1 kg ball rolls off a 33 m high cliff, and lands 23 m from the base of the cliff. Express the displacement and the gravitational force in terms of vectors and calculate the work done by the gravitational force. Note that the gravitational force is <0, , 0>, where is a positive number (+9.8 N/kg). (Let the origin be at the base of the cliff, with the direction towards where the ball lands, and the direction taken to be upwards.)

The cylinder will descend a distance of f/k before coming momentarily to rest.

When the cylinder is released, it will initially accelerate downwards due to the force of gravity. As it accelerates, the spring attached to it will begin to stretch, exerting an upward force on the cylinder. The cylinder will continue to descend until the force exerted by the spring is equal in magnitude to the force of kinetic friction between the cylinder and the walls of the tube. At this point, the net force on the cylinder will be zero, and it will come momentarily to rest.

To find the distance the cylinder descends before coming to rest, we can equate the force exerted by the spring with the force of kinetic friction:

kx = f

Where k is the force constant of the spring and x is the distance the cylinder has descended. Solving for x gives:

x = f/k

Therefore, the vertical distance the cylinder descends before it comes momentarily to rest is given by x = f/k.

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Answer:

11.2 Km/Sec

Explanation:

Escape velocity is the velocity that one has to achieve to go out of the gravitational effect of a body. Here they are talking about escaping from the Earth's gravity. To travel to Mars, first we will have to go far enough so that the impact of Earth's gravity is negligible.

The escape velocity ([tex]v_{e}[/tex]) can be calculate using the given formula:

[tex]v_{e} = \sqrt{\frac{2GM}{r}}[/tex]

where,

G = Gravitational Constant

M = Mass of the object from which we need to escape (Here it is the mass of the Earth)

r = distance from the center of that object (In this case it is equal to the radius of Earth)

For Earth, this comes out to be 11.2 KM/Sec

Incomplete question.The complete one is here

Spacecraft have been sent to Mars in recent years. Mars is smaller than Earth and has correspondingly weaker surface gravity. On Mars, the free-fall acceleration is only 3.8m/s2. What is the orbital period of a spacecraft in a low orbit near the surface of Mars?

Answer:

[tex]T=5900s=99min[/tex]

Explanation:

Given

[tex]r_{satelite}=r_{mars}=3.37*10^{6}m\\ g_{mars}=3.8m/s^{2}\\[/tex]

To find

orbital period of a spacecraft T

Solution

An the initial calculating is computing the angular velocity of satellite :

[tex]w=\frac{2\pi }{T}\\ w=\frac{2\pi }{110min}(1min/60s)\\ w=9.52*10^{-4}rad/s[/tex]

Computing T

[tex]T=\frac{2\pi }{w}\\ as\\w=\sqrt{\frac{a}{r} }\\ So\\T=\frac{2\pi }{\sqrt{\frac{a}{r} }} \\T=\frac{2\pi }{\sqrt{\frac{3.8m/s^{2} }{3.37*10^{6} m} }}\\T=5900s=99min[/tex]

Answer:

Explanation:

Acceleration without friction on an inclined plane = g sinθ

Acceleration with friction on inclined plane = g sinθ - μ g cosθ

s = 1/2 a t²

For motion on friction-less surface

s = 1/2 g sinθ t₁²

For motion on frictional surface

s = 1/2( g sinθ - μ g cosθ)  t₂²

t₂ = 2t₁

(  sinθ - μ  cosθ)  t₂² =  sinθ t₁²

(  sinθ - μ  cosθ)  4t₁² =  sinθ t₁²

(  sinθ - μ  cosθ)  4 =  sinθ

4sinθ - sinθ = 4μ  cosθ

3sinθ = 4μ  cosθ

3 / 4 tanθ =μ  

μ  = .75  tanθ

Answer:

Th = 50°C

Explanation:

See the attachment below.

K = 7.75

Tc = 13.5°C = 273+13.5 = 286.5K

Answer:

292.984 K or 19.98 °C

Explanation:

Using,

η = Tc/(Th-Tc)................ Equation 1

Where η = coefficient of performance, Tc = cold temperature of the refrigerator, Th = hot temperature of the refrigerator.

Make Th the subject of the equation

η (Th-Tc) = Tc

η Th-η Tc = Tc

Th = (Tc+ηTc)/η ....................... Equation 2

Given: Tc = -13.5°C+273 = 259.5 °C, η  = 7.75

Substitute into equation 2

Th = (259.5+7.75×259.5)/7.75

Th = (259.5+2011.125)/7.75

Th = 2270.625/7.75

Th = 292.984 °C or 19.98 °C

Hence the maximum hot reservoir temperature = 292.984 °C or 19.98 °C

Answer:23 N

Explanation:

Given

Weight of box [tex]W=60\ N[/tex]

Coefficient of static friction is [tex]\mu _s=0.5[/tex]

Applied force [tex]F=23\ N[/tex]

When Force is applied box is observed to be at rest i.e. static friction is overcoming the applied force.

