A block is moving at constant speed due to a horizontal force pulling to the right. The coefficient of kinetic friction, Hk, between the block and the surface is 0.20 and the mag- nitude of the frictional force is 100.0 N, what is the weight of the block? (a 400 N (b) 600 N (c) 500 N (d) 267 N

Answer:

17.92 Ω

Explanation:

R₀ = Initial resistance of the aluminum wire at 30.0°C = 7.00 Ω

R = resistance of the aluminum wire at 430.0°C = ?

α = temperature coefficient of resistivity for aluminum = 3.90 x 10⁻³ °C⁻¹

ΔT = Change in temperature = 430 - 30 = 400 °C

Resistance of the wire is given as

R = R₀ (1 + α ΔT)

R = (7) (1 + (3.90 x 10⁻³) (400))

R = 17.92 Ω

Answer:

[tex]d = 10\times t\sqrt{29}miles[/tex]

Explanation:

Given:

't' hour be the time taken for travel by  both the vehicles and 'd'  be the distance between then

then

Distance traveled by the car = 20 × t miles

and

Distance traveled by the truck  = 50 × t miles

now, using the Pythagoras theorem

[tex]d = \sqrt{(20t)^2+(50t)^2}[/tex]

or

[tex]d = \sqrt{400t^2+2500t^2}[/tex]

or

[tex]d = \sqrt{2900t^2}[/tex]

or

[tex]d = 10\times t\sqrt{29}[/tex]

thus, the equation relating the distance 'd' with the time 't' comes as

[tex]d = 10\times t\sqrt{29}miles[/tex]

Final answer:

To find the distance between the car and truck after t hours, apply the Pythagorean theorem. Calculate the hypotenuse of the right triangle formed by their paths. The distance apart in miles is represented by the expression 50√1.16t.

Explanation:

The question involves finding an expression for the distance d between two vehicles traveling perpendicularly away from an intersection, with one vehicle going east and another going south at different speeds. To solve this, we can apply the Pythagorean theorem which states that in a right-angled triangle, the square of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the other two sides. If the car travels east at 20 miles per hour and the truck travels south at 50 miles per hour, after t hours, the car will have traveled 20t miles and the truck 50t miles. These distances represent the two legs of a right triangle, and the distance d between the vehicles is the hypotenuse.

So, the distance d (in miles) at the end of t hours is given by:

d = √{(20t)² + (50t)²}

You can simplify this further to find:

d = t * √{20² + 50²}

d = t *√{400 + 2500}

d = t *√2900

d = t *50√1.16

d = 50√1.16t

Therefore, the distance apart in miles at the end of t hours is given by the expression 50√1.16t.

The magnitude of the gravitational acceleration (g) on this planet is equal to 12.52 [tex]m/s^2[/tex].

Given the following data:

Length = 53.0 cm to m = 0.53 m.

Number of cycle, n = 99.0 cycles.

Time = 128 seconds.

First of all, we would calculate the period for a full swing cycle as follows:

[tex]T=\frac{time}{n} \\\\T=\frac{128}{99.0}[/tex]

Period, T = 1.293 seconds.

Mathematically, the time taken (period) by a pendulum is given by this formula:

[tex]T=2\pi \sqrt{\frac{L}{g} }[/tex]

Making g the subject of formula, we have:

[tex]g=\frac{4\pi^2L}{T^2} \\\\g=\frac{4(3.142)^2 \times 0.53}{1.293^2} \\\\g=\frac{20.929}{1.672}[/tex]

g = 12.52 [tex]m/s^2[/tex].

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To determine the gravitational acceleration on an unfamiliar planet, the period of a pendulum was used. After calculating the period per swing, the pendulum formula T = 2π√(l/g) was rearranged to solve for g, yielding approximately 9.826 m/s² as the gravitational acceleration.

The student's question involves calculating the magnitude of the gravitational acceleration on an unfamiliar planet using the period of a simple pendulum. The pendulum's length is 53.0 cm and it completes 99.0 swings in 128 seconds. The formula to calculate the period of a pendulum (T) is T = 2π√(l/g), where l is the length of the pendulum and g is the acceleration due to gravity.

Therefore, the gravitational acceleration on the planet is around 9.826 m/s².

