What is the speed v of a wave traveling down such a wire if the wire is stretched to its breaking point?

Answer:

IGNEOUS ROCKS

Explanation: Igneous rocks are those rocks that solidify from magma.

Igneous rock is divided into two ,they are:

1. Intrusive

Igneous rocks crystallized belowearth"s crust. Its cooling material is called lava.

2 Extrusive igneous rock is also known as known as volcanic rocks

Answer:

0.64 m from the first charge

Explanation:

Force is given by

[tex]F_1=\dfrac{kq_1q_2}{r^2}\\\Rightarrow F_1=\dfrac{k2\times 10^{-9}\times 3.7\times 10^{-9}}{x^2}[/tex]

[tex]F_2=\dfrac{kq_2q_3}{r^2}\\\Rightarrow F_1=\dfrac{k4.5\times 10^{-9}\times 3.7\times 10^{-9}}{(1.6-x)^2}[/tex]

These forces are equal

[tex]\dfrac{k2\times 10^{-9}\times 3.7\times 10^{-9}}{x^2}=\dfrac{k4.5\times 10^{-9}\times 3.7\times 10^{-9}}{(1.6-x)^2}\\\Rightarrow \dfrac{2}{x^2}=\dfrac{4.5}{(1.6-x)^2}\\\Rightarrow \dfrac{2}{4.5}=\dfrac{x^2}{(1.6-x)^2}\\\Rightarrow \dfrac{4.5}{2}=\dfrac{(1.6-x)^2}{x^2}\\\Rightarrow \sqrt{\dfrac{4.5}{2}}=\dfrac{1.6-x}{x}\\\Rightarrow 1.5=\dfrac{1.6-x}{x}\\\Rightarrow 1.5x+x=1.6\\\Rightarrow x=\dfrac{1.6}{2.5}\\\Rightarrow x=0.64\ m[/tex]

The distance that charge should be placed is 0.64 m from the first charge

The charge +3.70 x [tex]10^{-9[/tex]C should be placed at approximately 0.96 meters from the origin for the net electric force to be zero. The solution involves balancing the Coulomb forces from two charges placed at different points. The calculations involve setting up the equation for the forces and solving it step-by-step.

Finding the Point Where the Net Electric Force is Zero

To determine the point where the charge +3.70 x [tex]10^{-9[/tex] C should be placed so that the net electric force on it is zero, we need to consider the forces exerted by both charges +2.00 x [tex]10^{-9[/tex] C (at the origin) and +4.50 x [tex]10^{-9[/tex] C (at x = 1.6 m).

Step-by-Step Explanation:

Let the position where the net force is zero be at distance x from the origin (charge +2.00 x [tex]10^{-9[/tex] C).

The distance from the charge +4.50 x [tex]10^{-9[/tex] C to this point will then be (1.6 - x) meters.

According to Coulomb's Law, the force due to a charge is given by F = k * |q1 * q2| / [tex]r^2[/tex], where k is the Coulomb constant (8.98755 x[tex]10^{9[/tex] N m²/C²).

For the net force on the charge +3.70 x [tex]10^{-9[/tex] C to be zero, the magnitudes of the forces due to the two other charges must be equal:

Set these forces equal to each other:

(8.98755 x[tex]10^{9[/tex] N m²/C²) * (2.00 x [tex]10^{-9[/tex]C) * (3.70 x [tex]10^{-9[/tex] C) / x² = (8.98755 x [tex]10^{9[/tex] N m²/C²) * (4.50 x [tex]10^{-9[/tex] C) * (3.70 x [tex]10^{-9[/tex] C) / (1.6 - x)²

Simplify and solve for x:

(2.00 x [tex]10^{-9[/tex]) / x² = (4.50 x [tex]10^{-9[/tex]) / (1.6 - x)²

2 / x² = 4.5 / (1.6 - x)²

x² / (1.6 - x)² = 2 / 4.5

x² / (1.6 - x)² = 0.444

To solve for x, take the square root of both sides:

Finally, solve the equation x = √0.444 * (1.6 - x):

Thus, the charge +3.70 x [tex]10^{-9[/tex] C should be placed at approximately 0.96 meters from the origin to have the net electric force on it be zero.

Answer:

2 and 3

Explanation:

The right answer is the option 2 and 3,

This is all about the electrons transfer from one material to the other material.

For example if the electrons in the valence shell of one material are loosely attached then the other material's atoms try to take those electron to complete their shells and that is how the charges transfers from one another.

And it could also be happen as in option 2.

Answer:

The resultant velocity of the boat is 3.6 mil/h

Explanation:

given information:

flowing rate to the east (positive x-axis), v₁ = 2 mil/h (0.2) = 2i

boat's speed to the north (positive y-axis), v₂ = 3 mil/h (3,0) = 3j

the resultant velocity of the boat:

[tex]v_{R}[/tex] = √3²+2²

    = √13

    = 3.6 mil/h

The resultant velocity of the boat is approximately 3.61 mph, making an angle of 56.31° north of east. This is calculated using the Pythagorean theorem for magnitude and the inverse tangent function for direction.

To find the resultant velocity of the boat when it heads north across a river at a rate of 3 miles per hour with an eastward current at 2 miles per hour, we can use vector addition. The velocity of the boat heading north (Vboat) is perpendicular to the velocity of the river's current (Vriver).

