A 1200-kg car, initially moving at 20 m/s, comes to a stop at a red light over a time of 3 s from the moment the driver hit the brakes. What is the impulse delivered to the car by the static friction force (assumed constant) between the road and the tires

Answer:

58.8 N

Explanation:

Let 'F₁' be the force by front arm and 'F₂' be the force by back arm.

Given:

Mass of the rod (m) = 3.00 kg

Length of the pole (L) = 4.50 m

Acceleration due to gravity (g) = 9.8 m/s²

Distance of 'F₁' from one end of pole (d₁) = 0.750 m

'F₂' acts on the end. So, distance between 'F₁' and 'F₂' = 0.750 m

Now, weight of the pole acts at the center of pole.

Now, distance of center of pole from 'F₁' is given as:

d₂ = (L ÷ 2) - d₁

[tex]d_2=\frac{4.50}{2}-0.75=1.5\ m[/tex]

Now, as the pole is held horizontally straight, the moment about the point of application of force 'F₁' is zero for equilibrium of the pole.

So, Anticlockwise moment  = clockwise moment

[tex]F_2\times d_1=mg\times d_2\\\\F_2=\frac{mg\times d_2}{d_1}[/tex]

Plug in the given values and solve for 'F₂'. This gives,

[tex]F_2=\frac{3.00\ kg\times 9.8\ m/s^2\times 1.5\ m}{0.75\ m}\\\\F_2=\frac{44.1}{0.75}=58.8\ N[/tex]

Therefore, the force exerted by the back arm on the pole is 58.8 N vertically down.

Final answer:

The velocity and acceleration vectors of a baseball can change during its flight. At the ball's maximum height, its velocity is parallel to its acceleration. There is no point where the ball's velocity is perpendicular to its acceleration.

Explanation:

In a game of baseball, the velocity and acceleration of a ball can change as it moves through the air. (a) Yes, there is a point during the flight of the ball where its velocity is parallel to its acceleration. This occurs when the ball reaches its maximum height. At this point, the ball stops moving upward and starts moving downward. Both the velocity and acceleration vectors are directed downward and therefore parallel to each other.

(b) No, there is no point during the flight of the ball where its velocity is perpendicular to its acceleration. The orientation of the velocity and acceleration vectors will always be either parallel or antiparallel to each other.

Answer:

intensity at a point on the circle at an angle of 4.45° from the center line is 1.77 W/m².

Explanation:

See attached picture.

The intensity at a point on the circle at an angle of 4.45° from the centerline is approximately 4.00 W/m².

To find the intensity at a point on the circle at an angle of 4.45° from the centerline, we need to consider the interference pattern created by the two transmitters.

1. Wavelength Calculation:

The wavelength [tex](\(\lambda\))[/tex] of the transmitted signal can be calculated using the frequency [tex](\(f\))[/tex] and the speed of light c.

[tex]\[ \lambda = \frac{c}{f} \][/tex]

Given:

- [tex]\( c = 3 \times 10^8 \)[/tex] m/s (speed of light)

- [tex]\( f = 1575.42 \times 10^6 \)[/tex] Hz (frequency)

[tex]\[ \lambda = \frac{3 \times 10^8}{1575.42 \times 10^6} \approx 0.1902 \text{ meters} \][/tex]

2. Phase Difference Calculation:

The phase difference [tex](\(\Delta \phi\))[/tex] between the two waves at a point on the circle is determined by the path difference [tex](\(\Delta L\))[/tex] travelled by the waves.

The path difference is given by:

[tex]\[ \Delta L = d \sin(\theta) \][/tex]

where:

- d is the distance between the two transmitters (5.18 mm = 0.00518 m)

- [tex]\(\theta\)[/tex] is the angle from the centerline (4.45°)

[tex]\[ \Delta L = 0.00518 \sin(4.45^\circ) \approx 0.00518 \sin(0.0776) \approx 0.00518 \times 0.0775 \approx 0.0004019 \text{ meters} \][/tex]

The phase difference is:

[tex]\[ \Delta \phi = \frac{2 \pi \Delta L}{\lambda} = \frac{2 \pi \times 0.0004019}{0.1902} \approx 0.0133 \text{ radians} \][/tex]

3. Intensity Calculation:

The intensity at a given point due to two sources interfering constructively or destructively can be found using:

[tex]\[ I = I_0 \left(1 + \cos(\Delta \phi)\right) \][/tex]

Here, [tex]\(I_0\)[/tex] is the intensity when both waves interfere constructively at the centerline [tex](\(\theta = 0^\circ\))[/tex], which is given as 2.00 W/m^2.

[tex]\[ I = 2.00 \left(1 + \cos(0.0133)\right) \][/tex]

Since [tex]\(\cos(0.0133) \approx 0.9999\)[/tex]:

[tex]\(\cos(0.0133) \approx 0.9999\)[/tex]

Thus, the intensity at a point on the circle at an angle of 4.45° from the centerline is approximately 4.00 W/m².

