The volume of a fluid in a tank is 0.25 m3 . of the specific gravity of the fluid is 2.0. Determine the mass of the fluid. Given the density of water is 1000 kg/m3. Express the answer in Kg.

Answer:

∆u =-111.8 kJ/kg

T_fin=-10.09℃

Explanation:

note:

solution is attached in word file due to some technical issue in mathematical equation. please find the attached documents.

Answer:

494.262996942 N

Upward

494.262996942 N

Downward

Explanation:

W = Weight of the firefighter = 712 N

g = Acceleration due to gravity = 9.81 m/s²

Mass is given by

[tex]m=\dfrac{W}{g}\\\Rightarrow m=\dfrac{712}{9.81}\\\Rightarrow m=72.5790010194\ kg[/tex]

Force is given by

[tex]T-W=-ma\\\Rightarrow T=W-ma\\\Rightarrow T=712-72.5790010194\times 3\\\Rightarrow T=494.262996942\ N[/tex]

The force on the firefighter is 494.262996942 N

directed upward

On the pole the force will be the same 494.262996942 N

But the direction will be downward

Final answer:

The magnitude of the vertical force on the firefighter from the pole is determined by the net force required for the given downward acceleration. The direction of the force the firefighter applies to the pole is downward while the reaction force from the pole is upward, with the magnitude being the same for both forces.

Explanation:

The question relates to the physics concept called Newton's second law of motion, which states that the acceleration of an object is directly proportional to the net force acting on the object and inversely proportional to the object's mass. The law is usually expressed by the equation F = ma, where F is the net force, m is the mass, and a is the acceleration.

To find the magnitude of the vertical force on the firefighter from the pole, we need to take into account both the downward force due to gravity (which is the firefighter's weight) and the additional force needed to produce the acceleration. If the firefighter weighs 712 N and has an acceleration of 3.00 m/s2 downward, we can calculate the force exerted on the pole using F = ma. The mass (m) of the firefighter can be obtained by dividing the weight (W) by the acceleration due to gravity (g), m = W/g. From F = ma, we get F = (W/g)a. Inserting the given values, we have F = (712 N/9.8 m/s2) × 3.00 m/s2. The resulting force is less than the weight of the firefighter because the net force is reduced by the downward acceleration. The direction of the force exerted by the firefighter on the pole is downward since the firefighter is sliding down.

The reaction force exerted on the firefighter by the pole, according to Newton's third law, is equal in magnitude but opposite in direction to the force exerted by the firefighter on the pole. So, the magnitude would be the same, and the direction would be upward.

Answer:

3.76 m/s

Explanation:

Instantaneous velocity: This can be defined as the velocity of an object in a non uniform motion. The S.I unit is m/s.

v' = dx(t)/dt..................... Equation 1

Where v' = instantaneous velocity, x = distance, t = time.

Given the expression,

x(t) = 28.0 m + (12.4 m/s)t - (0.0450 m/s³)t³

x(t) = 28 + 12.4t - 0.0450t³

Differentiating x(t) with respect to t.

dx(t)/dt = 12.4 - 0.135t²

dx(t)/dt = 12.4 - 0.135t²

When t = 8.00 s.

dx(t)/dt = 12.4 - 0.135(8)²

dx(t)/dt = 12.4 - 8.64

dx(t)/dt = 3.76 m/s.

Therefore,

v' = 3.76 m/s.

Hence, the instantaneous velocity = 3.76 m/s

Final answer:

To find the bird's instantaneous velocity at t = 8.00s, we differentiate its position function to get v(t) = 12.4 m/s - 0.135 m/s^2 × t^2, then substitute t = 8.00s to find v(8.00) = 3.76 m/s east.

Explanation:

The question asks for the instantaneous velocity of a bird flying due east when t = 8.00s, given the position function x(t) = 28.0 m + (12.4 m/s)t – (0.0450 m/s3)t3. To find the instantaneous velocity, we need to differentiate the position function with respect to time (t) to get the velocity function, v(t).

First, let's differentiate x(t):

Derivative of 28.0 m is 0 since it's a constant.

Derivative of (12.4 m/s)t is 12.4 m/s, as the derivative of t is 1.

Derivative of (-0.0450 m/s3)t3 is -0.135 m/s2 × t2, using the power rule for derivatives.

So, the velocity function is v(t) = 12.4 m/s - 0.135 m/s2 × t2. To find the instantaneous velocity at t = 8.00s, we plug in t = 8.00 into the velocity function:

v(8.00) = 12.4 m/s - 0.135 m/s2 × (8.002)

Calculating this gives us:

v(8.00) = 12.4 m/s - 0.135 m/s2 × 64.00 = 12.4 m/s - 8.64 m/s = 3.76 m/s

Therefore, the instantaneous velocity of the bird when t = 8.00s is 3.76 m/s east.