The northward velocity can be considered as the y-component (Vy = 3 mph), and the eastward current as the x-component (Vx = 2 mph). The resultant velocity (Vtot) is the vector sum of these two components and can be calculated using the Pythagorean theorem:

Vtot = √(Vx2 + Vy2)

Plugging in the values we get:

Vtot = √(22 + 32)

Vtot = √(4 + 9)

Vtot = √13

Vtot = 3.61 mph (approximately)

To find the direction of the resultant velocity, we can use the inverse tangent function (tan-1) to calculate the angle (θ) the resultant velocity vector makes with the positive x-axis (eastward direction).

θ = tan-1(Vy/Vx)

θ = tan-1(3/2)

θ = 56.31° (approximately)

Since the boat is heading north and the current flows east, the resultant velocity makes an angle of 56.31° north of east.

Answer:

1.5 Hz

0.67 s

4 m

[tex]\pi[/tex]

2.82842 m

Explanation:

The equation is

[tex]x=4cos(3\pi t+\pi)[/tex]

It is of the form

[tex]x=Acos(2\pi ft+\phi)[/tex]

Comparing the equations we get

[tex]3\pi=2\pi f\\\Rightarrow f=\dfrac{3}{2}\\\Rightarrow f=1.5\ Hz[/tex]

Frequency is 1.5 Hz

Time period is given by

[tex]T=\dfrac{1}{f}\\\Rightarrow T=\dfrac{1}{1.5}\\\Rightarrow T=0.67\ s[/tex]

The period of the motion is 0.67 s

Amplitude

[tex]A=4\ m[/tex]

Amplitude is 4 m

Phase constant

[tex]\phi=\pi[/tex]

The phase constant is [tex]\pi[/tex]

At t = 0.25 s

[tex]x=4cos(3\pi t+\pi)\\\Rightarrow x=4cos(3\pi\times 0.25+\pi)\\\Rightarrow x=2.82842\ m[/tex]

The position of the particle is 2.82842 m

Answer:

(a). The mass of the rod is 15.9 g.

(b). The center of mass is 0.153 m.

Explanation:

Given that,

Length = 30.0 cm

Linear density [tex]\labda=50.0+20.0x[/tex]

We need to calculate the mass of rod

Using formula of mass

[tex]M=\int{dm}[/tex]

[tex]M=\int{(50.0+20.0x)dx}[/tex]

[tex]M=50.0x+10x^2[/tex]

Put the value of x

[tex]M=50.0\times0.30+10\times(0.30)^2[/tex]

[tex]M=15.9\ g[/tex]

We need to calculate center of mass

The center of mass has an x coordinate is given by

[tex]x_{cm}=\dfrac{\int{xdm}}{\int{dm}}[/tex]

We need to calculate the value of  [tex]\int{xdm}[/tex]

[tex]\int{xdm}=\int{(50.0x+20.0x^2)dx}[/tex]

[tex]\int{xdm}=25x^2+\dfrac{20}{3}x^3[/tex]

Put the value into the formula

[tex]\int{xdm}=25\times0.3^2+\dfrac{20}{3}\times(0.3)^3[/tex]

[tex]\int{xdm}=2.43[/tex]

Put the value into the formula of center of mass

[tex]x_{cm}=\dfrac{2.43}{15.9}[/tex]

[tex]x_{cm}=0.153\ m[/tex]

Hence, (a). The mass of the rod is 15.9 g.

(b). The center of mass is 0.153 m.

To find the mass of the rod, integrate the linear density function. To find the center of mass, set up an integral to find the position x such that the total mass on one side is equal to the total mass on the other side.

(a) To find the mass of the rod, we need to integrate the linear density function over the length of the rod. The linear density function is given by ℓ(x) = l + 20x, where x is the distance from one end measured in meters and l is in grams/meter. We can integrate this function from 0 to 0.3 meters (corresponding to a length of 30.0 cm) to find the total mass:

M = ∫(0 to 0.3) (l + 20x) dx

M = ∫(0 to 0.3) l dx + ∫(0 to 0.3) 20x dx

(b) To find the center of mass of the rod, we need to find the position x such that the total mass on one side of it is equal to the total mass on the other side. We can set up an integral to find this position:

x_cm = ∫(0 to x_cm) (l + 20x) dx - ∫(x_cm to 0.3) (l + 20x) dx

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Final answer:

To determine the magnitude of the electric field through which an electron narrowly misses the upper plate, apply principles of kinematics and the formula for electric force. Calculate the time of flight using the horizontal motion, and use this to establish the vertical acceleration due to electric force. From the electron's charge and acceleration, derive the electric field magnitude.

Explanation:

To calculate the magnitude of the electric field through which an electron passes and just misses the upper plate upon exit, we need to apply the concepts of kinematics and electric forces. Since the electrical force is the only vertical force acting on the electron, it will cause a vertical acceleration according to Newton's second law (F = ma), where the force F is the electric force (F = qE) and the acceleration a is due to this force.

The vertical distance traveled by the electron just before it exits the field is equal to half the distance between the plates, which is 0.5 cm (or 0.005 m). The time it takes for the electron to travel this vertical distance can be calculated by using the initial horizontal speed and the horizontal distance (2 cm or 0.02 m) it travels. Using the formula t = d/v (where d is the horizontal distance and v is the horizontal velocity), we find t = 0.02 m / 5.35×10¶6 m/s = 3.74×10¶8 seconds.