Answer: [tex]3.66(10)^{33}kg[/tex]

Explanation:

We are told both planets describe a circular orbit around the star S. So, let's approach this problem begining with the angular velocity [tex]\omega[/tex] of the planet P1 with a period [tex]T=750years=2.36(10)^{10}s[/tex]:

[tex]\omega=\frac{2\pi}{T}=\frac{V_{1}}{R}[/tex] (1)

Where:

[tex]V_{1}=40.2km/s=40200m/s[/tex] is the velocity of planet P1

[tex]R[/tex] is the radius of the orbit of planet P1

Finding [tex]R[/tex]:

[tex]R=\frac{V_{1}}{2\pi}T[/tex] (2)

[tex]R=\frac{40200m/s}{2\pi}2.36(10)^{10}s[/tex] (3)

[tex]R=1.5132(10)^{14}m[/tex] (4)

On the other hand, we know the gravitational force [tex]F[/tex] between the star S with mass [tex]M[/tex] and the planet P1 with mass [tex]m[/tex] is:

[tex]F=G\frac{Mm}{R^{2}}[/tex] (5)

Where [tex]G[/tex] is the Gravitational Constant and its value is [tex]6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}[/tex]

In addition, the centripetal force [tex]F_{c}[/tex] exerted on the planet is:

[tex]F_{c}=\frac{m{V_{1}}^{2}}{R^{2}}[/tex] (6)

Assuming this system is in equilibrium:

[tex]F=F_{c}[/tex] (7)

Substituting (5) and (6) in (7):

[tex]G\frac{Mm}{R^{2}}=\frac{m{V_{1}}^{2}}{R^{2}}[/tex] (8)

Finding [tex]M[/tex]:

[tex]M=\frac{V^{2}R}{G}[/tex] (9)

[tex]M=\frac{(40200m/s)^{2}(1.5132(10)^{14}m)}{6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}}[/tex] (10)

Finally:

[tex]M=3.66(10)^{33}kg[/tex] (11) This is the mass of the star S

The induced electromotive force (emf) in the wire can be calculated using Faraday's law, which states that the induced emf is directly proportional to the time rate of change of magnetic flux through the circuit. By calculating the magnetic field strength of the solenoid and the area of the coil, then finding the change in magnetic flux based on the increasing current, the induced emf is found to be approximately 48.25 millivolts.

To calculate the induced electromotive force (emf) in the wire, we use Faraday's law of electromagnetic induction, which states that the induced emf in a closed circuit is directly proportional to the time rate of change of magnetic flux through the circuit. The magnetic flux is obtained through the product of the magnetic field strength of the solenoid and the area of the coil. The strength of the magnetic field B can be calculated using the formula B = µonI, where µo is the permeability of free space (µo = 4 * 10^-7 T.m/A), n is the number of turns per unit length and I is the current.

The radius of the solenoid is 2 cm, so it has 12000/0.02= 600000 turns per meter, giving a magnetic field of B = 4 * 10^-7 T.m/A * 600000 m^-1 * 40 A/s = 0.096 T. The area of the coil is given by ∏*r^2, so with a radius of 8 cm, the area is ∏*(0.08 m)^2 = 0.0201 m^2 (meters squared). The variation of the magnetic flux (∆Φ) is therefore B*∆A = 0.096 T * 0.0201 m^2 = 0.00193 T.m^2 (Tesla meters squared).

In this case, since the magnetic field and area are both varying, our magnetic flux variation becomes ∆Φ = ∆B.∆A. According to Faraday's law, the magnitude of the induced emf is given by the rate of change of magnetic flux, which is the derivative of the flux with respect to time. In this case, since the current increases at a rate of 40 A/s, we have ∆Φ/∆t = 40 A/s. Therefore, the induced emf is ∆Φ/∆t = 0.00193 T.m^2 / 40 s = 0.04825 V or 48.25 millivolts.

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To calculate an automobile's braking distance going up a 5° incline at 90 km/h requires more information than provided, such as the coefficient of friction and deceleration rate. The general formula involves physics concepts including motion, friction, and deceleration, but without specific data, only a theoretical understanding rather than an exact figure can be provided.

To determine the automobile's braking distance when going up a 5° incline at 90 km/h, we need to involve concepts of physics specifically related to motion on inclines, frictional forces, and deceleration. Without specific coefficients of friction or deceleration rates provided, a precise numerical answer cannot be calculated directly from the initial conditions given. Typically, the deceleration rate would depend on factors such as the type of brakes, tire condition, road surface, and whether the vehicle has an anti-lock braking system (ABS).