Answer:

Explanation:

A force is provided by the magnetic field which perpendicular to both the velocity of the charge and magnetic field

F = qvB  where B is the magnetic field in tesla, q is charge and v is velocity

The potential energy is transferred into kinetic energy

PE = Vq = 1/2 mv²

v = √(2Vq/ m)

charge on the proton = 1.602 × 10 ⁻¹⁹ C and mass of a proton = 1.673 × 10⁻²⁷

v = √ (( 2 × 1.602 × 10 ⁻¹⁹ C × 0.75 × 10³V ) / (1.673 × 10⁻²⁷)) = √( 1.4363 × 10¹¹ ) = 3.79 × 10⁵ m/s

The force of magnetic field produces centripetal force

qvB = mv² /R

where  R radius = 1.80mm = 0.0018 m / 2 = 0.0009 m

qvB = ( m / R ) × (2qV /m)

cancel the common terms

vB = 2V / R

3.79 × 10⁵ m/s × B = 2 × 0.75 × 10³V / 0.0009 = 1.667 × 10⁶

B = 1.667 × 10⁶ / 3.79 × 10⁵ m/s = 4.40 T

b) magnetic field needed for the electron

qvB = mv² /R where m is the mass of an electron = 9.11 × 10⁻³¹ Kg

qB = mv/R

qB = ( 9.11 × 10⁻³¹ Kg × 3.79 × 10⁵ m/s) / 0.0009

qB  = 3.8363 × 10 ⁻²²

B = 3.8363 × 10 ⁻²² / 1.602 × 10⁻¹⁹ kg = 0.0239 T

Answer:

The minimum work per unit heat transfer will be 0.15.

Explanation:

We know the for a heat pump the coefficient of performance ([tex]C_{HP}[/tex]) is given by

[tex]C_{HP} = \dfrac{Q_{H}}{W_{in}}[/tex]

where, [tex]Q_{H}[/tex] is the magnitude of heat transfer between cyclic device and    high-temperature medium at temperature [tex]T_{H}[/tex] and [tex]W_{in}[/tex] is the required input and is given by [tex]W_{in} = Q_{H} - Q_{L}[/tex], [tex]Q_{L}[/tex] being magnitude of heat transfer between cyclic device and low-temperature [tex]T_{L}[/tex]. Therefore, from above equation we can write,

[tex]&& \dfrac{Q_{H}}{W_{in}} = \dfrac{Q_{H}}{Q_{H} - Q_{L}} = \dfrac{1}{1 - \dfrac{Q_{L}}{Q_{H}}} = \dfrac{1}{1 - \dfrac{T_{L}}{T_{H}}}[/tex]

Given, [tex]T_{L} = 460 K[/tex] and [tex]T_{H} = 540 K[/tex]. So,  the minimum work per unit heat transfer is given by

[tex]\dfrac{W_{in}}{Q_{H}} = \dfrac{T_{H} - T_{L}}{T_{H}} = \dfrac{540 - 460}{540} = 0.15[/tex]

The minimum work required to operate a heat pump between two thermal reservoirs at 460 K and 540 K is determined using the performance coefficient equation, Kp = Qh/W = Th/(Th – Tc). The efficiency of a heat pump is determined by the energy it transfers through heat and the input work required. To improve efficiency, the temperature difference between the hot and cold reservoir should be maximized.

The question pertains to the work-power balance in a heat pump operating between two thermal reservoirs. It involves the concept of thermal efficiency and the performance coefficient (Kp) of the heat pump.

Given the heat pump's thermal energy reservoirs at 460 K and 540 K, one can find the minimum work using the equation Kp = Qh/W = Th/(Th – Tc). Here, Th and Tc are the temperatures of the hot and cold reservoirs, respectively, Qh is the heat delivered to the hot reservoir, and W is the work done.

Such a heat pump operates on the principle of heat transfer of energy from a low-temperature reservoir to a high-temperature one, which requires input work. The quality of a heat pump is judged by the energy transferred by heat into the hot reservoir and the input work required.

The conservation of energy is not violated in this process, as the heat pump might extract energy from the ambient air or ground, contingent on its settings. Also, to improve the efficiency of the heat pump, the temperature of the hot reservoir should be elevated, and the cold reservoir should be lowered as per the Carnot efficiency equation.

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Explanation:

The given data is as follows.

       C = [tex]20 \times 10^{-6} F[/tex]

        R = [tex]100 \times 10^{3}[/tex] ohm

        [tex]Q_{o} = 100 \times 10^{-6}[/tex] C

          Q = [tex]13.5 \times 10^{-6} C[/tex]

Formula to calculate the time is as follows.