Answer:

Final velocity of the elevator will be 4.453 m/sec

Explanation:

Let mass is m

Acceleration due to gravity is g m/sec^2

Distance s = 2.2 m

As the elevator is moving upward so net force on elevator

[tex]F=mg+ma[/tex]

So according to question

[tex]1.46mg=mg+ma[/tex]

0.46 mg = ma

a = 0.46 g

a = 0.46×9.8 = 4.508 [tex]m/sec^2[/tex]

Initial velocity of elevator is 0 m/sec

From third equation of motion

[tex]v_f^2=v_i^2+2as[/tex]

[tex]v_f^2=0^2+2\times 4.508\times 2.2[/tex]

[tex]v_f=4.453m/sec[/tex]

So final velocity of the elevator will be 4.453 m/sec

Answer:

75.5g

Explanation:

From the ionic equation, we can write

[tex]CU^{2+}+SO^{2-}_{4}\\[/tex]

next we find the number of charge

Note Q=it

for i=8.5A, t=3.75 to secs 3.75*60*60=13500secs

hence

[tex]Q=8.5*13500\\Q=114750C[/tex]

Since one faraday represent one mole of electron which equal 96500C

Hence the number of mole produced by 114750C is

114750/96500=1.2mol

The mass of copper produced is

[tex]mol=\frac{mass}{molar mass} \\mass=mole*molar mass\\mass=1.2*63.5\\mass=75.5g[/tex]

Hence the amount of copper produced is 75.5g

Answer:

75.5 g.

Explanation:

Dissociation equation:

Cu2+ + SO4^2- --> CuSO4

Q = I * t

Where,

I = current

= 8.5 A

t = time

= 3.75 hours

= 13500 s

Q = 8.5 * 13500

= 114750 C

1 faraday represent one mole of electron which equal 96500C

Number of mole of Cu2+

= 114750/96500

= 1.19 mol.

Mass = number of moles * molar mass

= 1.19 * 63.5

= 75.5 g.

The volumetric rate or flow rate of a fluid is defined as the amount of the volume of a fluid circulating on a surface per unit of time. In this case we have units given initially: Gallons and minutes. For the first part we will convert the minutes to seconds, and we will obtain the flow rate under that measure. For the second case we will convert the gallons to cubic meters and obtain the desired value. Recall the following conversion rates,

[tex]1 min = 60s[/tex]

[tex]1 U.S Gal = 0.00378541178 m^3[/tex]

If the flow rate is defined as the volume by time, the flow rate with the given values is

[tex]Q = \frac{V}{t}[/tex]

[tex]Q = \frac{50Gal}{8min}[/tex]

[tex]Q = 6.25 Gal/min[/tex]

PART A ) Converting to Gal/seconds, we have,

[tex]Q = 6.25 \frac{Gal}{min}(\frac{1min}{60s})[/tex]

[tex]Q = 0.10416Gal/s[/tex]

PART B) Converting Gal/seconds to [tex]m^3/s[/tex]

[tex]Q = 0.104116\frac{Gal}{s} (\frac{0.00378541178 m^3}{1 Gal})[/tex]

[tex]Q = 3.941*10^{-4}m^3/s[/tex]

The rate of filling the gasoline tank is 0.104 gallons per second or approximately 0.000383 cubic meters per second.

To calculate the rate at which the gasoline tank is being filled, we need to first convert the given quantities into the relevant units. Given that the gasoline tank is 50.0 gallons and it takes 8.00 minutes to fill it, the flow rate is 50/480 (since 8 min = 480 s) = 0.104 gallons/second. Since 1 US gallon = 231 cubic inches, this gives us a flow rate of 0.104 x 231 = 23.5 cubic inches per second.

To convert this to the rate in cubic meters per second, we use the fact that 1 inch = 0.254 cm and 1 cubic meter = 1,000,000 cubic cm. Therefore, 23.5 cubic inches = 23.5 x 0.254^3 cubic meters = approximately 0.000383 cubic meters per second (m^3/s).

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Answer:

[tex]E' = \frac{1}{8} E[/tex]

Explanation:

Given data:

first case

Distance of electric field from center of sphere is r_1 <R

Electric field at r_1< R

[tex]E = \frac{kQr_1}{R^3}[/tex]

second case

Distance of electric field from centre of sphere is r_1 < 2R

Electric field at r_1< 2R

[tex]E' = \frac{kQr_1}{8R^3}[/tex]

so, we have

[tex]E' = \frac{1}{8} E[/tex]

Answer:

725.2‬ N

Explanation:

Since it is not stated the scale, the person or both accelerated or experience weightlessness, the net force acting on the bathroom scale is the weight of the person acting downward as the person stands on the scale .

                       Weight = mass of a body × acceleration due to gravity

                                    = 74 kg × 9.8 m/s²

                                    = 725.2‬ N    

Answer:

(a). [tex]x=2.00\sin(3.00\pi t)[/tex], hence proved

(b). The maximum speed is 18.8 cm/s.

(c). The maximum  acceleration is 177.65 cm/s².

(d). The total distance is 12 cm.

Explanation:

Given that,

Amplitude = 2.00 cm

Frequency = 1.50 Hz

Given equation of position of the particle is

[tex]x=2.00\ sin(3.00\pit)[/tex]

(a). show that the position of the particle is given by

[tex]x=2.00\sin(3.00\pi t)[/tex]

We know the general equation of S.H.M

[tex]x=A\sin(\omega t[/tex]...(I)

At t =0, x = 0

On differentiating equation (I)

[tex]v=\dfrac{dx}{dt}[/tex]

[tex]v=A\omega\cos(\omega t)[/tex]

At t = 0, the particle moving to the right

[tex]V=A\omega[/tex] > 0

Given statement is true.

The equation of position is

[tex]x=A\sin(\omega t)[/tex]

here, [tex]\Omega= 2\pi f[/tex]

Put the value in the equation

[tex]x=2.00\sin(2\times1.50\pi t)[/tex]

[tex]x=2.00\sin(3.00\pi t)[/tex]

Hence proved.