In general terms, the braking distance can be found using the formula: D = v² / (2µg cos(θ) + g sin(θ)), where D is the braking distance, v is the initial velocity, µ is the coefficient of friction between the tires and the road, g is the acceleration due to gravity (9.8 m/s²), and θ is the incline angle.

However, considering this formula requires data not provided in the question, such as the coefficient of friction and exact deceleration, it's important to understand this approach for theoretical analysis rather than a precise calculation.

Answer:

(a): the car takes to stop 4.74 seconds.

(b): the car travels 54.78 meters in this time.

Explanation:

a= 4.87 m/s²

Vi= 23.1 m/s

Vf= Vi - a*t

t= Vi/a

t= 4.74 sec

d= Vi*t - (a*t²)/2

d= 54.78m

Answer:

Weight required = 194.51 N

Explanation:

The elongation is given by

            [tex]\Delta L=\frac{PL}{AE}[/tex]

Length , L= 1.6 m

Diameter, d = 1.1 mm

Area

   [tex]A=\frac{\pi d^2}{4}=\frac{\pi \times (1.1\times 10^{-3})^2}{4}=9.50\times 10^{-7}m^2[/tex]

Change in length, ΔL = 2.8 mm = 0.0028 m

Young's modulus of copper, E = 117 GPa = 117 x 10⁹ Pa

Substituting,

      [tex]\Delta L=\frac{PL}{AE}\\\\0.0028=\frac{P\times 1.6}{9.50\times 10^{-7}\times 117\times 10^9}\\\\P=194.51N[/tex]

Weight required = 194.51 N

Answer:

The resistance added to this coil is 0.04 ohm in parallel.

Explanation:

Given that,

Current I'= 0.5 A

Resistance R= 20 ohm

Deflection I= 1 mA

We need to calculate the resistance

Using ohm's law

[tex]V = I R[/tex]

Where,

V = voltage

I =current

R = resistance

V is constant so ,

Therefore,

I R=I'R'

[tex]R'=\dfrac{I}{I'}\times R[/tex]

[tex]R'=\dfrac{0.001}{0.5}\times20[/tex]

[tex]R'=0.04\ \Omega[/tex]

Hence, The resistance added to this coil is 0.04 ohm in parallel.

Answer:

D = [tex]\frac{ZM}{a^{3}N_A}[/tex]

Explanation:

the equation used for calculation of density of [tex]UO_2[/tex]

where D = density of [tex]UO_2[/tex]

M=molar mass of  [tex]UO_2[/tex]

a= lattice constant here  [tex]UO_2[/tex]  is  a body centered lattice and for body centered lattice a=0.547 nm

[tex]N_A[/tex]= Avogadro number which is equal to [tex]6.023\times 10^{23}[/tex]

Answer:

Required force equals 623.498 lb

Explanation:

We shall use newton's law of viscosity to calculate the shear force that acts on the cylinder

By Newton's law of viscosity we have

[tex]\tau =\mu \frac{dv}{dy}\\\\where \tau[/tex] is shear stress that acts on the internal surface

[tex]\mu[/tex] is dynamic viscosity of the fluid

[tex]\frac{dv}{dy}[/tex] is the velocity gradient that exists across the flow

The dynamic viscosity is calculated as follows

[tex]\mu =\rho \nu[/tex]

[tex]\mu =\rho \nu \\\rho[/tex] is density of the fluid

[tex]\nu =[/tex] kinematic viscosity of the fluid

By  no slip boundary condition the fluid in contact with the stationary cylinder shall not have any velocity while as the fluid in contact with the moving cylinder shall have velocity equal to that of the cylinder itself. This implies a velocity gradient shall exist across the gap in between the cylinders.

Applying values of the quantities we can calculate shear stress as follows

The density of fluid is [tex]\rho=G\times \rho_{w}[/tex]

G = specific gravity of fluid

[tex]\rho_{w}[/tex]is density of water

[tex]\tau =\rho \nu \frac{dv}{dy}\\\\\tau=62.42\times 0.92\times 0.006\times \frac{3}{\frac{0.125inches}{12inch}}\\\\\tau=99.23lb/ft^{2}[/tex]

This pressure shall oppose the motion of the internal cylinder hence the force of opposition = [tex]F=\tau\times Area[/tex]

Using the area of internal cylinder we get total force

F=[tex]2\pi rl\times \tau\\\\F=2\pi\times \frac{4}{12}ft\times 3ft\times \tau\\\\ F=623.498lb[/tex]

The force required to move the cylinder along the pipe can be calculated using the drag force formula. F = 6π(0.006)(1/3)(3) ≈ 0.113 ft-lbf

To calculate the force required to move the cylinder along the pipe at a constant velocity, we need to consider the drag force acting on the cylinder.