          [tex]Q_{t}  = Q_{o} [e^{\frac{-t}{\tau}][/tex]

       [tex]13.5 \times 10^{-6} = 100 \times 10^{-6} [e^{\frac{-t}{2}}][/tex]

               0.135 = [tex]e^{\frac{-t}{2}}[/tex]

         [tex]e^{\frac{t}{2}} = \frac{1}{0.135}[/tex]

                         = 7.407

           [tex]\frac{t}{2} = ln (7.407)[/tex]

                      t = 4.00 s

Therefore, we can conclude that time after the resistor is connected will the capacitor is 4.0 sec.

The required time to discharge the capacitor to a certain charge can be calculated by rearranging the exponential decay formula. The product of resistance and capacitance, known as the RC time constant, is a key value in this calculation. After substituting the values given into the formula, we get the required time.

The time it takes for a capacitor to discharge through a resistor can be calculated using the formula for the decay of charge on a capacitive circuit: Q = Q0 * e-t/RC, where Q is the charge at time t, Q0 is the initial charge, R is the resistance and C is the capacitance. To find the time at which the charge will be 13.5 uC, we rearrange the formula to solve for time t: t = -RC * ln(Q/Q0).

Using the values of the problem: R = 100 kΩ = 100,000 Ω, C = 20 uF = 20*10-6 F, Q0 = 100 uC = 100*10-6 C, and Q = 13.5 uC = 13.5*10-6 C, substitute these values into the equation: t = -100,000 * 20*10-6 * ln (13.5/100). Hence, the time it takes for a capacitor to discharge to a specified charge can be calculated using exponential decay formula based on the capacitor's unique RC time constant.

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The velocity at which the long jumper landed can be computed by separating the jump into horizontal and vertical components, using the initial take-off angle and the horizontal distance. The horizontal and vertical velocities are combined to find the magnitude of the total landing velocity, and the arctangent of their ratio gives the direction.

To determine the velocity at which the long jumper landed, we need to consider the projectile motion of the jump. There are two components of velocity to consider: the horizontal (vx) and the vertical (vy) at the point of landing.

First, the horizontal velocity (vx) can be found by dividing the total horizontal distance by the time of flight (t). The equation of horizontal motion is:

vx = d / t, where d is the horizontal distance.

The vertical velocity (vy) of the jumper when he lands will be the same magnitude but opposite in direction to the vertical velocity at take-off due to symmetry in projectile motion, assuming no air resistance. The vertical velocity at take-off can be calculated using the initial jump angle (Ө) and the initial velocity (v0).

Using the initial angle of 30°, the vertical component of the initial velocity at take-off (v0y) is:

v0y = v0 * sin(Ө).

Since the vertical motion is subject only to acceleration due to gravity (g), the final vertical velocity at landing (vy) will be v0y (but in the opposite direction).

The total landing velocity (v) is then found by combining these two components using the Pythagorean theorem:

v = √(vx^2 + vy^2).

The direction of the landing velocity is given by the angle made with the horizontal, found using the arctangent of the ratio of vy over vx:

θ = arctan(vy / vx).

These calculations would provide the magnitude and direction of the landing velocity.

Answer:

981.41 secs

Explanation:

Parameters given:

Voltage, V = 1.5V

Power, P = 0.204W

Number of electrons, n = 8.33 * 10^20

First, we calculate the current:

P = I*V

I = P/V

I = 0.204/1.5 = 0.136A

The total charge of 8.33 * 10^20 electrons is:

Q = 8.33 * 10^20 * 1.6023 * 10^(-19)

Q = 133.47 C

Current, I, is given as:

I = Q/t

=> t = Q/I

t = 133.47/0.136 = 981.41 secs

In physics, power is calculated using the formula P = IV, where P is power, I is current, and V is voltage. By applying this formula, along with the relationship between charge, current, and time, the time the flashlight was used can be calculated.

Power is determined by the rate at which energy is transferred, calculated as P = IV where P is power, I is current, and V is voltage. In this scenario, the power supplied is 0.204 W from a 1.50-V battery. The formula P = IV can be rearranged to find the current flowing, which is 0.136 A.

To determine the time the flashlight was used, we can utilize the relationship between charge, current, and time: Q = It. Given 8.33 x 10^20 electrons moved, we need to convert this to Coulombs by recognizing that 1 electron has a charge of approximately 1.6 x 10^-19 C. By dividing the total charge by the current, we find the time t to be approximately 1.23 x 10^19 seconds.

Answer:

note:

please find the attachment

Explanation:

A frayed cord means that the insulation of the cord is worn out which exposes the cord and when it touches any other conductor a short circuit happens which is basically the flow of very high amount of current.

If we talk about the example of frayed cord, when such short circuit happens the lamp will not turn on and a massive amount of fault current will flow throughout the path of short circuit.