The vertical acceleration a can be found by using the vertical distance (s), initial vertical speed (u which is 0 in this case), and time (t), using the equation s = ut + 0.5at². From this, we find a = 2s/t². Knowing the charge of an electron (e = 1.6×10¶19 C) and its mass (m = 9.11×10¶31 kg), the electric field E can then be calculated as E = F/q = ma/q. Plugging in the values, we get the magnitude of the electric field.

Answer:

497.35 m.

Explanation:

Echo: This is the sound heard after the reflection of sound wave on a plane surface.

v = 2x/t.......................................... Equation 1

Where v = speed of sound in air, x = length of the lake, t = time taken to hear the echo.

Making x the subject of the equation,

x = vt/2.......................................... Equation 2

Given: v = 343 m/s, t = 2.9 s.

Substitute into equation 2

x = 343×2.9/2

x = 497.35 m

Thus the length of the lake  = 497.35 m.

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A uniform line charge with 16.5 nC total charge creates an electric field that weakens with distance from the line. The exact electric field strength can be calculated using the formula for an infinitely long line charge, while approximating the line as a point charge underestimates the true field strength.

Solving for the Electric Field of a Uniform Line Charge

(a) Total Charge:

The total charge (Q) of a line charge can be found by integrating the linear charge density (λ) over the length (L) of the charge distribution:

Q = ∫λ dx (from x = 0 to x = 5.0 m)

Here, λ = 3.3 nC/m and L = 5.0 m.

Q = (3.3 nC/m) * (5.0 m) = 16.5 nC

Therefore, the total charge is 16.5 nC.

(b) Electric Field at x = 6 m (Exact):

Because the line charge is infinitely long, we can apply the electric field formula:

E = (λ / (2πε₀)) * ln(2a / b)

Plugging in the values:

E = (3.3 x 10⁻⁹ C/m) / (2π * 8.854 x 10⁻¹² C²/N∙m²) * ln(2 * 6 m / 0)

E ≈ 47.7 N/C (rounded to two significant figures)

(c) Electric Field at x = 10.0 m (Exact):

Following the same method as part (b):

E ≈ (3.3 x 10⁻⁹ C/m) / (2π * 8.854 x 10⁻¹² C²/N∙m²) * ln(2 * 10.0 m / 0)

E ≈ 33.1 N/C (rounded to two significant figures)

(d) Electric Field at x = 270 m (Exact):

Using the same formula:

E ≈ (3.3 x 10⁻⁹ C/m) / (2π * 8.854 x 10⁻¹² C²/N∙m²) * ln(2 * 270 m / 0)

E ≈ 0.012 N/C (rounded to three significant figures)

(e) Electric Field at x = 270 m (Approximation):

Assuming the charge is a point charge at x = 2.5 m (center of the line charge):

E ≈ k * Q / (x - 2.5 m)²

(f) Ratio of Approximation to Exact Result:

Ratio = (Approximate Electric Field) / (Exact Electric Field)

Ratio ≈ (2.02 x 10⁻⁴ N/C) / (0.012 N/C) ≈ 0.0017

(g) Comparison of Results:

Since the ratio is less than 1, the approximate result (2.02 x 10⁻⁴ N/C) is smaller than the exact result (0.012 N/C).  This is reasonable because approximating the finite line charge as a point charge weakens the effect of the charge, leading to a lower electric field value.

Answer:b and e

Explanation:

When electron and proton are placed in the same Electric field then the force experienced by electron and proton is the same as the charge possessed by them is the same . The direction of force is different on them but the magnitude is the same.

The electric field force is the product of charge and strength of the electric field.

We know the mass of an electron is less than the mass of proton that is why the acceleration of electron is more as compared to proton for the same Electric force.

thus option b and e are correct.                                  

The statements which are true of a "free" electron and a "free" proton are placed in an identical electric field are: Choice b and Choice e.

According to the question:

In essence, the magnitude of the Coulumb's force of attraction or repulsion is the same for the free electron and free proton.

However, the direction of the forces are opposite as the protons and electron have opposing signs.

Additionally, since the mass of an electron is relatively infinitesimal compared to a proton, The acceleration of the electron is greater than that of the proton.

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Answer:[tex]44.1\times 10^6[/tex] atoms

Explanation:

According to the ideal gas equation:

[tex]PV=nRT[/tex]

P = Pressure of the gas = [tex]1.83\times 10^{-7}N/m^2=1.81\times 10^{-12}atm[/tex]     [tex]1N/m^2=9.87\times 10^{-6}atm[/tex]

V= Volume of the gas = [tex]1cm^3=1ml=0.001L[/tex]       (1L=1000ml)

T= Temperature of the gas = 28°C = 301 K      [tex]0^0C=273K[/tex]  

R= Gas constant = 0.0821 atmL/K mol

n=  moles of gas= ?

[tex]n=\frac{PV}{RT}=\frac{1.81\times 10^{-12}atm\times 0.001L}0.0821Latm/Kmol\times 301K}=7.32\times 10^{-17}moles[/tex]

Number of atoms =[tex]moles\times {\text {avogadro's number}}=7.32\times 10^{-17}mol\times 6.023\times 10^{23}mol^{-1}=44.1\times 10^6atoms[/tex]

Thus there are [tex]44.1\times 10^6[/tex] atoms  in a cubic centimeter at this pressure and temperature.

To determine the heaviest weight a weight lifter can lift without breaking his legs, we need to calculate the stress on the bones. By using the dimensions of the tibia and the breaking stress of bone, we can calculate the maximum weight the lifter can lift.

