A river barge, whose cross section is approximately rectangular, carries a load of grain. The barge is 28 ft wide and 93 ft long. When unloaded its draft (depth of submergence) is 6 ft, and with the load of grain the draft is 9 ft. Determine: (a) the unloaded weight of the barge, and (b) the weight of the grain.

Answer:

The distance is 1.69 m.

Explanation:

Given that,

First charge [tex]q_{1}= 3.8\times10^{-6}\ C[/tex]

Second charge [tex]q_{2}=3.2\times10^{-6}\ C[/tex]

Distance = 3.25 m

We need to calculate the distance

Using formula of electric field

[tex]E_{1}=E_{2}[/tex]

[tex]\dfrac{kq_{1}}{x^2}=\dfrac{kq_{2}}{(d-x)^2}[/tex]

[tex]\dfrac{q_{1}}{q_{2}}=\dfrac{(x)^2}{(d-x)^2}[/tex]

[tex]\sqrt{\dfrac{q_{1}}{q_{2}}}=\dfrac{x}{d-x}[/tex]

[tex]x=(d-x)\times\sqrt{\dfrac{q_{1}}{q_{2}}}[/tex]

Put the value into the formula

[tex]x=(3.25-x)\times\sqrt{\dfrac{3.8\times10^{-6}}{3.2\times10^{-6}}}[/tex]

[tex]x+x\times\sqrt{\dfrac{3.8\times10^{-6}}{3.2\times10^{-6}}}=3.25\times\sqrt{\dfrac{3.8\times10^{-6}}{3.2\times10^{-6}}}[/tex]

[tex]x(1+\sqrt{\dfrac{3.8\times10^{-6}}{3.2\times10^{-6}}})=3.25\times\sqrt{\dfrac{3.8\times10^{-6}}{3.2\times10^{-6}}}[/tex]

[tex]x=\dfrac{3.25\times\sqrt{\dfrac{3.8\times10^{-6}}{3.2\times10^{-6}}}}{(1+\sqrt{\dfrac{3.8\times10^{-6}}{3.2\times10^{-6}}})}[/tex]

[tex]x=1.69\ m[/tex]

Hence, The distance is 1.69 m.

Answer:

[tex]\frac{E_{A}}{E_{B}}=4[/tex]

Explanation:

The electric field is defined as the electric force per unit of charge, this is:

[tex]E=\frac{F}{q}[/tex].

The electric force can be obtained through Coulomb's law, which states that the electric force between to electrically charged particles is inversely proportional to the square of the distance between them and directly proportional to the product of their charges. The electric force can be expressed as

[tex]F=\frac{kQq}{r^{2}}[/tex].

By substitution we get that

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

Now, letting [tex]E_{A}[/tex] be the electric field at point A, letting [tex]E_{B}[/tex] be the electric field at point B, and letting R be the distance from the charge to A:

[tex]E_{A}=\frac{kQ}{R^{2}}\\\\E_{B}=\frac{kQ}{(2R)^{2}}[/tex].

The ration of the electric fields is

[tex]\frac{E_{A}}{E_{B}}=\frac{\frac{kQ}{R^{2}}}{\frac{kQ}{(2R)^{2}}}\\\\\frac{E_{A}}{E_{B}}=\frac{\frac{1}{R^{2}}}{\frac{1}{(2R)^{2}}}\\\\\frac{E_{A}}{E_{B}}=\frac{\frac{1}{R^{2}}}{\frac{1}{(4)R^{2}}}\\\\\\\frac{E_{A}}{E_{B}}=\frac{1}{\frac{1}{(4)}}\\\\\frac{E_{A}}{E_{B}}=4[/tex]

This means that at half the distance, the electric field is four times stronger.

The ratio of the electric field strength at point A to the electric field strength at point B is 4.

Given to us

A positive charge Q and a point B is twice as far away from Q as point A.

According to Coulomb's law, the electric force between two electrical charges particles is inversely proportional to the square of the distance between them and it is directly proportional to the product of their charges.

[tex]F =\dfrac{k\ Q_1 Q_2}{r^2}[/tex]

We know that an electric field is written as,

[tex]F= EQ[/tex]

Merging the two-equation we get

[tex]E = \dfrac{k Q}{r^2}[/tex]

As we can write

[tex]\dfrac{E_A}{E_B} = \dfrac{\dfrac{k Q}{R^2}}{\dfrac{k Q}{(2R)^2}}[/tex]

[tex]\dfrac{E_A}{E_B} = 4[/tex]

Hence, the ratio of the electric field strength at point A to the electric field strength at point B is 4.

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Answer:

The object is moving to the right

Question:

See attachment for the complete question

Explanation:

Position vs time graph is a graphical representation of the distance covered by a particular motion plotted against the time taken.

According to the attached image;

- since the position increases with increase in time as seen on the graph, the motion is positive that means the object is moving to the right.

- the graph is not straight which means the velocity is changing (not constant speed) and also according to the shape of the curve the velocity is decreasing with time(deceleration)

The velocity is the change in position of the object with time. The given object is moving in the right direction.

It is a graphical representation of the distance covered by a moving object plotted against the time taken by the object.

Therefore, the given object is moving in the right direction.

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Answer:

[tex]F=33.25\ N[/tex]

[tex]a=89.8649\ m.s^{-2}[/tex]

Explanation:

Given:

Now the force on the spring on releasing the compression:

[tex]F=k. \Delta x[/tex]

[tex]F=175\times 0.19[/tex]

[tex]F=33.25\ N[/tex]

Now the acceleration due to this force:

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

[tex]a=\frac{33.25}{0.37}[/tex]

[tex]a=89.8649\ m.s^{-2}[/tex]

Final answer:

The force exerted on a 0.37 kg object when released from a compressed spring with a constant of 175 N/m is 33.25 N. The acceleration at this instant is 89.86 m/s².