Now what happens?

There are two possibilities;

Answer:

TRUE. The potential of a negatively charged conductor must be negative

Explanation:

Let's examine each statement

The positively charged proton moves in the direction of the electric field, the power and the electric field are related

                ΔU = - E ds

                [tex]U_{f}[/tex] - U₀ = - E ds

                E = (U₀ –U_{f}) / s

To have a positive electric field the initial potential must be greater than the final potential, so the proton moves from a greater potential to a smaller one.

This statement is FALSE

The second statement

The potential has the same sign as the elective charge.

This statement is TRUE

A proton tends to move to a region of higher potential is False, while The potential of a negatively charged conductor must be negative is True.

The proton is a positively charged particle and moves in the direction of the electric field lines.

In the case of a positive charge, the electric field lines are away from the charge, which will push the proton away.

We know that the potential is inversely proportional to distance.

Thus, as the proton moves away, it is going from a higher potential to a lower potential.

The same can be proven in the case of an electric field generated by a negative charge, by using proper sign convention.

So the statement that a proton tends to go from a region of low potential to a region of high potential is FALSE

The potential has the same sign as the electric charge.

So the statement that the potential of a negatively charged conductor must be negative is TRUE

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Answer:

Explanation:

Let the position of balance point be x distance from 1.5 µC on the right side . Balance point can not be on the left side or between the charges because in that case electric field by both the charges will be in the same direction .

For equilibrium of field

k x 4.75 µC / ( .35 +x )² -  k x  1.5  µC / x ² = 0

4.75 / ( .35 +x )² = 1.5 / x²

( .35 +x )² / x² = 4.75 / 1.5

( .35 +x )² / x² = 3.167

( .35 +x ) / x =  1.78

.35 + x = 1.78 x

.78 x = .35

x = .45 m ( + ve ) to the right of 1.50  µC

b ) When both the charges are positive , balance point will lie in between them because , electric field will be in opposite direction .

For equilibrium of field

k x 4.75 µC / ( .35 - x )² =   k x  1.5  µC / x ²

4.75 / ( .35 -x )² = 1.5 / x²

( .35 -x )² / x² = 4.75 / 1.5

( .35 -x )² / x² = 3.167

( .35 -x ) / x =  1.78

.35 -x = 1.78 x

2.78 x = .35

x = .126 m  ( - ve ) to the right of 1.50  µC

Answer:

Explanation:

Analysis of structure gives

a=gsinθ−μkgcosθ

Notice that all the expression are right but we want to know of we can simplify the expression further.

We want to analyse if we can still further simplify the expression,

Inspecting the Right hand side of the equation, we notice that the acceleration due to gravity is common to both side, so we can bring it out i.e.

So option a is wrong because the expression can be simplified further to

a=g(sinθ−μkcosθ)

Option b is right and the best option.

Since we are given that, g=9.8m/s²

We can as well substitute that to option a

So we will have

a=9.8metre/second²(sinθ−μkcosθ)

Also option C is correct but it is not best inserting the values of g directly without simplifying the expression first

So it will have been the best option if it was written as

a=9.8metre/second²(sinθ−μkcosθ)

So the best option is B.

Explanation:

As the given data is as follows.

         h = [tex]\frac{3}{5}d[/tex]

            = [tex]\frac{3}{5} \times (2r)[/tex]

Also, we know that r = [tex]\frac{4}{3}h[/tex]

and       Volume (V) = [tex]\frac{1}{3} \pir^{2}h[/tex]

                         = [tex]\frac{1}{3} \pi (\frac{4}{3}h)^{2} h[/tex]

                         = [tex]\frac{16}{27} \pi h^{3}[/tex]

And,  [tex]\frac{dV}{dt} = \frac{3 \times 16}{27} \pi h^{2} \frac{dh}{dt}[/tex]

         [tex]\frac{dV}{dt} = \frac{16}{9} \pi h^{2} \frac{dh}{dt}[/tex]

Putting the given values into the above formula as follows.

       [tex]\frac{dV}{dt} = \frac{16}{9} \pi h^{2} \frac{dh}{dt}[/tex]

       [tex]30 m^{3}/min = \frac{16}{9} \pi (2)^{2} \frac{dh}{dt}[/tex]    

          [tex]\frac{dh}{dt} = 1.343 m/min

or,                        = 134.3 cm/min     (as 1 m = 100 cm)

thus, we can conclude that the height changing at 134.3 cm/min when the pile is 2 m high.

Answer:

The critical angle is 41.8°.

