To calculate the electric field at a point due to a point charge, we use Coulomb's law and the expression for the electric field, E = kQ/r², where k is Coulomb's constant (8.99×10⁹ N⋅m²/C²), Q is the charge, and r is the distance from the charge. Placing a charge of -3.0×10⁻⁵ C at the origin, the electric field at the point x= 5.0 m on the x-axis is calculated as follows:

E =  kQ/r²
=  (8.99×10⁹)(-3.0×10⁻⁵) / (5.0)²
=  -1.08×10⁳ N/C

The electric field points towards the negative charge, which is towards the origin.

To determine the acceleration of a proton in an electric field, we use the formula F = qE, where F is the force, q is the charge of the proton (1.60×10⁻⁵ C), and E is the electric field strength. The acceleration a is then found using Newton's second law, F = ma:

F = qE
= (1.60×10⁻⁵)(0.50×10⁳)
= 8.00×10⁻⁴ N

a = F/m
= 8.00×10⁻⁴ / 1.67×10⁻⁲⁷
=  4.79×10ⁱ³ m/s²

To find how many times greater this is compared to the acceleration due to gravity, we simply divide the proton's acceleration by the acceleration due to gravity (9.81 m/s²).

Number of times greater = a / g
=  4.79×10ⁱ³ / 9.81
=
≈ 4.88×10ⁱ¹ times

Answer:

(a). The maximum height by first ball is [tex]1.8545\times10^{4}\ m[/tex]

The maximum height by second ball is [tex]5.1020\times10^{4}\ m[/tex]

(b). The total mechanical energy of the ball–Earth system at the maximum height for each ball is [tex]1.0\times10^{7}\ J[/tex]

Explanation:

Given that,

Mass of cannonball = 20.0 kg

Speed = 1000 m/s

Angle with horizontal= 37.08

Fired angle = 90.08

We need to calculate the speed of the ball

Using formula of speed

[tex]v_{y}=v\sin\theta_{H}[/tex]

Put the value into the formula

[tex]v_{y}=1000\times\sin37.08[/tex]

[tex]v_{y}=602.9\ m/s[/tex]

(a). We need to calculate the maximum height by first ball

Using conservation of energy

[tex]\dfrac{1}{2}mv^2=mgh[/tex]

[tex]h= \dfrac{v^2}{2g}[/tex]

Put the value into the formula

[tex]h=\dfrac{(602.9)^2}{2\times9.8}[/tex]

[tex]h=1.8545\times10^{4}\ m[/tex]

We need to calculate the maximum height by second ball

Using conservation of energy

[tex]\dfrac{1}{2}mv^2=mgh[/tex]

[tex]h= \dfrac{v^2}{2g}[/tex]

Put the value into the formula

[tex]h=\dfrac{(1000)^2}{2\times9.8}[/tex]

[tex]h=5.1020\times10^{4}\ m[/tex]

(b). We need to calculate the total mechanical energy of the ball–Earth system at the maximum height for each ball

Using formula of energy

[tex]E=\dfrac{1}{2}mv^2[/tex]

[tex]E=\dfrac{1}{2}\times20\times1000^2[/tex]

[tex]E=1.0\times10^{7}\ J[/tex]

Hence, (a). The maximum height by first ball is [tex]1.8545\times10^{4}\ m[/tex]

The maximum height by second ball is [tex]5.1020\times10^{4}\ m[/tex]

(b). The total mechanical energy of the ball–Earth system at the maximum height for each ball is [tex]1.0\times10^{7}\ J[/tex]