(b). We need to calculate the maximum speed

[tex]V=A\omega\cos(\omega t)[/tex]....(II)

At t = 0,

[tex]V_{max}=A\omega[/tex]

Put the value into the formula

[tex]V_{max}=2.00\times2\pi\times1.50[/tex]

[tex]V_{max}=6\pi[/tex]

[tex]V_{max}=18.8\ cm/s[/tex]

(c). We need to calculate the maximum  acceleration

Using equation (II)

[tex]V=A\omega\cos(\omega t)[/tex]

On differentiating

[tex]a=\dfrac{dV}{dt}[/tex]

[tex]a=-A\omega^2\sin(\omega t)[/tex]

[tex]a_{max}\ when\ \sin(\omega t)\ is\ -1[/tex]

[tex]a_{max}=-A\omega^2\times-1[/tex]

[tex]a=A\omega^2[/tex]

[tex]a_{max}=2\times(3\pi)^2\approx 177.65 cm/s^2[/tex]

(d). We need to calculate the total distance traveled between t = 0 and t = 1.00 s

Using equation (II)

[tex]V=A\omega\cos(\omega t)[/tex]

On integration

[tex]\int{V}=\int_{t}^{t'}{A\omega\cos(\omega t)}[/tex]

Put the vale into the formula

[tex]\int{V}=\int_{0}^{1}{A\omega\cos(\omega t)}[/tex]

[tex]D=\int_{0}^{1}|6\pi\cos\left(3\pi t\right)|dt[/tex]

[tex]D=12\ cm[/tex]

Hence, (b). The maximum speed is 18.8 cm/s.

(c). The maximum  acceleration is 177.65 cm/s².

(d). The total distance is 12 cm.

A particle in simple harmonic motion can be described by the equation x = (2.00 cm)sin(3.00πt). The maximum speed and maximum acceleration occur when the particle is at its maximum displacement from the equilibrium position. The total distance traveled between t = 0 and t = 1.00 s is 4.00 cm.

The position of the particle in simple harmonic motion is given by the equation x = (2.00 cm)sin(3.00πt), where x represents the displacement from the equilibrium position and t represents time in seconds.

(b) The maximum speed of the particle is equal to the amplitude of the motion multiplied by the angular frequency, vmax = (2.00 cm)(2π)(1.50 Hz). The earliest time at which the particle reaches this maximum speed is when it passes through its equilibrium position, t = 0.

(c) The maximum acceleration of the particle is equal to the amplitude of the motion multiplied by the angular frequency squared, amax = (2.00 cm)(2π)(1.50 Hz)2. The earliest time at which the particle reaches this maximum acceleration is when it is at its maximum displacement from the equilibrium position, which occurs at t = 0.

(d) To find the total distance traveled between t = 0 and t = 1.00 s, we calculate the area under the velocity versus time graph. Since the particle is in simple harmonic motion, the velocity varies sinusoidally, and the total distance traveled is equal to two times the amplitude of the motion, dtotal = 2(2.00 cm) = 4.00 cm.

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Answer:

Electric field, [tex]E=6.02\times 10^{-3}\ N/C[/tex] to the west direction.                  

Explanation:

Given that,

Acceleration of the electron, [tex]a=1.06\times 10^9\ m/s^2[/tex] (eastwards)

We need to find the magnitude and direction of the electric field. From Newton's law and electrostatic force,

ma = qE

[tex]E=\dfrac{ma}{q}[/tex]

[tex]E=\dfrac{9.1\times 10^{-31}\times 1.06\times 10^9}{1.6\times 10^{-19}}[/tex]

[tex]E=6.02\times 10^{-3}\ N/C[/tex]

The direction of electric field is in opposite direction of the acceleration of the electron. So, the electric field is acting in west direction.

The magnitude of the electric field accelerating an electron eastward is 6.04 × 10⁴ N/C, and its direction is westward since the electron has a negative charge.

To determine the magnitude and direction of the electric field that accelerates an electron eastward, we can use the formula F = qE, where F is the force, q is the charge of the electron, and E is the electric field strength. The force exerted on the electron can be found using Newton's second law, F = ma, where m is the mass of the electron (9.11 × 10⁻³¹ kg) and a is the acceleration (1.06 × 10⁹ m/s²). Thus, F = (9.11 × 10⁻³¹ kg)(1.06 × 10⁹ m/s²) = 9.66 × 10⁻¹¹ N. We then solve for E by rearranging the formula to E = F/q, with q being the charge of an electron (-1.60 × 10⁻¹¹ C). Thus, E = (9.66 × 10⁻¹¹ N) / (-1.60 × 10⁻¹¹ C) = -6.04 × 10⁴ N/C. The negative sign indicates that the electric field points westward since it accelerates the electron eastward, and the electron has a negative charge.