Explanation:

The critical angle is [tex]\theta_1[/tex] for which the angle of refraction [tex]\theta_2[/tex] is 90°. From Snell's law we have

[tex]n_1sin (\theta_1 ) = n_2 sin(\theta_2)[/tex]

[tex]n_1sin (\theta_1 ) = n_2 sin(90^o)[/tex]

[tex]sin (\theta_1 ) = n_2/n_1,[/tex]

[tex]\theta_1 = sin^{-1}(\dfrac{n_2}{n_1} ).[/tex]

Putting in [tex]n_2 =1.00,[/tex] and [tex]n_1 = 1.50[/tex] we get:

[tex]\theta_1 = sin^{-1}(\dfrac{1.00}{1.500} ),[/tex]

[tex]\boxed{\theta_1 = 41.8^o}[/tex]

Thus, the critical angle is 41.8°.

Final answer:

The critical angle for light propagating from a material with an index of refraction of 1.50 to a material with an index of refraction of 1.00 is approximately 41.8 degrees.

Explanation:

The critical angle θcrit for light propagating from a material with an index of refraction (n1) of 1.50 to a material with an index of refraction (n2) of 1.00 is found using Snell's Law and the concept of total internal reflection. The formula to calculate the critical angle is θcrit = sin−1(n2/n1). Plugging in the values for the indices of refraction, we have θcrit = sin⁻¹(1.00/1.50).

Performing the calculation, we get:

θcrit = sin⁻¹(0.6667) ≈ 41.8°

Therefore, the critical angle for light traveling from a medium with an index of refraction of 1.50 to a medium with an index of refraction of 1.00 is approximately 41.8 degrees.

Answer:

[tex]L=4.8*10^{17}m[/tex]

Explanation:

Given data

Power P=4.00×10²W

Radius r=6.5×10⁴m

Voltage V=120V

To find

Length of wire L

Solution

We know that resistance of wire can be obtained from

[tex]P=\frac{V_{2}}{R}\\ R=\frac{V_{2}}{P}[/tex]

We also know that R=pL/A solving the length noting that A=πr²

and using p=100×10⁻⁸Ω.m we find that

So

[tex]L=\frac{RA}{p}\\ L=\frac{\frac{(V^{2})}{P}(\pi r^{2}) }{p} \\L=\frac{V^{2}(\pi r^{2})}{pP}\\ L=\frac{(120V)^{2}\pi (6.5*10^{4} m)^{2} }{100*10^{-8}(4.00*10^{2} W) }\\ L=4.8*10^{17}m[/tex]

Final answer:

To calculate the length of Nichrome wire needed for a heating element using 400 W at 120 V, one determines the resistance and then uses it with the wire's cross-sectional area and resistivity. Approximately 41.45 meters of wire is required.

Explanation:

To estimate the necessary length of Nichrome wire for a laboratory heater, we start by calculating the resistance using the power and voltage supplied. The formula for power (P) in terms of voltage (V) and resistance (R) is P = V2 / R. Given P = 400 W and V = 120 V, the resistance can be calculated as follows:

R = V2 / P = (1202) / 400 = 36 Ω.

Next, we use the resistivity (ρ) of Nichrome and the cross-sectional area (A) of the wire to find the length (L). The formula R = ρL / A is applicable here, where ρ for Nichrome is approximately 1.10×10-6 Ω·m and A = πr2. The radius (r) given is 6.5×10-4 m, so:

A = π(6.5×10-4)2 = 1.33×10-6 m2.

Substituting the values into the formula, we get:

L = (R · A) / ρ = (36 · 1.33×10-6) / (1.10×10-6) ≠ 41.45 m.

Thus, approximately 41.45 meters of Nichrome wire is required for the heater to use 400 W of power at 120 V.

Answer:

The distance between the two slits is 1.2mm.    

Explanation:

The physicist Thomas Young establishes, through its double slit experiment, a relationship between the interference (constructive or destructive) of a wave, the separation between the slits, the distance between the two slits to the screen and the wavelength.

[tex]\Lambda x = L\frac{\lambda}{d} [/tex]  (1)

Where [tex]\Lambda x[/tex] is the distance between two adjacent maxima, L is the distance of the screen from the slits, [tex]\lambda[/tex] is the wavelength and d is the separation between the slits.  

If light pass through two slits a diffraction pattern in a screen will be gotten, at which each bright region corresponds to a crest, a dark region to a trough, as consequence of constructive interference and destructive interference in different points of its propagation to the screen.  

Therefore, d can be isolated from equation 1.

[tex]d = L\frac{\lambda}{\Lambda x} [/tex]  (2)

Notice that it is necessary to express L and [tex]\lambda[/tex] in units of millimeters.

[tex]L = 4m \cdot \frac{1000mm}{1m}[/tex] ⇒ [tex]4000mm[/tex]

[tex]\lambda = 600nm \cdot \frac{1mm}{1x10^{6}nm}[/tex] ⇒ [tex]0.0006mm[/tex]

[tex]d = (4000mm)\frac{0.0006mm}{2mm} [/tex]

[tex]d = 1.2mm[/tex]

Hence, the distance between the two slits is 1.2mm.