Explanation:

We are given a 0.37 kg object attached to a spring with a spring constant of 175 N/m, compressed by 0.19 m. The force exerted by the spring when it is released can be calculated using Hooke's Law, which states that the force exerted by a spring is proportional to the displacement from its equilibrium position:

F = -kx

Where:

Plugging in the values we get:

F = -(175 N/m) * (0.19 m)

F = -33.25 N

Since the negative sign indicates the direction of the force is opposite to the direction of displacement, we can say that the magnitude of the force is 33.25 N. To find the acceleration at this instant, we can use Newton's second law of motion:

a = F/m

a = 33.25 N / 0.37 kg

a = 89.86 m/s²

The acceleration of the object when it is released is 89.86 m/s².

Answer:

220.508 kPa

Explanation:

32 psi = 32 pound force per square inch = 32 lbf/in2 = 32 * (4.448 N/lbf) * (12 in/ft * 3.28 ft/m)^2

[tex] = 32*4.448*(12*3.28)^2 = 220508 N/m^2[/tex] or 220508 Pa or 220.508 kPa

So the maximum safe air pressure of a tire is 220.508 kPa

The correct conversion of 32 psi (gage) to kPa is approximately 220.6 kPa.

 To convert from psi (pounds per square inch) to kPa (kilopascals), one can use the conversion factor of 1 psi being equal to 6.89476 kPa. Therefore, to convert 32 psi to kPa, we multiply 32 by 6.89476:

[tex]\[ 32 \, \text{psi} \times 6.89476 \, \frac{\text{kPa}}{\text{psi}} = 220.632 \, \text{kPa} \][/tex]

 Rounding to a reasonable number of significant figures, we get approximately 220.6 kPa. This is the maximum pressure in kPa that the tire can safely withstand as indicated by the label.

Answer:

Part A:

[tex]r\geq 173.806 m[/tex]

Part B:

[tex]F=5289.5757 N[/tex]

Explanation:

Part A:

Airplane is moving in circular path so acceleration in circular path is given by:

[tex]a_c=\frac{V^2}{r}[/tex]

Where:

V is the velocity of object in circular path

r is the radius of circular path

In order to find radius r. Above equation will become:

[tex]r=\frac{V^2}{a_c}[/tex]

V=104 m/s

a_c=6.35*g=6.35*9.8=62.23 m/s^2

[tex]r\geq \frac{104^2}{62.23}\\ r\geq 173.806 m[/tex]

Part B:

Force on a object moving in a circular path is:

[tex]F=\frac{mV^2}{r}\\[/tex]

r=173.806 m

m=85 kg

V=104 m/s

[tex]F=\frac{85*(104)^2}{173.806} \\F=5289.5757 N[/tex]

Given:

(a)

We know,

→ [tex]a_c = mg[/tex]

       [tex]= 6.35\times g[/tex]

       [tex]= 6.35\times 9.8[/tex]

       [tex]= 62.23 \ m/s^2[/tex]

Now,

Acceleration in circular path will be:

→ [tex]a_c = \frac{V^2}{r}[/tex]

or,

→ [tex]r = \frac{V^2}{a_c}[/tex]

By substituting the values, we get

     [tex]= \frac{(104)^2}{62.23}[/tex]

     [tex]= 173.806 \ m[/tex] (radius)

(b)

The force on object will be:

→ [tex]F = \frac{mV^2}{r}[/tex]

      [tex]= \frac{85\times (104)^2}{173.806}[/tex]

      [tex]= 5289.58 \ N[/tex]

Thus the approach above is correct.

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Answer:

Spring Constant K=9290.9550 N/m

Explanation:

Formula we are going to use is:

[tex]2\pi f=\sqrt{\frac{K}{m}}[/tex]

Where:

f is the frequency

K is the spring constant

m is the mass

Given:

Mass of spring oscillator=m=0.206 kg

Resonance Frequency=f=33.8 Hz

Find:

Spring Constant=K=?

Solution:

From Above formula:

Taking Square on both sides

[tex]4(\pi)^2f^2=\frac{K}{m} \\K=4(\pi)^2f^2*m\\K=4(\pi)^2(33.8)^2*(0.206)\\K=9290.9550 N/m[/tex]

Spring Constant K=9290.9550 N/m

Answer:

 r = 0.1045 cm

Explanation:

The electric field of a point charge is given by the expression

          E = k q / r²

Let's look for the field created by each load

Q₁ charge

           E₁ = k Q₁ / (r-0)²

           E₁ = - 8.99 10⁹ 45.5 10⁻⁹ / r²

           E₁ = - 409.05 / r²

Q₂ charge

           E₂ = k Q₂ / (r-0.14)₂

           E₂ = - 8.99 10⁹ 5.5 10⁻⁹ / (r-0.14)²

           E₂ = -49.45 / (r-0.14)²

The electric field is a vector quantity, so we must use the sum of vectors to get the total field

           E_total = E₁ + E₂

The burden of proof is always considered positive as the two charges are negative the strength is attractive, consider three regions of interest.

- Left of the two charges in this case the two forces are directed to the right and therefore the field is not canceled

- to the right of the two charges, the forces go to the left and the field is not canceled for any distance

- Between the two charges, in this case each force goes in the opposite direction and there is a point for the field to cancel

             0 = E₁ - E₂

             E₁ = E₂

            -409.05 / r² = - 49.45 /(r-0.14)²

              (r-0.14)² = 49.45 / 409.05 r²

             r² - 2 0.14r + 0.14² = 0.12 r²

             r² (1-0.12) - 0.28 r + 0.0196 = 0

             0.88 r² - 0.28 r + 0.0196 = 0

             r² - 0.318 r + 0.0223 = 0

We solve the second degree equation

            r = [0.318 ±√(0.318² - 4 0.0223)] / 2

            r = [0.318 ± 0.109] 2

            r₁ = 0.209 / 2 = 0.1045 m

            r₂ = 0.2135 m

We see that the result in the area of ​​interest between charges is

             r = 0.1045 cm

- There are zones on the sides of the charge, but the force in these zones has a component that takes the test load to the part between the loads, therefore the field is never canceled

Answer:

4 half-lives will occur during this period of time.