Thus Static friction [tex]F_s[/tex]=applied force

[tex]F_s=23\ N[/tex]

Although it maximum value can go up to [tex](F_s)_{max}=\mu _sN[/tex]

[tex]F_s=0.5\times 60[/tex]

[tex]F_s=30\ N[/tex]

Answer:

The correct answer is the number 4. A wave on a pond is a mechanical wave which requires a medium to travel.

Explanation:

Mechanical waves are those that need a material medium to propagate. The waves of the sea and the waves that we produce on a guitar string, the sound, are examples of mechanical waves. Electromagnetic waves are energetic pulses capable of propagating in a vacuum. This way, a wave on a pond is a mechanical wave which requires a medium to travel.

Final answer:

A wave on a pond is a mechanical wave requiring water to propagate, while electromagnetic waves, such as light, don't require a medium and can travel through a vacuum. Therefore, statement 4 is correct. This reflects fundamental differences in how mechanical and electromagnetic waves propagate and their need for a medium.

Explanation:

To compare waves on a pond with electromagnetic waves, we should understand that a wave on a pond is a mechanical wave which requires a medium to travel (option 4). This is because mechanical waves are disturbances that propagate through a material medium and the pond water serves as this medium. On the other hand, electromagnetic waves do not need a medium; they can travel through a vacuum because they consist of oscillating electric and magnetic fields.

So, the correct statement would be that a wave on a pond is a mechanical wave which requires a medium to travel. Therefore, the waves on a pond are similar to sound waves since both are mechanical and need a medium. In contrast, light waves are an example of electromagnetic waves, which means they can travel through the vacuum of space without any medium.

Furthermore, all electromagnetic waves move at the same speed in a vacuum, which is known as the speed of light. This is a fundamental difference between mechanical and electromagnetic waves since the speed of mechanical waves depends on the medium in which they are traveling. The discovery that electromagnetic waves do not require a medium led to the abandonment of the aether theory, which was invented to provide a medium for light and other electromagnetic wave propagation.

Answer:

(a) the initially stationary spelunker is accelerated to a speed of 4.70 m/s - 6106 J

(b) he is then lifted at the constant speed of 4.70 m/s - 5488 J

(c) finally he is decelerated to zero speed. How much work is done on the 56.0 kg rescue by the force lifting him during each stage - 4869 J

Explanation:

knowing

d = 10 m

m = 56 kg

The work done by the applied force to pull the spelunker is given by

Wa + Wg = Kf - Ki

Wg = -mgd

First

Wa = mgd + 0.5 [tex]mv^{2}_{f}[/tex] - 0.5 [tex]mv^{2}_{i}[/tex]

Wa = (56*9.8*10) + (0.5*56*[tex]4.7^{2}[/tex])

Wa = 6106 J

Second

Kf = Ki

Wa = mgd + 0.5 [tex]mv^{2}_{f}[/tex] - 0.5 [tex]mv^{2}_{i}[/tex]

Wa = 56*9.8*10

Wa = 5488 J

Third

Kf = 0

Wa = mgd + 0.5 [tex]mv^{2}_{f}[/tex] - 0.5 [tex]mv^{2}_{i}[/tex]

Wa = (56*9.8*10) - (0.5*56*[tex]4.7^{2}[/tex])

Wa = 4869 J

Answer:

72 Nm²

Explanation:

work = F . d

work = (12i ,0j) . (6i ,-8j )

work = 72 J

The 12N force is a vector acting in the direction of the positive x axis, so its vector notation is 12i + 0j.

It's not exactly clear what the point's location is but I interpret it to be x= m*6, y= -m*8 (I hope the "m" wasn't a typo). The direction vector from the origin is then m*6i -m*8j

When a force vector F acts in a nonparallel direction d, the work is given by:

W = |F|*|d|*cos(theta) where theta is the angle between the vectors.

alternatively you can use the dot product of the two vectors to get:

W = (12N)*(m*6) + (0N)*(-m*8) = 72 Nm²

Final answer:

The work done on the block by the constant force of 12 N in the +x direction is 24 J.

Explanation:

The work done on the block can be calculated by multiplying the force applied to the block by the displacement of the block in the direction of the force. In this case, the force applied is 12 N in the +x direction and the displacement is (6 - 8) m = -2 m. Since the force and displacement are in opposite directions, the work done will be negative. The formula for work is given by:

Work = Force * Displacement * cos(theta)

where theta is the angle between the force and the displacement.

In this case, the angle between the force and the displacement is 180 degrees, so cos(theta) = -1. Substituting the values into the formula:

Work = 12 N * -2 m * (-1) = 24 J

Answer:

Normal Conversation: i=106i0

i(dB)=60

Power saw a 3 feet: i=1011i0

i(dB)=110

Jet engine at 100 feet: i=1018i0

i(dB)=180

Explanation:

if these are the same as edge, then these are the answers! :)

Normal Conversation: i=106i0

The intensity in decibels is 60,110,180, respectively.