The drag force can be calculated using the formula:

F = 6πηrv

Where F is the force, η is the viscosity of the oil, r is the radius of the cylinder, and v is the velocity of the cylinder.

In this case, the radius of the cylinder is half of the diameter, so the radius is 4 inches or 1/3 ft. The velocity is given as 3 fps.

Plugging in the values, we get:

F = 6π(0.006)(1/3)(3) ≈ 0.113 ft-lbf

Answer:

Metaphor

Explanation:

We can harness  visionary language by using metaphor.

In The framing theory when we compare two unlike things in figure of speech. The comparison influences us on unconscious level. The metaphor causes us to make an association. If we change the metaphor we change how the other thinks of the subject.

Complex metaphor forms the basis of narratives or stories.

Answer:

[tex]\tau_{max} = 0.024 Nm[/tex]

Explanation:

Dipole moment of the electric dipole is given by the equation

[tex]P = qd[/tex]

here we have

[tex]q = 1\mu C[/tex]

d = 2 cm[/tex]

[tex]P = (1\mu C)(0.02)[/tex]

[tex]P = 2\times 10^{-8} C-m[/tex]

now the maximum torque due to electric field is given as

[tex]\tau = \vec P \times \vec E[/tex]

[tex]\tau_{max} = PE sin90[/tex]

[tex]\tau_{max} = (2\times 10^{-8})(1.2\times 10^6)[/tex]

[tex]\tau_{max} = 0.024 Nm[/tex]

The maximum torque exerted on an electric dipole with a charge magnitude of 1µC and separated by 2 cm in an external electric field of 1.2 MN/C is 0.024 N·m.

The dipole moment (p) of an electric dipole is defined as the product of the charge q and the distance d between the charges. Calculating the dipole moment for a dipole with charges of magnitude q = 1µC (micro Coulombs) separated by a distance of d = 2 cm (0.02 meters), we have p = q * d. To find the maximum torque, we use the equation T = pE, where E is the external electric field magnitude. Given that E = 1.2 MN/C (Mega Newtons per Coulomb), the equation becomes T = (1µC * 0.02m) * 1.2MN/C, which gives us T = (1e-6 C * 0.02 m) * 1.2e6 N/C, leading to a maximum torque of 0.024 N·m.

Final answer:

The rectangular coil needs to be rotated at a specific frequency in order to generate a maximum potential of 110 V. The angular velocity of the coil can be calculated using the formula ω = V / (nBA), where V is the potential, n is the number of turns, B is the magnetic field strength, and A is the area of the coil.

Explanation:

To generate a maximum potential of 110 V, the rectangular coil needs to be rotated at a specific frequency. The potential generated by a rotating coil is given by the equation

V = nBAω

where V is the potential, n is the number of turns, B is the magnetic field strength, A is the area of the coil, and ω is the angular velocity. Rearranging the equation for ω, we have

ω = V / (nBA)

Substituting the given values, we get

ω = 110 V / (500 turns * 0.10 m * 0.25 m * 0.58 T)

Simplifying this expression will give you the required frequency at which the coil should be rotated to generate a maximum potential of 110 V.

Answer:

The radius of the loop is [tex]4.94\times10^{5}\ m[/tex]

Explanation:

Given that,

Current = 0.2kA = 200 A

Number of turns = 35

Magnetic field = 8.9 nT

We need to calculate the radius of one loop

Using formula of magnetic field

[tex]B=\dfrac{N\mu_{0}I}{2r}[/tex]

[tex]r=\dfrac{N\mu_{0}I}{2B}[/tex]

Where, I = current

N = number of turns

B = magnetic field

r = radius

Put the value into the formula

[tex]r=\dfrac{35\times4\pi\times10^{-7}\times200}{2\times8.9\times10^{-9}}[/tex]

[tex]r =494183.11\ m[/tex]

[tex]r=4.94\times10^{5}\ m[/tex]

Hence, The radius of the loop is [tex]4.94\times10^{5}\ m[/tex]

Answer:

Work done, W = 900 Joules

Explanation:

It is given that,

Force applies by the car on the van, F = 300 N

The van pulls a distance of, d = 3 m

We need to find the work was done by the car on the car. We know that the product of force and distance is equal to the work done by the object i.e.