Explanation:

Formula used :

[tex]a_o=a\times e^{-\lambda t}[/tex]

[tex]\lambda =\frac{0.693}{t_{1/2}}[/tex]

[tex]a=\frac{a_o}{2^n}[/tex]

where,

a = amount of reactant left after n-half lives and time t

[tex]a_o[/tex] = Initial amount of the reactant.

[tex]\lambda = [/tex] decay constant

[tex]t_{1\2}[/tex] = half life of an isotope

n = number of half lives

We have :

[tex]a_o=100.0 g[/tex]

a = ?

t = 552 days

[tex]t_{1/2}=138 days[/tex]

[tex]a=100.0 g\times e^{-\frac{0.693}{138}\times 552}[/tex]

[tex]a=6.254 g[/tex]

[tex]6.254 g=\frac{100.0 g}{2^n}[/tex]

[tex]2^n=\frac{100.0 g}{6.254 g}[/tex]

n = 4

4 half-lives will occur during this period of time.

By dividing the total time of 552 days by the half-life of Polonium-210 of 138 days, we find that 4 half-lives occur in this period.

The number of half-lives that occur during a given period of time can be found by dividing the total amount of time by the period of one half-life. Given that we are asked to find out the number of half-lives for a 552-day period and the half-life of Polonium-210 is 138 days, we can use this approach. So, the calculation would be 552 days ÷ 138 days/half-life = 4 half-lives. Therefore, 4 half-lives of Polonium-210 would occur in a 552-day period.

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Answer: I have attached the circuit diagram.

Explanation: NOTE that an independent voltage or current source is denoted in a circle. That is how you identify an independent source.

It simply means that the voltage and current source value does not depend on the value of voltage and current elsewhere in the circuit.

It highly utilised when using Mesh analysis on an electric circuit.

Please I have attached the circuit diagram connected in series.

Check it out!

Answer:

a) t = 4.16 s

b) x = 141.51 m

Explanation:

Given

v = 21.5 m/s

x0 = 52.0 m

a = 6.0 m/s²

a) Motorcycle

x = v0*t + (a*t²/2)

x = 21.5t + (6*t²/2)

x = 21.5t + 3t²   (I)

Car

x = x0 + v0*t

x = 52 + 21.5t  (II)

then we can apply I = II

21.5t + 3t² = 52 + 21.5t

⇒ 3t² = 52

⇒ t = 4.16 s

b) We can use I or II, then

x = 52 + 21.5*(4.16)

⇒ x = 141.51 m

To find the time it takes for the motorcycle to catch up with the car, we can use the equation of motion and the given information. The motorcycle starts accelerating at t1, so it covers a distance of 0 m. The distance covered by the car until the motorcycle catches up is equal to the initial distance between them, which is 52.0 m.

To find the time it takes for the motorcycle to catch up with the car, we first need to find the acceleration of the car. Since it is traveling at a constant speed, its acceleration is zero. The motorcycle's acceleration is given as 6.00 m/s². We can use the equation of motion, s = ut + 0.5at², to find the distance each vehicle covers from time t1 to t2. The motorcycle will cover a distance of 0 m, as it starts accelerating at t1. The car will cover a distance of (21.5 m/s)(t2 - t1). Since the motorcycle catches up with the car, the distance covered by the car is equal to the initial distance between them, which is 52.0 m.

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Answer:

A)    N = W cos θ , B)  fr = W sin θ, C) N = W cos θ

Explanation:

To find the required expressions, let's make a free body diagram of the block, see attached.

In the problem of inclined plane the reference system used the x axis is parallel to the plane and the y axis is perpendicular

In this reference system the only force that we must decompose is the body weight F_w = W) for this we use trigonometry

Note that the angle of the plane is equal to the angle between the axis and the weight, therefore

            sin θ = [tex]W_{x}[/tex] / W

            cos θ = [tex]W_{y}[/tex]  / W

            [tex]W_{x}[/tex]  = W sin θ

            [tex]W_{y}[/tex]  = W cos θ

B) Let us write Newton's second law for each axis, in this case as the block is still the acceleration is zero

X axis

             fr - [tex]W_{x}[/tex]  = 0

             fr = W sin θ

A) Y Axis

             N - [tex]W_{y}[/tex]  = 0

             N = W cos θ

C) N = W cos θ

Answer:

The tension in the string is 87.72 N.

Explanation:

Given that,

Mass of two each boxes = 10 kg

Suppose the inclined plane makes an angle is 60° with the horizontal and the coefficient of kinetic friction between the box and plane is 0.15.

We need to calculate the acceleration

Using balance equation

[tex]mg-T=ma[/tex]

Put the value into the formula

[tex]10g-T=10a[/tex]

[tex]98-T=10a[/tex]....(I)

Again using balance equation

[tex]T-mg\sin\theta-F_{f}=ma[/tex]

[tex]T-mg\sin\theta-\mu mg=ma[/tex]

Put the value into the formula

[tex]T-10g\sin60-\mu10\cos60=10a[/tex]

[tex]T-10\times9.8\dfrac{\sqrt{3}}{2}-0.15\times10\times9.8\cos\times\dfrac{1}{2}=10a[/tex]

[tex]T-84.8+7.35=10a[/tex]

[tex]T-77.45=10a[/tex]...(II)

From equation (I) and (II)

[tex]98-77.45=20a[/tex]

[tex]a=\dfrac{98-77.45}{20}[/tex]

[tex]a=1.0275\ m/s^2[/tex]

Put the value of acceleration in equation (I)

[tex]98-T=10\times1.028[/tex]

[tex]-T=10\times1.028-98[/tex]

[tex]T=87.72\ N[/tex]

Hence, The tension in the string is 87.72 N.

The tension in the string connecting the two 10-kilogram boxes is 98 N.