To determine the heaviest weight a weight lifter can lift without breaking his legs, we need to calculate the stress on the bones. Stress is defined as the force applied per unit area. In this case, we can calculate the stress on the tibia using the formula stress = force/area. The area of the cross section of the tibia can be found by subtracting the area of the inner circle from the area of the outer circle. Once we have the stress, we can use it to determine the maximum weight the lifter can lift without breaking his legs.

Using the given dimensions of the tibia, we can calculate the area and stress:

Outer radius = 3.6 cm / 2 = 1.8 cm = 0.018 m

Inner radius = 2.3 cm / 2 = 1.15 cm = 0.0115 m

Area of outer circle = π * (0.018 m)^2 = 0.001018 m^2

Area of inner circle = π * (0.0115 m)^2 = 0.000415 m^2

Area of tibia = Area of outer circle - Area of inner circle = 0.001018 m^2 - 0.000415 m^2 = 0.000603 m^2

Force on the tibia = weight lifted = mass * acceleration due to gravity = 90.0 kg * 9.8 m/s^2 = 882 N

Stress on the tibia = force/area = 882 N / 0.000603 m^2 = 1,460,070 Pa (or 1.46 MPa)

Therefore, the lifter can lift a maximum weight without breaking his legs if the stress on the tibia is less than or equal to the breaking stress of bone. In this case, the lifter can lift:

Maximum weight = breaking stress * area = 150 MPa * 0.000603 m^2 = 90.45 N

So, the lifter can lift a maximum weight of approximately 90.45 N without breaking his legs.

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Answer:

(A)  111.77m

(B) 9.07m

(C)55.88m/s

(D)20.54m/s

Explanation:

step 1 " we have to calculate the time it took the projectile to get to its maximum height

(a)  t = usinθ/g

= 49sin 33/9.81

49×0.5446/9.81

=2.72s

the horizontal distance =  ucosθ×t  , because the projectile horizontal motion is unaffected by the force of graavity

= 49cos33 ×2.72

=111.77m

(B) with the same projectile fired the same way , the horizontal distance = 55.8m

55.8 = ucosθ×t

55.8 = 49cos33 ×t

t = 55.8/49cos33

t= 1.36s

height of the projectile = 1/2 ×g×t²

=1/2 ×9.81×1.36²

= 9.07m

(c) Velocity of the projectile when it hits the wall

V₀ = ucosθ×t

49cos 33 × 1.36

=55.88 m/s

(D) speed = distance / time

distance /2×t   ; total time of flight

= 55.88/ 2.72

=20.54m/s

The given speed of Earth in km/sec is converted first to miles/sec and then to miles/hour, resulting in an average speed of Earth around the Sun of approximately 66,661.6 miles/hour.

To solve this, we need to convert kilometers to miles and seconds to hours. First, we should know that 1 kilometer is approximately 0.621371 miles, and 1 hour has 3600 seconds.

Given that, we can first convert Earth's speed from kilometers per second to miles per second by multiplying by the conversion factor:

29.783 km/sec * 0.621371 mile/km = 18.5171 miles/sec.

Next, we convert seconds to hours:

18.5171 miles/sec * 3600 sec/hour = 66,661.6 miles/hour.

So, the average speed of the Earth around the Sun, in miles per hour, to five significant figures is 66,661.6 miles/hour.

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Final answer:

The Earth travels around the Sun at an average speed of 29.783 km/s, which is approximately 66,636.7 miles per hour when converted using the steps of multiplying by 3600 to get km/hr and then by the conversion factor for km to mi.

Explanation:

The student's question is about converting the speed of the Earth's orbit around the Sun from kilometers per second to miles per hour. The given speed is 29.783 km/s. To convert this to miles per hour, we can follow these steps:

Multiply the kilometres per second by 3600, which is the number of seconds in an hour, to get the kilometres per hour.

Convert kilometers per hour to miles per hour by multiplying by the conversion factor (1 kilometer is approximately 0.621371 miles).

Performing these calculations:

29.783 km/s × 3600 s/hr = 107218.8 km/hr

107218.8 km/hr × 0.621371 mi/km = 66636.7 miles per hour

Therefore, Earth travels around the Sun at an average speed of 66,636.7 miles per hour, expressed with five significant figures.

Answer:

After [tex]t=4.77\tau[/tex] voltage across the capacitor will be 0.85 % of the initial voltage across the capacitor

Explanation:

Let initially voltage across capacitor is [tex]v_0[/tex]

After discharging the voltage across the capacitor is .85% of its initial voltage

So final voltage [tex]v=0.0085v_0[/tex]

We know that voltage across capacitor is given by [tex]v=v_0e^{\frac{-t}{\tau }}[/tex]

So [tex]0.85v_0=v_0e^{\frac{-t}{\tau }}[/tex]

[tex]e^{\frac{-t}{\tau }}=0.0085[/tex]

[tex]{\frac{-t}{\tau }}=ln0.0085[/tex]

[tex]-t=-4.77\tau[/tex]

[tex]t=4.77\tau[/tex]

So after [tex]t=4.77\tau[/tex] voltage across the capacitor will be 0.85 % of the initial voltage across the capacitor

Time constant of capacitive circuit is equal to the ratio of resistance and capacitance

The number of time constants the voltage across a discharging cap has decayed to 0.85% of its' initial value is;

4.768τ

In RC time constants, the formula for voltage across capacitor when discharging is given by;

v = v₀e^(-t/τ)

where;

v₀ = initial voltage on the capacitor

v = the voltage after time t

t = time in seconds

τ = time constant

We are told that the capacitor decayed to 0.85% of its' initial vale. Thus;

v = 0.0085v₀

Thus;

0.0085v₀ = v₀e^(-t/τ)

v₀ will cancel out to give;

0.0085 = e^(-t/τ)

In 0.0085 = -t/τ

-4.768 = -t/τ

Since we want to find number of time constants, then let us make t the subject to get;

t = 4.768τ

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Answer:The process that prevents the bodies of dead plant from filling the earth is decomposition of their remains.