Answer:

[tex]|R|=4.373km[/tex]

Explanation:

Given data

For first plate let it be p₁

[tex]p_{z1}=1162 m\\ p_{x1}=10.1km\\\alpha _{1}=34.2^{o}[/tex]

For second plate let it be p₂

[tex]p_{z2}=4162 m\\p_{x2}=9.5km\\\alpha _{2}=21.5^{o}[/tex]

To find

Distance R between them

Solution

To find distance between two plates first we need to find p₁ and p₂

Finding p₁

According to vector algebra

[tex]p_{1}=p_{x1}i+p_{y1}j+p_{z1}k\\ as\\tan\alpha =tan(34.2^{o} )=(p_{y1}/p_{x1})\\p_{y1}=10.1tan(34.2^{o} )\\p_{y1}=6.864km[/tex]

So we get

[tex]p_{1}=-10.1i-6.864j+1.162k[/tex]

Now to find p₂

[tex]p_{2}=p_{x2}i+p_{y2}j+p_{z2}k\\ as\\tan\alpha =tan(21.5^{o} )=(p_{y2}/p_{x2})\\p_{y2}=9.5tan(21.5^{o} )\\p_{y2}=3.74km[/tex]

So we get

[tex]p_{2}=-9.5i-3.74j+4.162k[/tex]

Now for distance R

According to vector algebra the position vector R between p₁ and p₂

[tex]R=p_{1}-p_{2}\\ R=(p_{x1}-p_{x2})i+(p_{y1}-p_{y2})j+(p_{z1}-p_{z2})k\\R=(-10.1-(-9.5))i+(-6.864-(-3.74))j+(1.162-4.162)k\\R=-0.6i-3.124j-3k\\|R|=\sqrt{(0.6)^{2}+(3.124)^{2}+(3)^{2} }\\ |R|=4.373km[/tex]

To find the distance between two planes, calculate the horizontal distances using the given angles and distances. Use the Pythagorean theorem to find the distance by applying the formula. The distance between the two planes is approximately 14.1 km.

To find the distance between the two planes, we can use the Pythagorean theorem. First, we calculate the horizontal distances using the given angles and distances. For the first plane, the horizontal distance is 10.1 km * cos(34.2°), and for the second plane, it is 9.5 km * cos(21.5°). Next, we use the Pythagorean theorem to find the distance between the two planes: d = √((9.5 km * cos(21.5°))^2 + (10.1 km * cos(34.2°))^2).

Calculating the values, we get d ≈ 14.1 km.

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Answer:

[tex]W = -2.76\times 10^{-9}~J[/tex]

Explanation:

The work done on the capacitor is equal to the difference in potential energy stored in the capacitor in two different cases.

The potential energy is given by the following formula:

[tex]U = \frac{1}{2}CV^2[/tex]

where C can be calculated using the plate separation and area.

[tex]C = \epsilon\frac{A}{d} = \epsilon\frac{0.0314}{0.01} = 3.14\epsilon[/tex]

Therefore, the potential energy in the first case is

[tex]U = \frac{1}{2}3.14\epsilon (20)^2 = 628\epsilon[/tex]

In the second case:

[tex]C_2 = \epsilon\frac{A}{d} = \epsilon\frac{0.0314}{0.02} = 1.57\epsilon\\U = \frac{1}{2}C_2 V^2 = \frac{1}{2}1.57\epsilon (20)^2 = 314\epsilon[/tex]

The permittivity of the air is very close to that of vacuum, which is 8.8 x 10^-12.

So, the difference in the potential energy is

[tex]W = U_2 - U_1 = \epsilon(314 - 628) = -314 \times 8.8 \times 10^{-12} = -2.76\times 10^{-9}~J[/tex]

To solve this problem we will apply the concepts related to power as a function of the change of energy with respect to time. But we will consider the energy in the body equivalent to kinetic energy. The change in said energy will be the difference between the two velocity data given by half of the mass. We will first convert the given units into an international system like this

Initial Velocity,

[tex]V_i = 60km/h (\frac{1000m}{1km})(\frac{1h}{3600s})[/tex]

[tex]V_i = 16.6667m/s[/tex]

Final Velocity,

[tex]V_f = 100km/h (\frac{1000m}{1km})(\frac{1h}{3600s})[/tex]

[tex]V_f = 27.7778m/s[/tex]

Now Power is defined as the change of Energy over the time,

[tex]P = \frac{E}{t}[/tex]

But Energy is equal to Kinetic Energy,

[tex]P = \frac{\frac{1}{2} m\Delta v^2}{t}[/tex]

[tex]P = \frac{\frac{1}{2} m(v_f^2-v_i^2)}{t}[/tex]

Replacing,

[tex]P = \frac{\frac{1}{2} (900)(27.7778^2-16.6667^2)}{4}[/tex]

[tex]P = 56kW[/tex]

Therefore the correct answer is A.

Final answer:

The additional power needed to achieve the acceleration is 62 kW.So,option (d) 62 kW is correct.

Explanation:

To calculate the additional power needed to achieve acceleration, we can use the formula for power: Power = Force x Velocity. We know the mass of the car (900 kg), the initial velocity (60 km/h), and the final velocity (100 km/h). We can convert the velocities to m/s and calculate the force required to accelerate the car. Then, we can multiply the force by the change in velocity to find the additional power needed.

In this case, the additional power needed is approximately 62 kW. Therefore, option (d) 62 kW is the correct answer.

Answer:

The magnitude of the electric field be 171.76 N/C so that the electron misses the plate.