The tension in the string connecting the two 10-kilogram boxes can be determined using Newton's second law and the equilibrium equation for the x-direction. Based on these equations, the tension in the shorter string is twice the tension in the longer string. Therefore, the tension in the string connecting the two boxes would be:

Tension = weight of the hanging mass

So, the tension in the string would be equal to the weight of each 10-kilogram box which is:

Tension = (10 kg) × (9.8 m/s²) = 98 N

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Answer: 20N

Explanation:

According to Newton's law of gravitation which states that every particle attracts every other particle in the universe with a force which is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centers

F = Gm1m2/r^2

Where

F = force between the masses

G universal gravitational constant

m1 and m2 = mass of the two particles

r = distance between the centre of the two mass

Therefore, weigh of an object on earth is inversely proportional to the square of its distance from the centre of the earth

W₁/W₂ = r₂²/r₁²

W₂ = W₁r₁²/r₂²

W₁ = 180 N

r₁ = R

r₂ = 3R

W₂ = (180 × R²)/(3R)²

W₂ = 180R²/9R²

W₂ = 180/9

W₂ = 20N

its weight at 3R is 20N

Final answer:

The weight of an object decreases with the square of the distance from the center of the mass it is orbiting. At 3R, the distance from the Earth's center, the gravitational force would be 1/9th of what it is on Earth's surface. Hence, an object that weighs 180 N on Earth would weigh 20 N at a distance of 3R from Earth's center.

Explanation:

To determine the weight of an object at a distance of 3R from the center of Earth, we need to understand how gravitational force decreases with distance. The gravitational force and, therefore, gravitational acceleration (g) at a distance r from a mass M is given by Newton's law of universal gravitation:

F = G(Mm/r²)

where G is the gravitational constant, M is the mass of Earth, m is the mass of the object, and r is the distance to the center of Earth. The weight (W) of an object is the product of its mass m and the gravitational acceleration g, so:

W = mg

The gravitational acceleration (g) on Earth's surface is 9.80 m/s². We can use the formula:

g' = GM/(3R)²

to find the gravitational acceleration at a distance of 3R. Since the mass of the object cancels out, we get:

g' = g/R²

And we know the object's weight on Earth's surface (W = 180 N), so:

W' = (W/R²) x 1/9

Substituting the values, we have:

W' = 180 N / 9 = 20 N

Therefore, at a distance of 3R from the center of the Earth, the object would weigh 20 N.

Answer:

a,b,d and e are correct.

Explanation:

a) Change of temperature without change of velocity is a conclusive evidence of an interaction. As the temperature of a body will only change if heat energy is flows in or out of the body with surrounding. So, option a) is correct.

b) Change of the direction without change of speed is an evidence of an interaction. The direction changes when the object is under a perpendicular acceleration acceleration which clearly means the particle is interacting or external force applied. So, option b is correct.

d) Change of shape or configuration without change of velocity is conclusive evidence of an interaction. The shape can change only if external stress is applied on the particle. So, Option d is correct.

e) Change of identity without change of velocity is a conclusive evidence of an interaction. The identity of an object can only change if it interacts with surroundings. So, option e is correct

Therefore the only incorrect option is c .

In this exercise we have to use our knowledge of physics to identify the best alternative that corresponds to each of the given questions:

the only incorrect alternative is the letter C

So analyzing each alternative, we find that:

a) Change of hotness outside change of speed exist a evidence of an interplay.

As the temperature of a main part of written work will only change if heat strength exist flows fashionable or exhausted the  accompanying encircling.

So, alternative A happen correct.

b) Change of the course outside change of speed happen an evidence of an interaction.

The management changes when the object happen secondary a at right angles to increasing speed acceleration which without any doubt wealth the piece exist communicate or outside force applied.

So, alternative B exist correct.

d) Change of shape or arrangement outside change of speed exist evidence of an interplay.

The shape can change only if outside stress happen used ahead of the atom.

So, alternative D exist correct.

e) Change of similarity outside change of speed happen a evidence of an interplay.

The person's individuality of an object can only change if it communicate accompanying environment.

So, alternative E exist correct

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Explanation:

I assume you're looking for the average angular velocity:

ω_avg = Δθ / Δt

ω_avg = (125° × π / 180°) / (1.3 s)

ω_avg = 1.6782 rad/s

Answer:

3.7

Explanation:

[tex]\epsilon_0[/tex] = Permittivity of free space = [tex]8.85\times 10^{-12}\ F/m[/tex]

r = Radius = [tex]5.1\times 10^{-2}\ m[/tex]

A = Area = [tex]\pi r^2[/tex]

V = Voltage = 1120 V

d = Distance of plate seperation = 0.23 mm

Charge is given by

[tex]Q=CV\\\Rightarrow Q=\dfrac{k\epsilon_0 A}{d}\times V\\\Rightarrow k=\dfrac{Qd}{\epsilon_0 AV}\\\Rightarrow k=\dfrac{1.3\times 10^{-6}\times 0.23\times 10^{-3}}{8.85\times 10^{-12}\times \pi (5.1\times 10^{-2})^2\times 1120}\\\Rightarrow k=3.69164\\\Rightarrow k=3.7[/tex]

The value of dielectric constant is 3.7

Final answer:

To find the dielectric constant (κ) for a parallel-plate capacitor, given its physical dimensions and charge-potential difference, we calculate the capacitance with and without the dielectric and then solve for κ using the relation between charge, capacitance, and potential difference.

Explanation:

The question involves finding the dielectric constant (κ) of the material placed between the plates of a parallel-plate capacitor given the radius of the plates, the separation between them, the charge on the capacitor, and the potential difference across it. The capacitance of a parallel-plate capacitor is given by C = ε_0 κ A / d, where ε_0 is the permittivity of free space, κ is the dielectric constant, A is the area of the plates, and d is the separation between them. The charge (Q) on a capacitor is related to the capacitance (C) and the potential difference (V) across it by Q = CV. From the given information, we can calculate the area of the plates from their radius, and using the given separation, charge, and potential difference, solve for the dielectric constant, κ.

Given:

Radius of the plates, r = 5.10×10−2 m

Separation between the plates, d = 0.23 mm = 0.23×10−3 m

Charge on the capacitor, Q = 1.3 μC = 1.3×10−6 C

Potential difference, V = 1120 V

We can first calculate the area (A) of the plates using the formula for the area of a circle, A = πr2, substitute the value of A and the given values into the formula for C, and then use the relationship Q = CV to solve for κ.