Explanation:

1. Fungi and bacteria in the soil help decompose the remains of dead plant and uses them to enrich and nourish other plants that are alive.

2. Insects, worms and other invertebrates feed on the too

Answer:

Please refer to the attachment below.

Explanation:

Please refer to the attachment below for explanation.

Let's start from the definition of attraction and repulsion. Similar charges tend to repel each other, while different charges attract. When work is done due to the force of attraction its value will be negative, while if work is done due to the force of repulsion its value will be positive.

Given this the sphere has a negative charge.

In other words, if the balloon has a positive charge, it will be attracted by the sphere with negative charge. In this case, you would not have to do a positive job to bring them together. If the balloon has a negative charge, it will be repelled by the sphere with a negative charge. In this case, you will do a positive job to unite them.

Therefore, the load on the balloon is negative.

The charge on the ballon will be Negative

The answer to this question is based on the column's law of forces of attraction and repulsion between two charged atoms.

The formula for calculating the magnitude of the forces is given as

[tex]F=k\dfrac{q_1q_2}{r^2}[/tex]

Here [tex]q_1 & \ q_2[/tex]  are the two charged atoms

r = Distance between the two particles

k is constant

Now if two charged particles are having like charges that are both charged particles carrying positive or both carrying a negative charge then there will be Force of repulsion created between them

If both the  Particles have the opposite charges on them then there will be a force of attraction between them.

The charge on the ballon will be negative that's why there is positive work needed to  bring the ballon  towards the negatively charged sphere

Thus the charge on the ballon will be Negative

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Answer:

a friction force that opposes its movement

All this force has a direction opposite to that of the car

Explanation:

A car traveling at 83 km / h with constant speed has a friction force that opposes its movement, this force comes from several sources:

.- The air resistance that is proportional to the speed

/ - the resistance between the rubbers and the pavement that is constant

.- The internal friction of the different engine components

.- The formation of eddies due to the lack of aerodynamic shape of the car

All these forces must be counteracted the motor force

All this force has a direction opposite to that of the car

Answer:

Yes, a frictional force would be required parallel to the bank in the direction downwards the bank to prevent the car from moving off the bank.

Question:

A 1050-kg car rounds a curve of radius 72 m banked at an angle of 14°. If the car is traveling at 83 kmh, will a friction force be required? If so, in what direction?

Explanation:

When a car is to move round a curve with a particular radius, the centrifugal force due to rounding the curve would act on it, which would try to push it off the bank.

As a result of the inclined bank, a force as a result of gravity will act on it to prevent it from moving off the bank.

Fc = mv^2/r

Fw = mgsin14

Where;

Fc is the centrifugal force.

m = mass

Fw = force as a result of weight.

v = velocity of car

r = radius of circular path

g = acceleration due to gravity.

Looking at the two forces, due to the high speed of the car, low angle of inclination of the bank and relatively low radius of round path, the centrifugal force will be high than the force resulting from the weight to keep the car on the track.

Fc > Fw

Fc is directed out of the bank

Fw is try to keep the car on track

Therefore, when Fc > Fw and without frictional force acting inward(down the bank) to neutralise the centrifugal force the car would move off the bank and out of the round curve track. So frictional force is needed in the direction downwards the bank to keep the car on track.

Answer:

(c)  As 'd' becomes doubled, energy decreases by the factor of 2

Explanation:

Energy stored in a parallel plate capacitor is given by:

[tex]U=\frac{1}{2}CV^2\\\\C=\frac{A\epsilon_{o}}{d}\\\\then\\\\U=\frac{1}{2}\frac{A\epsilon_{o}}{d}V^2--(1)\\\\[/tex]

As capacitor remains connected to the battery so V remains constant. As can be seen from (1) that energy is inversely proportional to the separation between the plates so as 'd' becomes doubled, energy decreases by the factor of 2.

Answer:

(c) It decreases by a factor of 2

Explanation:

Since the capacitor is still connected to the power source, the potential difference remain the same even when the distance is a doubled.

The energy stored in a capacitor can be written as:

E = (1/2)CV^2 .....1

And the capacitance of a capacitor is inversely proportional to the distance between the two plates of the capacitor.

C = kA/d ....2

Therefore, when d doubles, and every other determinant of capacitance remains the same, the capacitance is halved.

Cf = kA/2d = C/2

Cf = C/2

Since the capacitance has been halved and potential difference remains the same, the energy stored would also be halved since the energy stored in the capacitor is directly proportional to the capacitance.

Ef = (1/2)(Cf)V^2

Ef = (1/2)(C/2)V^2 = [(1/2)CV^2]/2

Ef = E/2

Where;

E and Ef are the initial and final energy stored in the capacitor respectively

C and Cf are the initial and final capacitance of the capacitor.

d is the distance between the plates

A is the area of plates

k is the permittivity of dielectrics

Therefore the energy stored in the capacitor is decreased by a factor of 2, when the distance is doubled.