Explanation:

As data is incomplete here, so by seeing the complete question from the search the data is

vx_0=1.1 x 10^6

ax=0 As acceleration is zero in the horizontal axis so

Equation of motion in horizontal direction is given as

[tex]s_x=v_x_0 t[/tex]

[tex]t=\frac{s_x}{v_x}\\t=\frac{2 \times 10^{-2}}{1.1 \times 6}\\t=1.82 \times 10^{-8} s[/tex]

Now for the vertical distance

vy_o=0

than the equation of motion becomes

[tex]s_y=v_y_0 t+\frac{1}{2} at^2\\s_y=\frac{1}{2} at^2\\0.5 \times 10^{-2}=\frac{1}{2} a(1.82 \times 10^{-8})^2\\a=3.02 \times 10^{13} m/s^2[/tex]

Now using this acceleration the value of electric field is calculated as

[tex]E=\frac{F}{q}\\E=\frac{ma}{q}\\E=\frac{m_ea}{q_e}\\[/tex]

Here a is calculated above, m is the mass of electron while q is the charge of electron, substituting values in the equation

[tex]E=\frac{9.1\times 10^{-31} \times 3.02 \times 10^{13} }{1.6 \times 10^{-19}}\\E=171.76 N/C[/tex]

So the magnitude of the electric field be 171.76 N/C so that the electron misses the plate.

If the electron misses the upper plate, the magnitude of the electric field is equal to 171.88 Newton per coulomb.

Given the following data:

Scientific data:

To calculate the magnitude of the electric field:

First of all, we would determine the time taken by this electron to travel through the plates.

Mathematically, time is given by this formula:

[tex]Time = \frac{distance}{speed} \\ \\ Time = \frac{0.02}{1.10 \times 10^6} \\ \\ Time = 1.82 \times 10^{-8}\;m/s[/tex]

Next, we would find the acceleration of the electron in the vertical direction by using this formula:

[tex]a=\frac{2y}{t^2} \\ \\ a=\frac{2 \times 0.005}{(1.82 \times 10^{-8})^2}\\ \\ a=\frac{0.01}{3.31 \times 10^{-16}}\\ \\ a=3.02 \times 10^{13}\;m/s^2[/tex]

Mathematically,  the electric field is given by this formula:

[tex]E=\frac{ma}{q}[/tex]

Where:

Substituting the given parameters into the formula, we have;

[tex]E=\frac{9.1 \times 10^{-31} \times 3.02 \times 10^{13}}{1.6 \times 10^{-19}}\\ \\ E=\frac{2.75 \times 10^{-17}}{1.6 \times 10^{-19}}[/tex]

Electric field, E = 171.88 N/C.

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Complete Question:

An electron is projected with an initial speed v0 = [tex]1.6 \times 10^6[/tex] m/s into the uniform field between the parallel plates. The distance between the plates is 1 cm and the length of the plates is 2 cm. Assume that the field between the plates is uniform and directed vertically downward, and that the field outside the plates is zero. The electron enters the field at a point midway between the plates. E = N/C

(a) If the electron just misses the upper plate as it emerges from the field, find the magnitude of the electric field.

Answer:

 λ₂ = 1.8 m

Explanation:

given,

wavelength of the string 1 = 0.90 m

frequency of the string 1 = 600 Hz

wavelength of string 2 = ?

frequency of the string 2 = 300 Hz

we now,

[tex]f\ \alpha\ \dfrac{1}{\lambda}[/tex]

now,

[tex]\dfrac{f_1}{f_2}=\dfrac{\lambda_2}{\lambda_1}[/tex]

[tex]\dfrac{600}{300}=\dfrac{\lambda_2}{0.9}[/tex]

λ₂ = 2 x 0.9

 λ₂ = 1.8 m

Hence, the wavelength of the second string is equal to  λ₂ = 1.8 m

Answer:

a,b and c are true.

Explanation:

Following are true statements

a. Electric field lines and Equipotential surfaces are always mutually perpendicular is  a true statement.

b. When all charges are at rest, the surface of a conductor is always an equipotential surface.

c. An equipotential surface is a three-dimensional surface on which the electric potential is the same at every point.

Following are False statements

d. The potential energy of a test charge increases as it moves along an equipotential surface.

e. The potential energy of a test charge decreases as it moves along an equipotential surface.

Reason: A t any point in an equipotential surface, the potential is same throughout. There is no increase or decrease in potential energy as the test charge moves in an equipotential environment.

Statements a, b, and c are correct: Electric field lines and equipotential surfaces are always mutually perpendicular; when all charges are at rest, the surface of a conductor is always an equipotential surface; and an equipotential surface is a three-dimensional surface on which the electric potential is the same at every point. Statements d and e are not correct because the potential energy of a test charge does not change as it moves along an equipotential surface.

The following statements are true about electrical fields and potential:

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Answer:

0.00461031264 m/s

Explanation:

G = Gravitational constant = 6.67 × 10⁻¹¹ m³/kgs²

M = Mass of the Earth =  6 × 10²⁴ kg

r = Distance between Earth and Sun = [tex]1.5\times 10^{11}\ m[/tex]

t = Time taken = 3 days

Acceleration is given by

[tex]a=\dfrac{GM}{r^2}\\\Rightarrow a=\dfrac{6.67\times 10^{-11}\times 6\times 10^{24}}{(1.5\times 10^{11})^2}\\\Rightarrow a=1.77867\times 10^{-8}\ m/s^2[/tex]

Velocity of the star

[tex]v=u+at\\\Rightarrow v=0+1.77867\times 10^{-8}\times 3\times 24\times 60\times 60\\\Rightarrow v=0.00461031264\ m/s[/tex]

The Sun's speed will be 0.00461031264 m/s

Answer:

W = 17,700 lb

Explanation:

given,

Weight limit of the Bridge = 50,000 lb

Weight of empty truck = 19800 lb

Weight on the back of the truck = 12500 lb

Now,

Total weight of truck + cargo

 = Weight of empty truck + Weight on the back of the truck

 = 19800 + 12500

 = 32300 lb

Weight of cargo which is still allowed.