[tex]W=F\times d[/tex]

[tex]W=300\ N\times 3\ m[/tex]

W = 900 Joules

So, the work done by the car on the van is 900 Joules. Hence, this is the required solution.

Answer:

Percentage increase in the fundamental frequency is

d)-14.02%

Explanation:

As we know that fundamental frequency of the wave in string is given as

[tex]f_o = \frac{1}{2L}\sqrt{\frac{T}{\mu}}[/tex]

now it is given that tension is increased by 30%

so here we will have

[tex]T' = T(1 + 0.30)[/tex]

[tex]T' = 1.30T[/tex]

now new value of fundamental frequency is given as

[tex]f_o' = \frac{1}{2L}\sqrt{\frac{1.30T}{\mu}}[/tex]

now we have

[tex]f_o' = \sqrt{1.3}f_o[/tex]

so here percentage change in the fundamental frequency is given as

[tex]change = \frac{f_o' - f_o}{f_o} \times 100[/tex]

% change = 14.02%

-- pushing on a brick wall

-- standing on your little brother's back so that he can't get up

-- taking a nap while on the job

-- squeezing anything that doesn't yield to your squeeze, such as a glass bottle or your girl friend

-- watching TV

-- solving math problems in your head

-- making pictures out of clouds in the sky

Answer:

a) Displacement of the object from t = 0 to t = 3 is 1.5 m

b)  Distance of the object from t = 0 to t = 3 is 1.83 m

Explanation:

Velocity, v(t) = t² - 3t + 2

a) Displacement is given by integral of v(t) from 0 to 3.

   [tex]s=\int_{0}^{3}(t^2-3t+2)dt=\left [ \frac{t^3}{3}-\frac{3t^2}{2}+2t\right ]_0^3=\frac{3^3}{3}-\frac{3^3}{2}+6=1.5m[/tex]

b) t² - 3t + 2 = (t-1)(t-2)

   Between 1 and 2,  t² - 3t + 2 is negative

   So we can write t² - 3t + 2 as -(t² - 3t + 2)

   Distance traveled

             [tex]s=\int_{0}^{1}(t^2-3t+2)dt+\int_{1}^{2}-(t^2-3t+2)dt+\int_{2}^{3}(t^2-3t+2)dt\\\\s=\left [ \frac{t^3}{3}-\frac{3t^2}{2}+2t\right ]_0^1-\left [ \frac{t^3}{3}-\frac{3t^2}{2}+2t\right ]_1^2+\left [ \frac{t^3}{3}-\frac{3t^2}{2}+2t\right ]_2^3\\\\s=\frac{1^3}{3}-\frac{3\times 1^2}{2}+2-\left ( \frac{2^3}{3}-\frac{3\times 2^2}{2}+4\right )+\frac{1^3}{3}-\frac{3\times 1^2}{2}+2+\frac{3^3}{3}-\frac{3\times 3^2}{2}+6-\left ( \frac{2^3}{3}-\frac{3\times 2^2}{2}+4\right )=1.83m[/tex]

The displacement of the object from t = 0 to t = 3 is -4.5 m, indicating the object moved 4.5 meters in the opposite direction from its initial position. The distance the object traveled during the same period can be found by taking the integral of the absolute value of the velocity function from 0 to 3, and adding the magnitudes for time intervals when the velocity was positive and when it was negative.

The given function indicates the velocity of an object moving along a straight line as a function of time - v(t) = t^2 - 3t + 2. It's a quadratic function so one way to find the displacement from t = 0 to t = 3, is to integrate the velocity function. The integral of v(t) from 0 to 3 gives the total change in position, or displacement, which would be the integral ∫ from 0 to 3 of (t^2 - 3t + 2) dt = [t^3/3 - 1.5t^2 + 2t] from 0 to 3 = 3^3/3 - 1.5 * 3^2 + 2*3 - (0 - 0 + 0) = 3 - 13.5 + 6 = -4.5 m.