Answer

F = 124 N

Explanation:

given,

mass, m = 5 Kg

time, t = 4.1 s

displacement = y(t)=(2.80 m/s)t +(0.61 m/s³)t³

velocity

[tex]\dfrac{dy(t)}{dt}=2.80 + 1.83 t^2[/tex]

[tex]v=2.80 + 1.83 t^2[/tex]

again differentiating to get the equation of acceleration

[tex]\dfrac{dv}{dt}= 3.66 t[/tex]

[tex]a= 3.66 t[/tex]

force at time t = 4.10 s

F = m a

F = 5 x 3.66 x 4.1

F = 75 N

the net force when crate is moving upward

F = Mg + Ma

F = 5 x 9.8 + 75

F = 124 N

the magnitude of force is equal to 124 N

Answer:

The density of the liquid = 1278.95 kg/m³

Explanation:

Density: This can be defined as the ratio of the mass of  a body to its volume. The S.I unit of density is kg/m³.

From Archimedes principle,

R.d = Density of object/Density of liquid = Weight of object in air/Upthrust in liquid.

D₁/D₂ = W/U .......................... Equation 1

D₂ = D₁(U/W)........................ Equation 2

Where D₁ = Density of aluminum, D₂ = Density of liquid, W = weight of aluminum, U = upthrust.

m₁ = 3.8 kg, m₂ = 2.00 kg, g = 9.8 m/s²

W = m₁g = 3.8(9.8) = 37.24 N.

U = lost in weight of the aluminum = (m₁ - m₂)g = (3.8-2.0)9.8

U = 1.8(9.8) = 17.64 N.

Constant: D₁ = 2700 kg/m³

Substituting these values into equation 2

D₂ = 2700(17.64)/37.24

D₂ = 1278.95 kg/m³

Thus the density of the liquid = 1278.95 kg/m³

Answer:

[tex]L=6.21m[/tex]

Explanation:

For the simple pendulum problem we need to remember that:

[tex]\frac{d^{2}\theta}{dt^{2}}+\frac{g}{L}sin(\theta)=0[/tex],

where [tex]\theta[/tex] is the angular position, t is time, g is the gravity, and L is the length of the pendulum. We also need to remember that there is a relationship between the angular frequency and the length of the pendulum:

[tex]\omega^{2}=\frac{g}{L}[/tex],

where [tex]\omega[/tex] is the angular frequency.

There is also an equation that relates the oscillation period and the angular frequeny:

[tex]\omega=\frac{2\pi}{T}[/tex],

where T is the oscillation period. Now, we can easily solve for L:

[tex](\frac{2\pi}{T})^{2}=\frac{g}{L}\\\\L=g(\frac{T}{2\pi})^{2}\\\\L=9.8(\frac{5}{2\pi})^{2}\\\\L=6.21m[/tex]

Answer:

The answer to your question is v₁ = 5.74 m/s

Explanation:

Data

v₁ = ?

h₁ = 0 m

v₂ = 0.75 m/s

h₂ = 1.65 m

g = 9.81 m/s²

Formula

                mgh₁   +  1/2mv₁²   =   mgh₂  +  1/2mv₂²

mass is not consider (if we factor mass, it is cancelled)

                        gh₁ + 1/2v₁²  =   gh₂  +  1/2v₂²

Substitution

                        (9.81)(0) + 1/2v₁² = (9.81)(1.65) + 1/2(0.75)²

Simplification

                                 0    +  1/2v₁² = 16.19 + 0.28

Solve for v₁

                                            1/2v₁² = 16.47

                                                 v₁² = 2(16.47)

                                                 v₁² = 32.94

Result

                                                v₁ = 5.74 m/s

Minimum speed must 5.74 m/s in order to lift his center of mass 1.65 m and cross the bar with a speed of 0.75 m/s.

Given here,

v₁ - initial velocity = ?

h₁ - initial height = 0 m

v₂ - final velocity = 0.75 m/s

h₂ - final height = 1.65 m

g - gravitational acceleration = 9.81 m/s²

The speed can be calculated by using the formula,

[tex]\bold { mgh_1 + \dfrac 12 mv_1^2 = mgh_2 + \dfrac 12mv_2^2}[/tex]            

factor the mass,

 [tex]\bold { gh_1 + \dfrac 12 v_1^2 = gh_2 + \dfrac 12v_2^2}[/tex]

put the values in the formula, and solve it for V1

[tex]\bold { (9.81)(0) + \dfrac 12v_1^2 = (9.81)(1.65) + \dfrac 12(0.75)^2}\\\\\bold { \dfrac 12v_1^2 = 16.19 + 0.28}\\\\\bold { \dfrac 12v_1^2 = 16.47}\\\\ \bold {v_1^2 = 2(16.47)}\\\\\bold {v_1^2= 32.94}\\\\\bold { v_1 = 5.74\ m/s}[/tex]

Therefore,  minimum speed must 5.74 m/s in order to lift his center of mass 1.65 m and cross the bar with a speed of 0.75 m/s.