Answer:

The projectile is in air for 13.03 seconds.

Explanation:

Given that,

Angle of projection of the projectile, [tex]\theta=55^{\circ}[/tex]

Initial speed of the projectile, u = 78 m/s

To find,

We need to find the time of flight of the projectile.

Solution,

It is defined as the time taken by the projectile when it is in air. It is given by the formula as :

[tex]T=\dfrac{2u\ \sin\theta}{g}[/tex]

[tex]T=\dfrac{2\times 78\ \sin(55)}{9.8}[/tex]

T = 13.03 seconds

So, the projectile is in air for 13.03 seconds.

Answer:

The work and heat transfer for this process is = 270.588 kJ

Explanation:

Take properties of air from an ideal gas table.  R = 0.287 kJ/kg-k

The Pressure-Volume relation is PV = C

T = C for isothermal process

Calculating for the work done in isothermal process

W = P₁V₁ [tex]ln[\frac{P_{1} }{P_{2} }][/tex]

   = mRT₁[tex]ln[\frac{P_{1} }{P_{2} }][/tex]      [∵pV = mRT]

   = (5) (0.287) (272.039) [tex]ln[\frac{2.0}{1.0}][/tex]

   = 270.588 kJ

Since the process is isothermal, Internal energy change is zero

ΔU = [tex]mc_{v}(T_{2} - T_{1} ) = 0[/tex]

From 1st law of thermodynamics

Q = ΔU  + W

   = 0 + 270.588

   = 270.588 kJ

To find the work and heat transfer in this process, we can apply the First Law of Thermodynamics. We can determine the work done using the equation pV = constant and calculate the heat transfer using the equation Q = CpdT.

To find the work and heat transfer in this process, we need to apply the First Law of Thermodynamics. The First Law states that the change in internal energy of a system is equal to the heat transfer into the system minus the work done by the system. In this case, since the process is isobaric (constant pressure), we can determine the work done by integrating the equation pV = constant. This will give us the equation W = p(Vf - Vi), where W is the work done, p is the pressure, and Vf and Vi are the final and initial volumes respectively.

Since the process is isobaric, the heat transfer can be calculated using the equation Q = CpdT, where Q is the heat transfer, Cp is the specific heat at constant pressure, and dT is the change in temperature. In this case, since the temperature change is not given explicitly, we can assume it to be the same as the change in internal energy, which gives us dT = dU/Cp.

By substituting the given values into the equations, we can calculate the work and heat transfer.

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When pushing the body it is necessary to break the frictional force generated by the floor. Once this frictional force is overcome, the body will begin to move. Ideally, if a constant velocity is maintained or close to this value, the acceleration that will be exerted will tend to be zero and therefore, by Newton's second law the value of the Force will also tend to minimum values.

Remember that this law tells us that

[tex]F= ma[/tex]

[tex]F= m \frac{\Delta v}{t}[/tex]

Therefore the best strategy is A. keep pushing the box forward at a steady speed

To minimize the average force applied to a box being pushed across a rough floor, you should keep pushing the box forward at a steady speed. This way, you're only balancing the frictional force, and there's no need for an extra force to accelerate the box.

To minimize the average force applied to the box on a rough floor, you would opt for method a. Keep pushing the box forward at a steady speed. This option will ensure constant velocity, meaning that the net force on the box is zero. In this case, the force you're applying is just balancing the frictional force the box experiences due to the rough floor.

Contrastingly, methods b and c involve changing the box's velocity, which necessitates an acceleration. According to Newton's second law (F=ma), a force is required for acceleration. Thus, these methods will require a greater average force compared to method a.

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Answer:

Explanation:

capacitance of each capacitor

C₀= Q₀ / V₀

V₀ = Q₀ / C₀

New total capacitance = C₀ ( 1 + K )

Common potential

= total charge / total capacitance

= 2 Q₀ / [ C₀ ( 1 + K ) ]

2 V₀ / ( 1 + K )

b )

Common potential = 2 x V₀ / ( 1 + 7.8 )

= .227  V₀

charge on capacitor with dielectric

= .227  V₀ x 7.8 C₀

= 1.77 V₀C₀

= 1.77 Q₀

Ratio required = 1.77

The new potential difference V across the capacitors and the ratio of the charge on the capacitor with the dielectric after it is inserted as compared with the initial charge can be calculated using specific equations.

(a) What is the new potential difference V across the capacitors.

To find the new potential difference V across the capacitors, we need to consider the effect of adding a dielectric. The potential difference V is given by the equation:

V = Vo/2k

where Vo is the initial potential difference and k is the dielectric constant. In this case, since the dielectric constant is 7.8, we can substitute the values and calculate the new potential difference V.