Answer:

[tex]\displaystyle v'=4\ km/h[/tex]

Explanation:

Linear Momentum

The total linear momentum of an isolated system (with no external forces) is conserved regardless of the internal mutual interactions of its parts. Recall the momentum is obtained by multiplying the speed by the object's mass. We'll refer to the railroad diesel engine as mass 2 and the flatcar as mass 1.

We have the following data

[tex]\displaystyle m_2=4\ m_1[/tex]

[tex]\displaystyle v_1=0[/tex]

[tex]\displaystyle v_2=5\ km/h[/tex]

Since both objects remain coupled after their encounter, the final speed is common to both:

[tex]\displaystyle v_1'=v_2'=v'[/tex]

Let's sketch the principle of conservation of linear momentum as follows

[tex]m_1v_1+m_2v_2=m_1v_1'+m_2v_2'[/tex]

Using the mentioned conditions for the speeds

[tex]\displaystyle m_1\ v_1+m_2\ v_2=(m_1+m_2)\ v'[/tex]

Solving for v'

[tex]\displaystyle v'=\frac{m_1\ v_1+m_2\ v_2}{m_1+m_2}[/tex]

[tex]\displaystyle v'=\frac{m_1.0+4m_1\ v_2}{m_1+4m_1}[/tex]

[tex]\displaystyle v'=\frac{4\cancel{m_1} \ v_2}{5 \cancel{m_1}}[/tex]

[tex]\displaystyle v'=\frac{4}{5}v_2[/tex]

[tex]\displaystyle v'=\frac{4}{5}\ 5km/h[/tex]

[tex]\boxed{\displaystyle v'=4\ km/h}[/tex]

To find the speed at which the two vehicles coast after coupling together, we can apply the principle of conservation of momentum. The railroad diesel engine weighs 4 times as much as the flatcar, so we can use this information to calculate the final velocity of the coupled vehicles.

To solve this problem, we can apply the principle of conservation of momentum. The momentum before the coupling is equal to the momentum after the coupling.

Let's assume the mass of the flatcar is m kg. According to the given information, the mass of the railroad diesel engine is 4 times the mass of the flatcar, so the mass of the engine is 4m kg.

Before the coupling, the momentum of the engine is (4m)(5 km/h) = 20m km/h, and the momentum of the flatcar is (m)(0) = 0.

After the coupling, the combined mass is (4m+m) = 5m kg, and their common velocity is v km/h. Using the principle of conservation of momentum, we can write the equation:

(20m) + (0) = (5m)(v)

Simplifying the equation, we have:

20m = 5mv

Dividing both sides by 5m, we get:

4 = v

Therefore, the two coast after they couple together at a speed of 4 km/h.

Answer:

There are [tex]2.12\times 10^{19}\ electrons[/tex] that must be removes from an electrically neutral silver dollar.

Explanation:

In this case, we need to find the number of electrons must be removes from an electrically neutral silver dollar to give it a charge of +3.4 C. Let n number of electrons removed. It is a case of quantization of electric charge. It is given by :

q = ne

[tex]n=\dfrac{q}{e}[/tex]

e is the charge on an electron

[tex]n=\dfrac{3.4\ C}{1.6\times 10^{-19}\ C}[/tex]

[tex]n=2.12\times 10^{19}\ electrons[/tex]

So, there are [tex]2.12\times 10^{19}\ electrons[/tex] that must be removes from an electrically neutral silver dollar.

Answer:

The kinetic energy of the merry-goround after 3.62 s is  544J

Explanation:

Given :

Weight w = 745 N

Radius r =  1.45 m

Force =  56.3 N

To Find:

The kinetic energy of the merry-go round after 3.62  = ?

Solution:

Step 1:  Finding the Mass of merry-go-round

[tex]m = \frac{ weight}{g}[/tex]

[tex]m = \frac{745}{9.81 }[/tex]

m = 76.02 kg

Step 2: Finding the Moment of Inertia of solid cylinder

Moment of Inertia of solid cylinder I =[tex]0.5 \times m \times r^2[/tex]

Substituting the values

Moment of Inertia of solid cylinder I  

=>[tex]0.5 \times 76.02 \times (1.45)^2[/tex]

=> [tex]0.5 \times 76.02\times 2.1025[/tex]

=> [tex]79.91 kg.m^2[/tex]

Step 3: Finding the Torque applied T

Torque applied T = [tex]F \times r[/tex]

Substituting the values

T = [tex]56.3 \times 1.45[/tex]

T = 81.635 N.m

 Step 4: Finding the Angular acceleration

Angular acceleration ,[tex]\alpha = \frac{Torque}{Inertia}[/tex]

Substituting the values,

[tex]\alpha = \frac{81.635}{79.91}[/tex]

[tex]\alpha = 1.021 rad/s^2[/tex]

 Step 4: Finding the Final angular velocity

Final angular velocity ,[tex]\omega = \alpha \times t[/tex]

Substituting the values,

[tex]\omega = 1.021 \times 3.62[/tex]

[tex]\omega = 3.69 rad/s[/tex]

Now KE (100% rotational) after 3.62s is:

KE = [tex]0.5 \times I \times \omega^2[/tex]

KE =[tex]0.5 \times 79.91 \times 3.69^2[/tex]

KE = 544J

Answer:

[tex]P_g=\rho_w.g.h[/tex] (gauge pressure)

[tex]P_{abs}=P_{atm}+\rho_w.g.h[/tex] (absolute pressure)

Explanation:

The gauge pressure is the pressure measured with reference to the atmospheric pressure where as the absolute pressure is the pressure measured including the atmospheric pressure taking absolute vacuum as the reference.

So when gauge measuring the pressure at the bottom of a lake the pressure only due to the water column above that point is measured which can be calculated as:

[tex]P_g=\rho_w.g.h[/tex]

where:

[tex]\rho_w=[/tex] density of water

[tex]g=[/tex] acceleration due to gravity

[tex]h=[/tex] height of the water column

Since the atmospheric pressure at the surface of the sea level is defined as the atmospheric pressure.