On the other hand, distance is a scalar quantity and does not account for the direction, only the magnitude of movement. As such the object's total distance travelled from t = 0 to t = 3 may be calculated by finding the integral of the absolute value of the velocity from 0 to 3. In this case, v(t) is positive from t = 0 to t = 1 and negative from t = 1 to t = 3 (as substantiated by equating the velocity function to 0). Thus, the total distance traveled by the object is the sum of distances in segments (0,1) and (1,3), obtained as the sum of integrals ∫ from 0 to 1 of (t^2 - 3t + 2) dt + ∫ from 1 to 3 of (-t^2 + 3t - 2) dt.

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(a) The temperature of the drink after 45 minutes is 13.85°C.

(b) The temperature of the drink will be ( 16°C ) after ( 67.75) minutes.

To solve this problem, we can use Newton's Law of Cooling, which states that the rate of change of the temperature of an object is proportional to the difference between its own temperature and the ambient temperature. The formula can be written as:

[tex]\[ \frac{dT}{dt} = -k(T - T_a) \][/tex]

(a) To find the temperature of the drink after 45 minutes, we first need to determine the constant (k). We can do this using the given data points:

1. At ( t = 0 ), ( T = 5°C).

2. At ( t = 25) minutes, ( T = 10°C), and [tex]\( T_a[/tex] = 20°C.

Using these values, we can write the equation as:

[tex]\[ 10 = 20 - (20 - 5)e^{-25k} \][/tex]

Solving for (k):

[tex]\[ e^{-25k} = \frac{20 - 10}{20 - 5} \] \[ e^{-25k} = \frac{10}{15} \] \[ e^{-25k} = \frac{2}{3} \] \[ -25k = \ln\left(\frac{2}{3}\right) \] \[ k = -\frac{\ln\left(\frac{2}{3}\right)}{25} \][/tex]

Now that we have (k), we can find the temperature after 45 minutes:

[tex]\[ T = 20 - (20 - 5)e^{-45k} \] \[ T = 20 - 15e^{45\left(\frac{\ln\left(\frac{2}{3}\right)}{25}\right)} \] \[ T = 20 - 15e^{1.8\ln\left(\frac{2}{3}\right)} \] \[ T = 20 - 15\left(\frac{2}{3}\right)^{1.8} \] \[ T \approx 20 - 15(0.41) \] \[ T \approx 20 - 6.15 \] \[ T \approx 13.85C \][/tex]

So, the temperature of the drink after 45 minutes is approximately ( 13.85°C ).

(b) To find when the temperature of the drink will be (16°C), we use the same formula and solve for ( t ):

[tex]\[ 16 = 20 - (20 - 5)e^{-kt} \] \[ e^{-kt} = \frac{20 - 16}{20 - 5} \] \[ e^{-kt} = \frac{4}{15} \] \[ -kt = \ln\left(\frac{4}{15}\right) \] \[ t = -\frac{\ln\left(\frac{4}{15}\right)}{k} \] \[ t = -\frac{\ln\left(\frac{4}{15}\right)}{-\frac{\ln\left(\frac{2}{3}\right)}{25}} \] \[ t = 25\frac{\ln\left(\frac{4}{15}\right)}{\ln\left(\frac{2}{3}\right)} \][/tex]

Using a calculator, we find:

[tex]\[ t \approx 25\frac{\ln(0.2667)}{\ln(0.6667)} \] \[ t \approx 25 \times 2.71 \] \[ t \approx 67.75 \][/tex]

So, the temperature of the drink will be ( 16°C ) after ( 67.75) minutes.

Final answer:

The charge that flows from the battery after inserting a piece of mica into a 2400 pF air-gap capacitor connected to a 6.4 V battery is 76,800 picoCoulombs.

Explanation:

When a piece of mica is placed between the plates of an air-gap capacitor connected to a battery, the charge stored in the capacitor changes due to the increased capacitance. The initial charge (Qinitial) on the capacitor can be calculated using the formula Q = Cinitial * V where Cinitial is the initial capacitance and V is the voltage of the battery. With a capacitance of 2400 pF (picoFarads) and a battery voltage of 6.4 V, the initial charge is Qinitial = 2400 pF * 6.4 V = 15,360 pC (picoCoulombs).

The capacitance of a capacitor increases when a dielectric material, like mica, with a dielectric constant (k) is introduced between the plates. The new capacitance (Cnew) is Cnew = Cinitial * k. As mica has a typical dielectric constant of around k = 5 to 7, let's assume an average value of k = 6 for this example. The new capacitance is Cnew = 2400 pF * 6 = 14,400 pF.