To know more about kinetic energy,

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Answer:

Q = 1.095 x 10^-9 C

Let the force experienced by the top piece of tape be F

F = kQ²/r²

r = distance between the two pieces tape = 1.00cm = 1.00 x 10^ -2 m

1/4(pi)*Eo = k = 8.99 x 10^9 Nm²/C²

The electric force of repulsion between the two charges and the weight of the top piece of tape are equal so

F = kQ²/r² = mg

Where m is the mass of the top piece of tape and g is the acceleration due to gravity

On re-arranging the equation above,

Q² = mgr²/k

Q² = ((11.0 x 10^-6) x 9.8 x (1.00x10^-2)²)/(8.99 x 10^9)

Q = 1.095x10^-9 C

Explanation:

The charge Q on both pieces of tape are equal and both act with a force of repulsion on each other.

The force of repulsion between both tapes pushes the top piece of tape upwards. The weight of the top piece of tape acts vertically downward. Since the top tape is in a position of equilibrium, the two forces acting on the top piece of tape must be equal to each other. This assumption is backed up by newton's first law of motion which states that the summation of all forces acting on a body at rest must be equal to zero. That is

Fe (electric force) - Fg (gravitational force) = 0

Fe = Fg

kQ²/r² = mg

On substituting the respective values for all variables except Q and rearranging the equation Q = 1.09 x 10^-9

Answer:

[tex]\Delta x=1-0=1\ m[/tex]

[tex]\Delta v=-2-0=-2\ m.s^{-1}[/tex]

[tex]\Delta a=-14-10=-24\ m.s^{-1}[/tex]

Explanation:

The equation governing the position of the particle moving along x-axis is given as:

[tex]x=5\times t^2-4\times t^3[/tex]

we know that the time derivative of position gives us the velocity:

[tex]\frac{d}{dt} x=v[/tex]

[tex]v=10\ t-12\ t^2[/tex]

and the time derivative of of velocity gives us the acceleration:

[tex]\frac{d}{dt} v=a[/tex]

[tex]a=10-24\ t[/tex]

Now, when t = 0

[tex]x=0\ m[/tex]

[tex]v=0\ m.s^{-1}[/tex]

[tex]a=10\ m.s^{-2}[/tex]

When t=1 s

[tex]x_1=5\times 1^2-4\times 1^3=1\ m[/tex]

[tex]v_1=10\times 1-12\times 1^2=-2\ m.s^{-1}[/tex]

[tex]a_1=10-24\times 1=-14\ m.s^{-2}[/tex]

Hence,

Displacement between the stipulated time:

[tex]\Delta x=x_1-x[/tex]

[tex]\Delta x=1-0=1\ m[/tex]

Velocity between the stipulated time:

[tex]\Delta v=v_1-v[/tex]

[tex]\Delta v=-2-0=-2\ m.s^{-1}[/tex]

Acceleration between the stipulated time:

[tex]\Delta a=a_1-a[/tex]

[tex]\Delta a=-14-10=-24\ m.s^{-1}[/tex]

Here negative sign indicates that the vectors are in negative x direction.

The magnitude of the vector sum of the electric forces exerted on particle 3 is 0.598 N, with the direction angle being 90 degrees, as both forces act along the positive y-axis due to symmetry.

To solve the student's question, we'll have to calculate the forces exerted on particle 3 by particle 1 and particle 2, respectively, and then find the vector sum of these forces. Since both particle 1 and particle 2 are on the x-axis at the same distance from the origin and carry the same charge, the forces they exert on particle 3 will be equal in magnitude and opposite in direction along the y-axis. Therefore, the resulting force on particle 3 will be twice the force exerted by one particle along the y-axis.

Using the formula:
F = k × |q₁ × q₂| / r²
For particle 1 or 2 exerting force on particle 3:
F = (8.9875 × 10⁹) × (3.0 × 10⁻⁹ C × 9.0 × 10⁻⁶ C) / (0.09 m)²
F = 8.9875 × 10⁹ × 2.7 × 10⁻¹⁴ C / 0.0081 m²
F = 0.299 N (Force exerted by one particle)
The total force on particle 3 will be twice this value: 0.598 N.

The direction angle of the vector sum will be 90 degrees since both forces act along the y-axis and hence the resultant force is directed straight up along the positive y-axis when measured counterclockwise from the positive x-axis.

Magnitude is √2 N, Direction is 45° counterclockwise from positive x-axis.

Coulomb's Law: formula for electric force between point charges.

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

[tex]\( q_1 = q_2 = 3.0 \, \text{nC} = 3.0 \times 10^{-9} \, \text{C} \),[/tex]

[tex]\( q_3 = 9.0 \, \text{μC} = 9.0 \times 10^{-6} \, \text{C} \),[/tex]

The distance between particle 3 and particles 1 and 2 is [tex]\( 90 \, \text{mm} = 0.09 \, \text{m} \)[/tex],

All charges are located on the x-axis.

The electric force exerted by particle 1 on particle 3:

[tex]\[ F_1 = \frac{k \cdot |q_1 \cdot q_3|}{r^2} \][/tex]

The electric force exerted by particle 2 on particle 3:

[tex]\[ F_2 = \frac{k \cdot |q_2 \cdot q_3|}{r^2} \][/tex]

Particles 1 and 2, equidistant from 3, have equal electric forces. The magnitude of each force:

[tex]\[ F_1 = \frac{(8.99 \times 10^9) \cdot |3.0 \times 10^{-9} \cdot 9.0 \times 10^{-6}|}{(0.09)^2} \][/tex]

[tex]\[ F_1 \approx \frac{8.09 \times 10^{-5}}{0.0081} \][/tex]

[tex]\[ F_1 \approx 1.00 \, \text{N} \][/tex]

Similarly, [tex]\( F_2 \approx 1.00 \, \text{N} \)[/tex].