W = weight limit - weight of the system at present

W = 50000 - 32300

W = 17,700 lb

Weight Truck can still carry is equal to W = 17,700 lb

Answer:

The magnitude of electric field at the center of each ring is 129.96 N/C

Explanation:

As per the question:

The diameter of the ring , d = 10 cm = 0.1 m

Radius, [tex]r = \frac{d}{2} = \frac{0.1}{2} = 0.05\ m[/tex]

Separation between the rings, d = 20.0 cm = 0.20 m

Charge on a ring, q = +20 nC = [tex]20\times 10^{- 9}\ C[/tex]

Now,

The electric field at the center of either ring is given by:

[tex]E = \frac{1}{4\pi \epsilon_{o}}\frac{qd}{(d^{2} + r^{2})^{\frac{3}{2}}}[/tex]

where

[tex]\frac{1}{4\pi \epsilon_{o}} = 9\times 10^{9}[/tex]

Thus

[tex]E = 9\times 10^{9}\times \frac{20\times 10^{- 9}\times 0.20}{(0.20^{2} + 0.05^{2})^{\frac{3}{2}}}[/tex]

E = 129.96 N/C

Final answer:

The amplitude of the oscillator is 1.989 m.

Explanation:

The amplitude of an oscillator can be determined using the initial displacement and initial velocity of the system. In this case, the initial displacement is given as 1.00 m and the initial velocity as 1.72 m/s. The amplitude, also known as the maximum displacement, is equal to the square root of the sum of the squares of the initial displacement and initial velocity.

Using the formula:

X = √(x₀² + v₀²)

Where X is the amplitude, x₀ is the initial displacement, and v₀ is the initial velocity.

Substituting the given values into the formula:

X = √(1.00² + 1.72²)

= √(1 + 2.9584)

= √3.9584

= 1.989 m

The amplitude of the oscillator is 1.989 m.

Answer:

Both choices B and C are valid

Explanation:

Sound wave are Mechanical wave. Air (or viscus fluid) is the medium of propagation. Sound is produced by the back and forth vibration of the object. Consider the vibration of object is from left to right then this back and forth vibrations of object displaces the molecules of the medium both rightward and leftward (to and fro) to the direction of the energy transport forming compression and rarefaction. This shows that the motion of the molecules of the medium is both parallel (and anti-parallel) to the direction of the sound wave propagation.

Final answer:

The motion of the particles of a medium in a sound wave is parallel to the direction of the wave motion.

Explanation:

A sound wave is a longitudinal wave, which means that the motion of the particles of the medium is parallel to the direction of the wave motion. In other words, the particles of the medium vibrate or oscillate back and forth in the same direction that the sound wave is traveling. This can be compared to compressing and stretching a coiled spring. As the wave propagates through the medium, it creates zones of compression and rarefaction, causing the air molecules to move in the same direction as the sound wave.

Answer:

[tex]F=3.2345*10^{-10}N[/tex]

Explanation:

Given data

Distance r=7.5×10⁻⁹m

Charge of electron -e= -1.6×10⁻¹⁹C

Charge of proton e=1.6×10⁻¹⁹C

To find

Electric force F

Solution

From Coulombs law we know that:

[tex]F=K\frac{q_{1}q_{2} }{r^{2} }[/tex]

q₁ is charge of electron

q₂ is the charge of gold nucleus which contains 79 positively charge protons and 118 neutral neutrons.  

The Charge of single proton e=1.6×10⁻¹⁹C

79 proton charge q₂=79×1.6×10⁻¹⁹=1.264×10⁻¹⁷C

So

[tex]F=\frac{1}{4\pi *8.85*10^{-12} } \frac{-1.6*10^{-19}*1.264*10^{-17}}{(7.5*10^{-9})^{2} }\\ F=3.2345*10^{-10}N[/tex]

The magnitude of the electric force exerted by the gold nucleus on the electron [tex]3.235\times10^{-10}\rm N[/tex].

Electric force is the force of attraction or repulsion between two charged particles. It can be given as,

[tex]F=K\dfrac{q_1\times q_2}{r^2}[/tex]

Here [tex]k[/tex] is coulomb's constant, [tex]q[/tex] is charge on the objects and [tex]r[/tex]  is the distance between two objects.

Given information-

The number of proton in gold atom is 79.

The number of neutrons in gold atom is 118.

The distance of the electron from the nucleus is [tex]7.5 \times 10^{-9} \rm m[/tex].[tex]r[/tex]

The charge on electron is [tex]-1.6\times 10^{-19} C[/tex] and the charge on the proton is [tex]1.6\times 10^{-19} C[/tex].

Put the values in the above equation as,

[tex]F=8.98\times10^9\times\dfrac{(79\times1.6\times10^{-19})\times1.6\times10^{-19}}{(7.5\times10^{-19})^2}\\F=3.235\times10^{-10}\rm N[/tex]

Hence the magnitude of the electric force exerted by the gold nucleus on the electron [tex]3.235\times10^{-10}\rm N[/tex].

Learn more about the electric force here;

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Answer:

a) The acceleration of the rocket is 249 m/s².

b) The acceleration of the rocket is 25 times the acceleration of a free-falling body (25 g),

c) The distance traveled in 0.900 s was 101 m.

d) The figures are not consistent. The acceleration of the rocket was 20 g.