So, the absolute pressure:

[tex]P_{abs}=P_{atm}+\rho_w.g.h[/tex]

where:

[tex]P_{atm}=[/tex] the atmospheric pressure at the sea level

Therefore, the gage pressure at the bottom of the lake is [tex]\( 294300 \, \text{Pa} \)[/tex], and the absolute pressure at the bottom of the lake is [tex]\( 395625 \, \text{Pa} \).[/tex]

To solve the problem, we can use the formula for gage pressure in a fluid, [tex]( P_g = \rho g h \))[/tex], where ρ is the density of the fluid g is the acceleration due to gravity, and h is the depth below the surface of the fluid.

Given that the density of the lake water is [tex]\( 1000 \, \text{kg/m}^3 \)[/tex], the acceleration due to gravity is [tex]\( 9.81 \, \text{m/s}^2 \)[/tex], and the depth of the lake is [tex]\( 30 \, \text{m} \),[/tex] we can calculate the gage pressure:

[tex]\[ P_g = (1000 \, \text{kg/m}^3) \times (9.81 \, \text{m/s}^2) \times (30 \, \text{m}) = 294300 \, \text{Pa} \][/tex]

To find the absolute pressure[tex](\( P_{abs} \)),[/tex] we add the atmospheric pressure [tex](\( P_{atm} = 101325 \, \text{Pa} \)):[/tex]

[tex]\[ P_{abs} = P_g + P_{atm} = 294300 \, \text{Pa} + 101325 \, \text{Pa} = 395625 \, \text{Pa} \][/tex]

The final velocity of the rock is 15 m/s upwards. The rock's initial velocity is upwards at 5 m/s. The acceleration due to gravity is downwards at 9.8 m/s². Hence the correct option is A.

The initial velocity of the rock is upwards at 5 m/s. This is denoted by u.

The acceleration due to gravity is downwards at 9.8 m/s². This is denoted by a.

The time taken for the rock to reach its final velocity is 2 seconds. This is denoted by t.

The final velocity of the rock is denoted by v.

We can use the following equation to calculate the final velocity of the rock:

v = u + at

Substituting the values of u, a, and t, we get:

= 5 + 9.8 * 2

= 15 m/s

Therefore, the final velocity of the rock is 15 m/s upwards. Hence the correct option is A.

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The rock's speed two seconds after being thrown upward is zero. Gravity causes the object to slow down and eventually come to a stop before starting to fall back down.

The rock's speed two seconds after being thrown upward is zero (D).

When an object is thrown upwards, its speed decreases due to the force of gravity pulling it downward. Gravity causes the object to slow down and eventually come to a stop before starting to fall back down. In this case, since the rock's speed is zero after two seconds, it means it has reached its highest point and is now starting to fall back down.

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Answer:

15086 mm³/s

Explanation:

Given

dr / dt = 3 mm/s

v = 4/3 π r³

dv / dt = (4 / 3)  π 3r²  dr/dt

= 4πr² x 3

= 12πr²

= 12 x 22/7 x 20²

= 15086 mm³/s

The volume of a sphere with a radius increasing at 3 mm/s will be increasing at a rate of approximately 15072 mm³/s when the diameter is 40 mm. This is calculated using the derivative of the sphere's volume formula.

Given that the radius of a sphere is increasing at a rate of 3 mm/s, we need to find how fast the volume is increasing when the diameter is 40 mm. First, we note that the diameter is 40 mm, so the radius is 20 mm.

The formula for the volume of a sphere is:

V = (4/3)πr³

To find the rate at which the volume is increasing, we take the derivative of the volume with respect to time (t):

dV/dt = 4πr²(dr/dt)

We know from the problem that dr/dt (the rate at which the radius is increasing) is 3 mm/s, and r (the radius at the given moment) is 20 mm. Plugging these values into the derivative formula, we get:

dV/dt = 4π(20)²(3)

dV/dt = 4π(400)(3) = 4800π mm³/s

Evaluating this value numerically and rounding to the nearest whole number, we get:

dV/dt ≈ 15072 mm³/s

Final answer:

The initial acceleration of a proton and an electron released 5.50×10−10 mm apart is calculated using Coulomb's law and Newton's second law. The proton's acceleration is approximately 4.91×10¹⁴ m/s², and the electron's acceleration is approximately 8.99×10¹⁶ m/s².

Explanation:

To find the initial acceleration of a proton and an electron when they are released 5.50×10−10 mm apart, we use Coulomb's law to calculate the force and then apply Newton's second law to find the acceleration. First, convert the distance to meters, so 5.50×10−10 mm is 5.50×10−12 m. Coulomb's law gives us the force as F = k×|q1×q2|/r2, where k is Coulomb's constant (9.00×109 N×m2/C2), q1 and q2 are the charges of the proton and electron (1.60×10−19 C), and r is the separation distance. Substituting the values gives F = 8.20×10−10 N. Then, use Newton's second law, a = F/m, where m is the mass of the particle. For the proton, m = 1.67×10−27 kg, and for the electron, m = 9.11×10−31 kg. Therefore, the initial acceleration of the proton is ap ≈ 4.91×1014 m/s2 and the initial acceleration of the electron is ae ≈ 8.99×1016 m/s2.

The initial acceleration of the proton is approximately [tex]\(4.56 \times 10^{19} \, \text{m/s}^2\)[/tex] and the initial acceleration of the electron is approximately [tex]\(8.35 \times 10^{21} \, \text{m/s}^2\)[/tex].

To calculate the initial acceleration of the proton and electron when they are 5.50×10^(-10) mm apart, we can use Coulomb's law, which states that the force between two point charges is directly proportional to the product of the charges and inversely proportional to the square of the distance between them.

The formula for Coulomb's law is:

[tex]\[ F = \frac{{k \cdot |q_1 \cdot q_2|}}{{r^2}} \][/tex]

Where:

- F is the force between the charges,

- k is Coulomb's constant [tex](\(8.9875 \times 10^9 \, \text{N m}^2/\text{C}^2\))[/tex],

- [tex]\( q_1 \)[/tex] and [tex]\( q_2 \)[/tex] are the magnitudes of the charges (in this case, the charge of a proton and an electron, which are [tex]\(1.6 \times 10^{-19} \, \text{C}\) and \(-1.6 \times 10^{-19} \, \text{C}\)[/tex], respectively),

- r is the distance between the charges.