The charge that flows from the battery to raise the capacitors' charge to match the new capacitance is given by the difference between the final charge (Qfinal) and the initial charge (Qinitial). The final charge is Qfinal = Cnew * V = 14,400 pF * 6.4 V = 92,160 pC. Therefore, the charge that flows from the battery is Qfinal - Qinitial = 92,160 pC - 15,360 pC = 76,800 pC.

Answer:

Total charge, q = 1800 C

Explanation:

It is given that,

Power of light bulb, P = 60 watts

It carries a current, I = 0.5 A

Time, t = 1 hour = 3600 seconds

We need to find the total charge passing through it in one hour. We know that current through an electrical appliance is defined as the charge flowing per unit time i.e.

[tex]I=\dfrac{q}{t}[/tex]

[tex]q=I\times t[/tex]

[tex]q=0.5\ A\times 3600\ s[/tex]

q = 1800 C

So, the total charge passing through it is 1800 C. Hence, this is the required solution.

The current of 0.5 amperes means that 0.5 coulombs of charge pass every second. Therefore, over an hour (3600 seconds), a total charge of 1800 Coulombs would pass. So, the answer to the question is E) 1800 C.

The question is about understanding the relationship between electric current, time and charge. The electric current is the charge passing per unit time. In this case, we know that our current is 0.5 amperes, which conveys that 0.5 coulombs of electric charge is passing every second.

 Therefore, when you want to find out how much charge passes in an hour, you would multiply by the number of seconds in an hour which is equal to 3600 (60 seconds per minute x 60 minutes).

 Hence, the total charge passing for an hour is 0.5 amperes x 3600 seconds = 1800 Coulombs, so the answer is (E) 1800 C.

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Answer:

-58.876 kJ

Explanation:

m = mass of air = 1 kg

T₁ = Initial temperature = 15°C

T₂ = Final temperature = 97°C

Cp = Specific heat at constant pressure = 1.005 kJ/kgk

Cv = Specific heat at constant volume = 0.718 kJ/kgk

W = Work done

Q = Heat = 0 (since it is not mentioned we are considering adiabatic condition)

ΔU = Change in internal energy

Q = W+ΔU

⇒Q = W+mCvΔT

⇒0 = W+mCvΔT

⇒W = -mCvΔT

⇒Q = -1×0.718×(97-15)

⇒Q = -58.716 kJ

Answer:

The Pressure is 0.20 MPa.

(b) is correct option.

Explanation:

Given that,

Change in volume = 9.05%

{tex]\dfrac{\Delta V}{V_{0}}=0.0905[/tex]

We know that.

The bulk modulus for water

[tex]B=0.20\times10^{10}\ N/m^2[/tex]

We need to calculate the pressure difference

Using formula bulk modulus formula

[tex]B=\Delta P\dfrac{V_{0}}{\Delta V}[/tex]

[tex]\Delta P=B\dfrac{\Delta V}{V_{0}}[/tex]

[tex]\Delta P=0.2\times10^{10}\times0.0905[/tex]

[tex]\Delta P=0.2\times10^{6}\ Pa[/tex]

[tex]\Delta P=0.20 MPa[/tex]

Hence, The Pressure is 0.20 MPa.

Answer:

The value of the linear coefficient of thermal expansion is : α=1.01 *10⁻⁵ (ºC)⁻¹

Explanation:

Li = 0.2m

ΔL = 0.2 mm = 0.0002m

T1 = 21ºC

T2 = 120ºC

ΔT =99ºC

α =ΔL/(Li*ΔT)

α =0.0002m /(0.2m * 99ºC)

α = 1.01 *10⁻⁵   (ºC)⁻¹

The correct value of the linear coefficient of thermal expansion for the material is [tex]1.01 \times 10^{-5} \ (^\circ{C})^{-1}}\)[/tex].

To determine the linear coefficient of thermal expansion [tex](\(\alpha\))[/tex] for the material, we can use the formula:

[tex]\[\alpha = \frac{\Delta L}{L_0 \Delta T}\][/tex]

where:

[tex]\(\Delta L\)[/tex] is the change in length,

[tex]\(L_0\)[/tex] is the original length, and

[tex]\(\Delta T\)[/tex] is the change in temperature.

Given:

The original length [tex]\(L_0\)[/tex] is 0.20 m,

The elongation [tex]\(\Delta L\)[/tex] is 0.20 mm, which we need to convert to meters to match the units of [tex]\(L_0\)[/tex].