(b) If the dielectric constant is 7.8, calculate the ratio of the charge on the capacitor with the dielectric after it is inserted as compared with the initial charge.

The ratio of the charge on the capacitor with the dielectric after it is inserted as compared with the initial charge can be calculated using the equation:

Q_with_dielectric / Q_initial = k

where Q_with_dielectric is the charge on the capacitor with the dielectric and Q_initial is the initial charge. Given that k is 7.8, we can substitute the values and calculate the ratio.

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Answer:

your question is incomplete as the options are not given. I guess following is the complete question.

Which of these atoms is most likely to share electrons with other atoms?

a) chlorine (7 valence electrons)

b) calcium (2 valence electrons)

c) argon (8 valence electrons)

d) carbon (4 valence electrons)

e) potassium (1 valence electron)

The correct option is d) carbon (4 valence electrons)

Explanation:

Carbon has four electrons in its valence shell. In order to complete the 8 electrons in its valence shell carbon has to make four covalent bonds by sharing its four electrons with the other atom. Carbon atom will neither gain the electrons nor it losses the electrons to follow the octet rule. So in the above mentioned options carbon is the atom that will share maximum electrons.

→ Chlorine has 7 electrons, it will gain 1 electron. It will not do the sharing.

→ Calcium has 2 electrons, it will lost these 2 electrons to complete its shell.

→ Argon has already a completed shell. It will not react with other atom.

→ Potassium has only 1 valence electron which it will lose to complete its shell.

Answer: F

Explanation:

Answer:

[tex]F_{avg}=2096.71N[/tex]

Explanation:

As from the given data that:

m/Δt=53.9kg/s

The numerator is the mass and denominator is the change in time

Solve for change in velocity we have:

Δv=v₂-v₁

[tex]=0-38.9m/s\\=-38.9m/s[/tex]

Δv= -38.9m/s

From Newtons second law we know that:

F=ma

We can write this equation as:  

Favg=(m/Δt)Δv

Substitute the given values

So

[tex]F_{avg}=(53.9kg/s)(-38.9m/s)\\F_{avg}=-2096.71N[/tex]

Using the average velocity formula.The answer will be positive, the negative only implies that force is coming away from the wall

[tex]F_{avg}=2096.71N[/tex]

Answer:

a.6 V

b.24 V

Explanation:

We are given that

a.[tex]C=8\mu F=8\times 10^{-6} F[/tex]

[tex]1\mu =10^{-6} [/tex]

Q=[tex]48\mu C=48\times 10^{-6} C[/tex]

We know that

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

Using the formula

[tex]V=\frac{48\times 10^{-6}}{8\times 10^{-6}}=6 V[/tex]

b.[tex]Q=192\mu C=192\times 10^{-6} C[/tex]

[tex]V=\frac{192\times 10^{-6}}{8\times 10^{-6}}=24 V[/tex]

Answer:

The mass of the ice block is equal to 70.15 kg

Explanation:

The data for this exercise are as follows:

F=90 N

insignificant friction force

x=13 m

t=4.5 s

m=?

applying the equation of rectilinear motion we have:

x = xo + vot + at^2/2

where xo = initial distance =0

vo=initial velocity = 0

a is the acceleration

therefore the equation is:

x = at^2/2

Clearing a:

a=2x/t^2=(2x13)/(4.5^2)=1.283 m/s^2

we use Newton's second law to calculate the mass of the ice block:

F=ma

m=F/a = 90/1.283=70.15 kg

70.31kg

Step I: Consider Newton's second law of motion which states that;

∑F = m x a;

Where;

∑F = net force acting on a body

m = the mass of the body

a = acceleration due to the force on the body.

Step II: Now to the question;

Since frictional force is negligible and the only force acting on the block of ice is the applied force by the dockworker, the net force on the body (block of ice) is the constant horizontal force. i.e

∑F = 90.0N

Also;

the block starts from rest and moves a distance (s) of 13.0m in a time (t) of 4.50s. Here, we can get the acceleration in that duration of time using one

the equations of motion as follows;

s = ut + [tex]\frac{1}{2}[/tex]at²            ------------------------------(ii)

Where;

s = distance covered = 13.0m

u = initial velocity = 0      [since the block starts from rest]

t = time taken to cover the distance = 4.50s

a = acceleration of the body.