First, we need to convert the distance from milli meters to meters:

[tex]\[ r = 5.50 \times 10^{-10} \, \text{mm} = 5.50 \times 10^{-13} \, \text{m} \][/tex]

Now, we can calculate the force between the charges using Coulomb's law:

[tex]\[ F = \frac{{8.9875 \times 10^9 \times |(1.6 \times 10^{-19}) \times (-1.6 \times 10^{-19})|}}{{(5.50 \times 10^{-13})^2}} \][/tex]

[tex]\[ F = \frac{{8.9875 \times 10^9 \times 2.56 \times 10^{-38}}}{{3.025 \times 10^{-25}}} \][/tex]

[tex]\[ F \approx 7.61 \times 10^{-8} \, \text{N} \][/tex]

Since force F = ma, we can rearrange to find acceleration [tex]\( a = \frac{F}{m} \)[/tex]. For both the proton and electron, we use their respective masses, which are approximately [tex]\(1.67 \times 10^{-27} \, \text{kg}\) and \(9.11 \times 10^{-31} \, \text{kg}\)[/tex], respectively.

For the proton:

[tex]\[ a_p = \frac{7.61 \times 10^{-8}}{1.67 \times 10^{-27}} \][/tex]

[tex]\[ a_p \approx 4.56 \times 10^{19} \, \text{m/s}^2 \][/tex]

For the electron:

[tex]\[ a_e = \frac{7.61 \times 10^{-8}}{9.11 \times 10^{-31}} \][/tex]

[tex]\[ a_e \approx 8.35 \times 10^{21} \, \text{m/s}^2 \][/tex]

So, the initial acceleration of the proton is approximately [tex]\(4.56 \times 10^{19} \, \text{m/s}^2\)[/tex] and the initial acceleration of the electron is approximately [tex]\(8.35 \times 10^{21} \, \text{m/s}^2\)[/tex].

These values represent the accelerations they experience due to the electrostatic force when they are [tex]5.50\times10^{-10}[/tex] mm apart.

Complete question :- Determine the initial acceleration of a proton and an electron when they are initially [tex]5.50\times10^{-10}[/tex] mm apart, a typical atomic distance. Provide the answer in meters per second squared.

The operating coefficient or performance coefficient of a heat pump is the ratio between the heating or cooling provided and the electricity consumed. The higher coefficients are equivalent to lower operating costs. The coefficient can be greater than 1, because it is a percentage of the output: losses, other than the thermal efficiency ratio: input energy. Mathematically can be written as,

[tex]\text{Coefficient of performance} = \frac{\text{Heat extracted}}{\text{Electrical energy supplied}}[/tex]

[tex]COP = \frac{Q}{E}\rightarrow Q =COP\times E[/tex]

Replacing,

[tex]Q = 2.1\times 1.82kJ[/tex]

[tex]Q = 3.822kJ[/tex]

Therefore the heat is 3.822kJ

Explanation:

Work done by gravity is given by the formula,

           W = [tex]\rho A (h_{1} - h)g (h - h_{2})[/tex] ......... (1)

It is known that when levels are same then height of the liquid is as follows.

           h = [tex]\frac{h_{1} + h_{2}}{2}[/tex] ......... (2)

Putting value of equation (2) in equation (1) the overall formula will be as follows.

       W = [tex]\frac{1}{4} \rho gA(h_{1} - h_{2})^{2})[/tex]

           = [tex]\frac{1}{4} \times 1.23 g/cm^{3} \times 9.80 m/s^{2} \times 3.89 \times 10^{-4} m^{2}(1.76 m - 0.993 m)^{2})[/tex]

           = 0.689 J

Thus, we can conclude that the work done by the gravitational force in equalizing the levels when the two vessels are connected is 0.689 J.

Answer:

Gemini

Explanation:

Zodiac Constellation or Sun sign is the constellation in which the Sun resides on the date of birth of a person. Throughout the year the Sun crosses across 12 constellations thus there are 12 Sun-signs. Though astronomically the Sun crosses across 13 constellations so there should be 13 zodiacs but most of the astrologers do not accept this. On the date of June 20, the Sagittarius which is a summer constellation and a zodiac can be seen high up in the sky in the night time. At this time the Sun will be in a constellation which is almost opposite to Sagittarius in the celestial sphere. That constellation is Gemini. Thus for a person born on 20 June, zodiac will be Gemini.

Your zodiac constellation according to modern Western astrology would be Gemini, not Sagittarius, if you see Sagittarius high in the sky around your birthday on June 20. This is due to the precession of constellations, which has caused a shift of about 1/12th of the zodiac, or roughly the width of one sign, since the astrological signs were first designated over 2000 years ago.

If you see Sagittarius high in your night sky on June 20 and today is your birthday, which would mean your birthday is also around June 20, your zodiac constellation is not Sagittarius but the one that precedes it. Due to precession, the sun signs are about a month behind the real constellations. Since Sagittarius is the zodiac for those born around December 22 to January 19, if we go back one sign from Sagittarius, we find Scorpio, which is for those born around October 23 to November 21. However, due to precession, you would actually be a Gemini by the sun sign astrology, which is designated for those born around May 21 to June 20. Another important aspect to consider here is the difference between astronomy and astrology: while astronomy is a scientific study of celestial bodies and the universe, astrology is a belief system that suggests a relationship between the positions of celestial bodies and events on Earth. Even though astrology assigns a sign to the dates when it was first set up, the actual constellations have since shifted.