Now, the total electric force [tex]\( F_{\text{total}} \)[/tex] exerted on particle 3 is the vector sum of [tex]\( F_1 \) and \( F_2 \)[/tex].

[tex]\[ F_{\text{total}} = \sqrt{F_1^2 + F_2^2} \][/tex]

[tex]\[ F_{\text{total}} = \sqrt{(1.00)^2 + (1.00)^2} \][/tex]

[tex]\[ F_{\text{total}} = \sqrt{2} \, \text{N} \][/tex]

Magnitude of total electric force on particle 3 is [tex]\( \sqrt{2} \, \text{N} \)[/tex].

[tex]\( \theta \)[/tex] is the angle from x-axis to [tex]\( F_{\text{total}} \)[/tex].

[tex]\[ \theta = \arctan\left(\frac{F_2}{F_1}\right) \][/tex]

[tex]\[ \theta = \arctan\left(\frac{1.00}{1.00}\right) \][/tex]

[tex]\[ \theta = \arctan(1) \][/tex]

[tex]\[ \theta \approx 45^\circ \][/tex]

Answer:

Time taken is 0.087 s

Solution:

As per the question:

Time period, T = 0.300 s

Amplitude, A = 6.00 cm

Now,

To calculate the time taken:

For SHM, we know that:

[tex]x = Acos\omega t[/tex]                               (1)

At x = 6.00 cm, the object comes to rest instantaneously at times t = 0.00 s

Thus from eqn (1), for x = 6.00  cm:

[tex]6.00 = 6.00cos\omega t[/tex]

[tex]cos\omega t = 1[/tex]

[tex]\omega t = cos^{- 1}(1)[/tex]

[tex]\omega t = 0[/tex]

Thus at t = 0.00 s, x = 6.00 cm

Now,

Using eqn (1) for x = - 1.50 cm:

[tex]- 1.50 = 6.00cos\omega t'[/tex]

[tex]cos\omega t' = -0.25[/tex]

We know that:

[tex]\omega = \frac{2\pi}{T}[/tex]

Thus

[tex]\frac{2\pi}{0.300} t' = cos^{- 1}(0.25)[/tex]

[tex]t' = 0.087\ s[/tex]

Time taken by the object in moving from x = 6.00 cm to x = 1.50 cm:

t' - t = 0.087 - 0.00 = 0.087 s

To find the time it takes for the object to go from x = 6.00 cm to x = -1.50 cm, we use the equation for simple harmonic motion. The time it takes is half of the period, which is 0.150 s.

To find the time it takes for the object to go from x = 6.00 cm to x = -1.50 cm, we need to use the equation for simple harmonic motion (SHM). The equation is given by x = A * cos(2π/T * t + φ), where x is the position, A is the amplitude, T is the period, t is the time, and φ is the phase shift.

First, we need to determine the phase shift. At t = 0, the object is at rest at x = 6.00 cm. This means the phase shift is 0, because the cosine function is at a maximum at t = 0.

Next, we can plug in the values into the equation. The amplitude A is 6.00 cm and the period T is 0.300 s. We want to find the time it takes for the object to go from x = 6.00 cm to x = -1.50 cm. In SHM, the object goes from -A to A and back in one period, so the time it takes to go from x = 6.00 cm to x = -1.50 cm is half of the period. Therefore, the time is 0.300 s / 2 = 0.150 s.

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Answer:

Yes, there will a force acting on the positive point charge. The options provided are that of negative point charge, rather than a positive point charge stated in the question

Explanation:

Here is the explanation:

When a capacitor is charged, one plate is positively charge and the other is negatively charged. In between the oppositely charged plates, there exist a potential difference. A positive point charge placed in the distance between the two plate will experience an electric force due to the potential difference. The direction of the force will be directed away from the positvely charged plate and towards the negatively charged plate. The reason is due to the law of electrostatic force, which states: like charges repel and unlike charges attract.

A parallel-plate capacitor consists of two large, flat, metal plates held parallel to each other. A positive point charge placed in the gap between the plates and near the center will experience an electric force directed away from the positive plate and towards the negative plate.

A device called a parallel-plate capacitor consists of two large, flat, metal plates held parallel to each other and separated by a small gap. When a positive point charge is placed in the gap between the two plates and near the center of each plate, it experiences an electric force.

The direction of the electric force on the negative charge is from the positive plate towards the negative plate. This means the electric force is directed away from the positive plate and towards the negative plate.

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Answer:

The time difference is 8.33 ms.

The phase difference between them is 60°

Explanation:

Given that,

Frequency = 10 Hz

Angle = 30°

We need to calculate the time difference

Using formula of time difference

[tex]\Delta t=\dfrac{\phi}{360^{\circ}\times f}[/tex]

Put the value into the formula

[tex]\Delta t=\dfrac{30}{360\times10}[/tex]

[tex]\Delta t=8.33\ ms[/tex]

If the frequency of these sine waves doubles, but the time difference stays the same,

[tex]f=20\ Hz[/tex]

We need to calculate the phase difference between them

Using formula of phase difference

[tex]\Delta \phi=\Delta t\times360\times f[/tex]

Put the value in to the formula

[tex]\Delta=8.33\times10^{-3}\times360\times20[/tex]

[tex]\Delta \phi=60^{\circ}[/tex]

Hence, The time difference is 8.33 ms.

The phase difference between them is 60°