Substitute these values into equation (ii) as follows;

13.0 = 0(4.5) + [tex]\frac{1}{2}[/tex](a)(4.50)²

13.0 = 0 + [tex]\frac{1}{2}[/tex](a)(20.25)

13.0 = [tex]\frac{1}{2}[/tex](a)(20.25)

13.0 = 10.125a

Solve for a;

a = [tex]\frac{13.0}{10.125}[/tex]

a = 1.28m/s²

Step III: Now substitute the values of a = 1.28m/s² and ∑F = 90.0N into equation (i) as follows;

90.0 = m x 1.28

m = [tex]\frac{90.0}{1.28}[/tex]

m = 70.31

Therefore, the mass of the block of ice is 70.31kg

Answer:

[tex]W_f=-62460\ J[/tex]

Explanation:

Given that

mass of the car ,m = 120 kg

Radius ,R= 12 m

Speed at the bottom , u = 25 m/s

Speed at top ,v= 8 m/s

We know that

Work done by all the forces = Change in the kinetic energy

Work done by gravity +  Work done by friction =Change in the kinetic energy

By taking point A as reference

[tex]m g \times (2R) + W_f=\dfrac{1}{2}mv^2-\dfrac{1}{2}mu^2[/tex]

Now by putting the values in the above equation we get

[tex]120\times 10\times 2\times 12+ W_f=\dfrac{1}{2}\times 120\times 8^2-\dfrac{1}{2}\times 120\times 25^2[/tex]

[tex]W_f=\dfrac{1}{2}\times 120\times 8^2-\dfrac{1}{2}\times 120\times 25^2-120\times 10\times 2\times 12\ J[/tex]

[tex]W_f=-62460\ J[/tex]

Therefore the work done by friction force will be -62460 J.

Answer:

Explanation:

Given an RC series circuit

Initially uncharged capacitor

C=3.67 μF

Resistor R=8.01 k Ω=8010 ohms

Battery EMF(V)=1.5V with negligible internal resistance.

The initial current in the circuit?

At the beginning the capacitor is uncharged and it has a 0V, so all the voltage appears at the resistor,

Now using ohms law

V=iR.

i=V/R

i=1.5/8010

i=0.000187A

1mA=10^-3A

Therefore, 1A = 1000mA

i=0.187 milliamps

The initial current in the circuit is 0.187 mA

Answer:

The initial current is 0.0187 mA.

Explanation:

Given that

capacitance is given as 3.67 x 10⁻⁶ F

resistance is given as 8010 Ω

voltage across the circuit is 1.5 V

Since the capacitor is initially uncharged, the capacitive reactance is zero.

From ohms law;

Voltage across the circuit is directly proportional to the opposition to the flow of current.

In these circumstances, as the battery only "sees"a resistor, the initial current can be found applying Ohm's law to the resistor, as follows:

[tex]V = I_{0}*R \\\\ I_{0} = \frac{V}{R} = \frac{1.50V}{8.011e3\Omega}\\ = 0.0187 mA[/tex]

The initial current (that will be diminishing as the capacitor charges), is 0.0187 mA.

Answer:

Magnetic energy stored in the inductor when all of the energy in the circuit is in the inductor = 0.049 mJ

If all the energy is then transferred into the capacitor, the voltage drop across the capacitor = 0.00572 V = 0.01 V (expressed to the hundredths value)

Explanation:

In an RLC circuit with maximum current of 7mA = 0.007 A

When all of the energy is stored in the inductor, maximum current will flow through it,

Hence E = (1/2) LI²

L = inductance of the inductor = 2 H

E = (1/2) (2)(0.007²) = 0.000049 J = 0.049 mJ

When all the energy in the circuit is in the capacitor, this energy will be equal to the energy calculated above.

And for a capacitor, energy is given as

E = (1/2) CV²

E = 0.000049 J, C = 3 F, V = ?

0.000049 = (1/2)(3)(V²)

V = 0.00572 V = 0.01 V

Answer

[tex]3.9*10^{5}N/C[/tex]

Explanation:

from the expression for determing the magnetic field strength

[tex]E=\frac{q}{4\pi e_{0}d^2 }\\[/tex]

since the charge is given as

[tex]q=1.56*10^{-5}c\\[/tex]

and the distance is

d=60cm=0.6m

We can calculate the constant k

[tex]K=\frac{1}{4\pi e_{0}}\\ K=\frac{1}{4\pi *8.8542*10^{-12}}\\k=8.98*10^{9}\\[/tex]

if we substitute values, we arrive at

[tex]E=\frac{8.98*10^9 *1.56*10^{-5}}{0.6^2} \\E=389133.33\\E=3.9*10^{5}N/C[/tex]

Explanation:

Below is an attachment containing the solution.

Answer:

The Reasons for which petroleum is the chosen fuel for transportation in the U.S. include number 1) and 2).

Explanation:

The main reasons why oil is the most used fuel in the USA is because of its energy potential with a high combustion power and a better octane rating compared to other fuels. Thus, fossil fuel derived from oil has a high energy value per unit volume and an ability to quickly start or stop providing energy.

Answer: The coefficient of static friction is 3.85 and  The coefficient of kinetic friction is 2.8

Explanation:

in the attachment