Since 1 m = 1000 mm, 0.20 mm = [tex]\(0.20 \times 10^{-3}\)[/tex] m,

The change in temperature [tex]\(\Delta T\)[/tex] is from 21°C to 120°C, so [tex]\(\Delta T = 120^\circ{C} - 21^\circ{C} = 99^\circ{C}\)[/tex].

Now we can plug these values into the formula:

[tex]\[\alpha = \frac{0.20 \times 10^{-3} \ m}{0.20 \ m \times 99^\circ{C}}\][/tex]

[tex]\[\alpha = \frac{0.20}{0.20 \times 99} \times 10^{-3} \ (^\circ{C})^{-1}\][/tex]

[tex]\[\alpha = \frac{1}{99} \times 10^{-3} \ (^\circ{C})^{-1}\][/tex]

[tex]\[\alpha \approx 1.0101 \times 10^{-5} \ (^\circ{C})^{-1}\][/tex]

Rounding to two significant figures, we get:

[tex]\[\alpha \approx 1.01 \times 10^{-5} \ (^\circ{C})^{-1}\}[/tex]

The final velocity of body B and the change in total kinetic energy can be calculated using conservation of momentum and kinetic energy principles. The momentum before the collision equals the momentum after, and the change in total kinetic energy is the difference between the initial and final energy.

Using physics principles, notably the law of conservation of momentum, we can calculate the final velocity of B and the change in the total kinetic energy. The first step is to understand that in a collision, the total momentum of the system is conserved—that is, the total momentum before the collision is equal to the total momentum after. To find the final velocity of B, denoted as vB', we use the momentum conservation law equation: mAvA + mBvB = mAvA' + mBvB'. Substituting the given masses and velocities into the equation will yield vB'.

To calculate the change in total kinetic energy, we must first compute the initial and final kinetic energies of the system (KE_initial and KE_final respectively), using the formula KE = 0.5 m v². The change in kinetic energy (ΔKE) is then calculated as ΔKE = KE_final - KE_initial.

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Answer:

Option A is the correct answer.

Explanation:

When the frequency of the driving force equals the natural frequency of the system, the system is said to be in resonance. At resonance the system vibrates in maximum amplitude.

Marching soldiers are cautioned to break stride on a bridge is because of resonance, if the frequency of soldiers stride is equal to frequency of bridge, the bridge will vibrate with maximum amplitude. This will in turn collapse bridge.

Option A is the correct answer.

Answer:

  A.  20 Watts

Explanation:

Power is the rate of doing work.

  P = Fd/t = (15 N)(8 m)/(6 s) = 20 N·m/s = 20 W

20 watts. If the power of the man who pushes the box 8m with a force of 15n is 6 seconds, it will force 20 watts in 6 seconds.

First, you do is multiply and then divide to give us the final answer.

[tex]\displaystyle 8m\times15n\div6s[/tex]

[tex]\displaystyle 8\times15=120\div6=20[/tex]

[tex]\Large\boxed{20}[/tex], which is our answer.

Final answer:

To bring a 950-kg car to rest from 90 km/h over 120 m requires an average force of 2473.96 N. If the car hits a concrete abutment and stops within 2 m, the force exerted is much higher, at 148,437.50 N. This illustrates the impact of stopping distance on the force experienced by a vehicle.

Explanation:

Work-Energy Theorem Application

We'll first convert the speed from km/h to m/s by multiplying by 1000/3600. Therefore, 90.0 km/h is 25 m/s. Using the work-energy theorem, we know the work done to stop the car is equal to the change in kinetic energy.

Plugging in the values: W = KE = 1/2mv²

W = 1/2(950 kg)(25 m/s)²

W = 1/2(950 kg)(625 m²/s²)

W = 296,875 J

Since work is also equal to force times distance (W = Fd), the force needed can be found by dividing the work by the distance.

F = W/d = 296,875 J/120 m = 2473.96 N

This is the average force required to stop the car over 120 m. Now let's calculate the force if the car hits a concrete abutment and stops in 2.00 m:

F = 296,875 J/2 m = 148,437.50 N

The force exerted on the car in this case is significantly higher, showing the importance of cushioning distance in reducing impact forces.

Answer:

A) southward

Explanation:

Applying the rule of the right hand for the determination of the magnetic field produced by the current flowing through the cable yields the following results: between the wires it is to the north, the west of the wires is to the south and east of the wires , which asks the problem is to the south.