Explanation:

Hi there!

a) To calculate the acceleration of the rocket let's use the equation of velocity of the rocket:

v = v0 + a · t

Where:

v = velocity of the rocket.

v0 = initial velocity.

a = acceleration.

t = time.

We know that at t = 0.900 s, v = 224 m/s. The initial velocity, v0, is zero because the rocket starts from rest.

v = v0 + a · t

Solving for a:

(v - v0) / t = a

224 m/s / 0.900 s = a

a = 249 m/s²

The acceleration of the rocket is 249 m/s²

b) The acceleration of gravity is ≅ 10 m/s². The ratio of the acceleration of the rocket to the acceleration of gravity will be:

249 m/s² / 10 m/s² = 25

So, the acceleration of the rocket is 25 times the acceleration of gravity or 25 g.

c) The equation of traveled distance is the following:

x = x0 + v0 · t + 1/2 · a · t²

Where:

x = position of the rocket at time t.

x0 = initial position.

v0 = initial velocity.

t = time.

a = acceleration.

Since x0 and v0 are equal to zero, then, the equation of position gets reduced to:

x = 1/2 · a · t²

x = 1/2 · 249 m/s² · (0.900 s)²

x = 101 m

The distance traveled in 0.900 s was 101 m.

d) Now, using the equation of velocity, let's calculate the acceleration. We know that at 1.40 s the velocity of the rocket is zero and that the initial velocity is 283 m/s.

v = v0 + a · t

0 m/s = 283 m/s + a · 1.40 s

-283 m/s / 1.40 s = a

a = -202 m/s²

The figures are not consistent because 40 g is equal to an acceleration of 400 m/s² and the magnitude of the acceleration of the rocket was ≅20 g.

Answer:

a. They both reach at the same time.

Explanation:

On a frictionless incline, the only force that moves the person downwards is the x-component of the persons weight. (x-direction is the direction along the incline.)

[tex]F = mg\sin(\theta)[/tex]

Here, θ is the angle of the incline above horizontal.

This force is equal to 'ma' according to Newton's Second Law.

Comparing the weights of the two persons gives

[tex]F_1 = 85g\sin(\theta) = 85a_1\\F_2 = 75g\sin(\theta) = 75a_1\\a_1 = g\sin(\theta)\\a_2 = g\sin(\theta)[/tex]

Since the accelerations of both persons are the same, they reach the bottom at the same time.

The crucial point here is that the acceleration on a frictionless incline is independent from the mass of the object. If there were friction on the surface, then the person with smaller mass would reach the bottom first.

Final answer:

The magnitude of the average velocity equals the average speed when the direction of motion doesn't change and the speed is constant. These conditions are met in straightforward travel without directional change or speed variations. For round trips or trips involving direction changes, the average speed may differ from the magnitude of the average velocity.

Explanation:

The conditions under which the magnitude of the average velocity equals the average speed occur when the motion does not involve a change in direction. The average speed is calculated by dividing the total distance traveled by the elapsed time, while the magnitude of the average velocity is the total displacement divided by the elapsed time. For these two quantities to be the same, the direction of travel must remain constant, meaning there is no reversal or change in direction.

When you take a road trip and do not change direction, and your speed is consistent, then your average speed is equal to the magnitude of the average velocity. However, if you ended up back at your starting point, despite having moved, your displacement would be zero, and hence so would your average velocity, even though your average speed is greater than zero.

If you're calculating the ratio of the total distance as shown on the car's odometer to the time of the trip, you're calculating average speed. The speedometer of a car measures instantaneous speed, not velocity, because it does not provide information about direction.

To find the spring constant (k) of the spring, the potential energy of the falling safe is equated with the elastic potential energy of the spring at maximum compression. The calculated spring constant is approximately 20641 N/m.

To determine the spring constant of the spring, we can use the conservation of energy principle. Since the safe falls from a height onto the spring and compresses it, we can equate the potential energy of the safe at the height to the elastic potential energy stored in the spring at maximum compression.

The potential energy (PE) of the safe when it is above the spring is given by PE = mgh, where m is the mass (1100 kg), g is the acceleration due to gravity (9.81 m/s2), and h is the height (2.4 m). The elastic potential energy (EPE) stored in the spring when compressed is given by [tex]EPE = (1/2)kx^2[/tex], where k is the spring constant we need to find, and x is the compression distance (0.52 m).

Equating these two energies, [tex]mgh = (1/2)kx^2[/tex], we solve for the spring constant (k). Plugging in the values gives us:

[tex]1100 kg * 9.81 m/s^2 * 2.4 m = (1/2)k * (0.52 m)^2[/tex]

Solving for k, we get:

[tex]k = (1100 kg * 9.81 m/s^2 * 2.4 m) / (0.5 * (0.52 m)^2)[/tex]

After calculating, we find that the spring constant k is approximately 20641 N/m.

The spring constant k of the spring is 232614 N/m.

Use energy conservation: lost gravitational energy equals stored elastic potential energy in spring.