Answers:

a) [tex]1.05 rad/s[/tex]

b) [tex]1.38 J[/tex]

Explanation:

a) Final angular velocity :

Before solving this part, we have to stay clear that the angular momentum [tex]L[/tex] is conserved, since we are dealing with circular motion, then:

[tex]L_{o}=L_{f}[/tex]

Hence:

[tex]I_{i} \omega_{i}=I_{f} \omega_{f}[/tex] (1)

Where:

[tex]I_{i}[/tex] is the initial moment of inertia of the system

[tex]\omega_{i}=0.73 rad/s[/tex] is the initial angular velocity

[tex]I_{f}[/tex] is the final moment of inertia of the system

[tex]\omega_{f}[/tex] is the final angular velocity

But first, we have to find [tex]I_{i}[/tex] and [tex]I_{f}[/tex]:

[tex]I_{i}=I_{s}+2mr_{i}^{2}[/tex] (2)

[tex]I_{f}=I_{s}+2mr_{f}^{2}[/tex] (3)

Where:

[tex]I_{s}=8 kgm^{2}[/tex] is the student's moment of inertia

[tex]m=2 kg[/tex] is the mass of each object

[tex]r_{i}=1 m[/tex] is the initial radius

[tex]r_{f}=0.28 m[/tex] is the final radius

Then:

[tex]I_{i}=8 kgm^{2}+2(2 kg)(1 m)^{2}=12 kgm^{2}[/tex] (4)

[tex]I_{f}=8 kgm^{2}+2(2 kg)(0.28 m)^{2}=8.31 kgm^{2}[/tex] (5)

Substituting the results of (4) and (5) in (1):

[tex](12 kgm^{2}) (0.73 rad/s)=8.31 kgm^{2}\omega_{f}[/tex] (6)

Finding [tex]\omega_{f}[/tex]:

[tex]\omega_{f}=1.05 rad/s[/tex]  (7) This is the final angular speed

b) Change in kinetic energy:

The rotational kinetic energy is defined as:

[tex]K=\frac{1}{2}I \omega^{2}[/tex] (8)

And the change in kinetic energy is:

[tex]\Delta K=\frac{1}{2}I_{f} \omega_{f}^{2}-\frac{1}{2}I_{i} \omega_{i}^{2}[/tex] (9)

Since we already calculated these values, we can solve (9):

[tex]\Delta K=\frac{1}{2}(8.31 kgm^{2}) (1.05 rad/s)^{2}-\frac{1}{2}(12 kgm^{2}) (0.73 rad/s)^{2}[/tex] (10)

Finally:

[tex]\Delta K=1.38 J[/tex] This is the change in kinetic energy

To find the final angular speed of the rotating student who pulls objects closer, we apply the conservation of angular momentum, which allows us to solve for the new angular speed. To calculate the change in kinetic energy, we compare the initial and final kinetic energies of the system, considering the changes in moment of inertia and angular speed.

The question involves the concept of conservation of angular momentum and relates to the change in angular speed when a student on a rotating stool moves two 2 kg objects closer to the axis of rotation. Initially, the objects are 1 m away from the rotation axis and the system rotates with an angular speed of 0.73 rad/sec. The moment of inertia (I) of the student plus the stool is 8 kg·m2, and the student pulls the objects to a radius of 0.28 m.

Using conservation of angular momentum, we can say that the initial angular momentum (Linitial) equals the final angular momentum (Lfinal). The formula for angular momentum is L = I·ω, where ω is the angular speed. Therefore:

Linitial = (I + 2·m·rinitial2)·ωinitial

Lfinal = (I + 2·m·rfinal2)·ωfinal

Plugging in the values:

8 kg·m2 + 2(2 kg)(1 m)2)· 0.73 rad/s = (8 kg·m2 + 2(2 kg)(0.28 m)2)·ωfinal

Solving for ωfinal, we find the final angular speed of the student.

For the change in kinetic energy (KE), we calculate the initial and final kinetic energies using KE = (1/2)Iω2, and then find the difference between them to get the change in kinetic energy.

According to the principles of physics, the magnitude of the electric field at the center of the square depends on the charge and location of the beads. Oppositely charged beads cancel out, while same-charged beads add up, resulting in a larger field.

The configurations need to be analyzed based on the direction of the electric field each bead creates at the center of the square. As known, the electric field generated by a positive charge points away from it, while for a negative charge it points towards it. In configurations where positive and negative charges are opposite each other, their fields at the center will cancel each other out, leaving a smaller net field. In a scenario where all beads are positively charged or all are negatively charged, the electric field at the center will be maximized due to the fields all adding up. Field magnitude is always proportional to the amount of charge - in this case +q or −q.

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a) The expression in terms of A , B and t for the velocity of the particle as a function of time is v ( t ) = A + 2Bt.

b) The particle's velocity is zero at time t = 0.379 seconds.

Given data:

a)

The position function is represented as [tex]x ( t ) = At + Bt^2[/tex].

Velocity is the derivative of position with respect to time. To find the velocity function, we need to differentiate the position function.

So, [tex]v(t)=\frac{d(x(t))}{dt}[/tex].

[tex]v(t)=\frac{d(At+Bt^2)}{dt}[/tex]

[tex]v(t)=A+2Bt[/tex].

So, the expression is [tex]v(t)=A+2Bt[/tex].

Now, the value of A = -4.1 m/s and B = 5.4 m/s².

Substituting the values in the expression:

v(t) = -4.1 + 10.8t

b)

The velocity of the particle is 0.

So, v(t) = 0

-4.1 + 10.8t = 0

t = 0.379 seconds

Hence, the time is 0.379 seconds where the particle velocity is 0.

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The velocity of the particle as a function of time is obtained by differentiating the position function, yielding v(t) = A + 2Bt. Setting this equal to zero and solving for t will give the time at which the particle's velocity is zero, which for given values of A and B, is about 0.38 s

The motion of a particle is described by the function x(t) = A t + B t2. To find the velocity of the particle as a function of time, we will differentiate the function x(t) with respect to time (t). This is because velocity is the rate of change of the position of an object.

The derivative of x(t) yields: v(t) = A + 2Bt.

To find the time at which the particle's velocity is zero, simply set the velocity function v(t) equal to zero and solve for t: 0 = A + 2Bt.

After doing the math with the provided values for A and B, we get t = -A / (2B) = -(-4.1 m/s) / (2*(5.4 m/s2)) = 0.38 s.

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