Gravitational Potential Energy (GPE) lost by the safe:

Height dropped by safe: [tex]\( h = 2.4 \, \text{m} + 0.52 \, \text{m} = 2.92 \, \text{m} \)[/tex]

Mass of safe (m): 1100 kg

Acceleration due to gravity [tex](\( g \)): \( 9.81 \, \text{m/s}^2 \)[/tex]

GPE lost = [tex]\( mgh = 1100 \times 9.81 \times 2.92 \)[/tex]

Elastic Potential Energy (EPE) stored in the spring:

Compression of spring [tex](\( x \)): 0.52 m (52 cm)[/tex]

EPE stored = [tex]\( \frac{1}{2} k x^2 \)[/tex]

Set lost gravitational potential energy equal to spring's stored elastic energy.

[tex]\[ mgh = \frac{1}{2} k x^2 \][/tex]

Rearrange the formula to solve for k:

[tex]\[ k = \frac{2mgh}{x^2} \][/tex]

Substitute the values:

[tex]\[ k = \frac{2 \times 1100 \times 9.81 \times 2.92}{(0.52)^2} \][/tex]

[tex]\[ k = \frac{2 \times 1100 \times 9.81 \times 2.92}{0.2704} \][/tex]

[tex]\[ k = \frac{62898.672}{0.2704} \][/tex]

[tex]\[ k \approx 232613.6 \, \text{N/m} \][/tex]

Answer:

The potential at point P is -111 Volt.

Explanation:

Given that,

Distance = 0.033 m

Work = 111 ev

Suppose what is the potential at point P?

We need to calculate the potential at point P

Using formula of potential

[tex]W=qV[/tex]

[tex]V=\dfrac{W}{q}[/tex]

Where, W = work

q = charge

Put the value into the formula

[tex]V= \dfrac{111\times1.6\times10^{-19}}{-1.6\times10^{-19}}[/tex]

[tex]V=-111\ V[/tex]

Hence, The potential at point P is -111 Volt.

The question is about the process of bringing an electron from rest to rest near a fixed charge q at a specific distance, and the energy required for this process. The answer explains the equation for the energy required per unit charge and the concept of electron volts (eV) as an energy unit.

The question is asking about the process of bringing an electron from rest at an infinite distance to rest at a point located a distance of 0.033 m from a fixed charge q using an external agent. The process requires 111 eV of energy to perform the necessary work. The equation for the energy required per unit charge is given, which relates the potential energy to the charge and distance. The concept of electron volts (eV) as an energy unit is also mentioned.

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Answer:

The general equation of movement in fluids is obtained from the application, at fluid volumes, of the principle of conservation of the amount of linear movement. This principle establishes that the variation over time of the amount of linear movement of a fluid volume is equal to that resulting from all forces (of volume and surface) acting on it. Expressed in This equation is called the Navier-Stokes equation.

The equation is shown in the attached file

Explanation:

The derivative of velocity with respect to time determines the change in the velocity of a particle of the fluid as it moves in space. It also includes convective acceleration, expressed by a nonlinear term that comes from convective inertia forces). With this equation, Stokes studied the motion of an infinite incompressible viscous fluid at rest at infinity, and in which a solid sphere of radius r makes a rectilinear and uniform translational motion of velocity v. It assumes that there are no external forces and that the movement of the fluid relative to a reference system on the sphere is stationary. Stokes' approach consists in neglecting the nonlinear term (associated with inertial forces due to convective acceleration).

Answer:

a)The initial velocity of the puck is 20 ft/s.

b)The coefficient of friction is 0.062.

Explanation:

Hi there!

a)For this problem let's use the equations of position and velocity of an object moving in a straight line with constant acceleration:

x = x0 + v0 · t + 1/2 · a · t²

v = v0 + a · t

Where:

x = position of the puck after a time t.

x0 = initial position.

v0 = initial velocity.

a = acceleration.

t = time.

v = velocity of the puck at a time t.

Let's place the origin of the frame of reference at the point where the puck is hit so that x0 = 0.

We know that at t = 10 s the velocity of the puck is zero (v = 0) and its position is 100 ft (x = 100 ft):

100 ft = v0 · 10 s + 1/2 · a · (10 s)²

0 = v0 + a · 10 s

We have a system of two equations with two unknowns, so, we can solve the system.

Solving for v0 in the second equation:

0 = v0 + a · 10 s

v0 = -a · 10 s

Replacing v0 in the first equation:

100 ft = (-a · 10 s) · 10 s + 1/2 · a · (10 s)²

100 ft = -50 s² · a

100 ft / -50 s² = a

a = -2.0 ft/s²

Then the initial velocity of the puck will be:

v0 = -a · 10 s

v0 = -(-2.0 ft/s²) · 10 s

v0 = 20 ft/s

The initial velocity of the puck is 20 ft/s.

b) The friction force is calculated as follows:

Fr = N · μ

Where:

Fr = friction force.

N = normal force.

μ = coefficient of friction.

Since the only vertical forces acting on the puck are the weight of the puck and the normal force and since the puck is not being accelerated in the vertical direction, then, the normal force is equal to the weight of the puck. The weight (W) is calculated as follows:

W = m · g

Where "m" is the mass of the puck and "g" is the acceleration due to gravity (32.2 ft/s²).

Then the friction force can be calculated as follows:

Fr = m · g · μ

Since the acceleration of the puck is provided only by the friction force, then, due to Newton's second law:

Fr = m · a

Where "m" is the mass of the puck and "a" its acceleration. Then:

Fr = m · g · μ

Fr = m · a

m · g · μ = m · a

μ = a/g

μ = 2.0 ft/s² / 32.2 ft/s²

μ = 0.062

The coefficient of friction is 0.062.