⇒ normal conversation: i =60(dB)

⇒power saw 3 feet: i =110(dB)

⇒jet engine at 100 feet: i = 180(dB)

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The pilot's maximum endurance speed in a circle of 85 m diameter without blacking out can be calculated using the centripetal acceleration formula, taking into account the acceleration limit of 7 g, where 1 g equals 9.80 m/s².

The question requires the application of concepts from physics, specifically the idea of centripetal force and acceleration when a body moves in a circular path.

Given that the maximum acceleration a pilot can endure without blacking out is 7 g, and knowing that 1 g is equivalent to 9.80 m/s², we can calculate the maximum speed under these conditions using the formula for centripetal acceleration, which is `a = v² / r`, where `a` is the acceleration, `v` is the speed, and `r` is the radius of the circle.

The diameter of the circle is given as 85 meters, which means the radius `r` will be half of that, 42.5 meters. The maximum acceleration of 7 g translates to 7 × 9.80 m/s², which is 68.6 m/s². We can rearrange the formula to solve for `v`, leading to `v = √(a × r)`.

Substituting the values provided, we get `v = √(68.6 m/s² × 42.5 m)

v = √2923.175 m²/s² ≈ 54.1 m/s

Therefore, the maximum speed the pilot can tolerate while spinning in the horizontal circle is approximately 54.1 m/s, which is equivalent to about 192 km/h.

Answer:

  B = -0.215 s⁻¹,  A₀ = 1.537º

Explanation:

To be able to make this graph we must linearize the data, the best procedure is to calculate the logarithm of the values

           A = A₀ [tex]e^{-Bt}[/tex] + C

The constant creates a uniform displacement of the graph if we start the graph at the angle of 1, the constant disappears, this is done by subtracting 1 from each angle, so the equation is

            (A-1) = A₀  e^{-Bt}

We do the logarithm

           Log (A-1) = log Ao –Bt log e

Make this graph because paper commercially comes in logarithm 10, if we use graph paper if we can calculate directly in base logarithm e, let's perform this calculation

              Ln (A-1) = Ln A₀ - Bt

To graph the points we subtract 1 from each angle and calculate the logarithm, the data to be plotted are

              θ’= ln (θ -1)

θ(º)    t(s)        θ'(º)

6.2      2     1.6487

4.7      4     1.3083

3.2      6    0.7885

2.4      8    0.3365

1.8     10   -0.2231

We can use a graph paper and graph on the axis and the primary angle (θ') and on the x-axis time, mark the points and this graph is a straight line, we see that the point for greater time has a linearity deviation , so we will use the first three for the calculations

To find the line described by the equation

              y -y₀ = m (x -x₀)

             m = (y₂ -y₁) / (x₂ -x₁)

Where m is the slope of the graph e (x₀, y₀) is any point, let's start as the first point of the series

            (x₀, y₀) = (2, 1.65)

           (x₂, y₂) = (2, 1.65)

           (x₁, y₁) = (6, 0.789)

We use the slope equation

           m = (1.65 - 0.789) / (2-6)

           m = -0.215  

The equation is

           y - 1.65 = -0.215 (x- 2)

           y = -0.215 x +0.4301

We buy the two equations and see that the slope is the constant B

            B = -0.215 s⁻¹

The independent term is

            b = ln A₀

            A₀ = [tex]e^{b}[/tex]

           A₀ = [tex]e^{0.43}[/tex]

           A₀ = 1.537º

Answer:

Explanation:

Length, l = 35.5 cm = 0.355 m

y = 2 x

Slope, y / x = 2

tan θ = 2

θ = 63.4°

i = 22.5 A

B = 0.318 i Tesla

[tex]\overrightarrow{l}=0.355\widehat{k}+0.355 Cos63.4\widehat{i}+0.355 Sin63.4\widehat{j}[/tex]

[tex]\overrightarrow{l}=0.16\widehat{i} + 0.32\widehat{j}+0.355\widehat{k}[/tex]

[tex]\overrightarrow{B}=0.318\widehat{i}[/tex]

The magnetic force is given by

[tex]\overrightarrow{F}=i(\overrightarrow{l}\times \overrightarrow{B})[/tex]

[tex]\overrightarrow{F}=22.5\left ( 0.16\widehat{i}+0.32\widehat{j}+0.355\widehat{k} \right )\times 0.318\widehat{i}[/tex]

[tex]\overrightarrow{F}=2.54\widehat{j}-2.29\widehat{k}[/tex]

Magnitude of force

[tex]F=\sqrt{2.54^{2}+2.29^{2}}[/tex]

F = 3.42 N

The angle is Ф

tanФ = -2.29/2.54

Ф = 42° below y axis

The total force experienced by the wire in the magnetic field has a magnitude of 145 N and is directed -30 degrees relative to the z-axis, according to the right-hand rule and Lorentz force law.

In order to determine the direction of the total force on the wire, you would need to use the right-hand rule. First, you know the current is flowing down the z-axis and out along the y=2x line in the xy-plane. You also know that the magnetic field, B, is given by (.318 i)T which lies along the x-axis.

According to the Lorentz force law, the force exerted on a current-carrying conductor in a magnetic field is given by F = I * (L x B), where I represents the current, L is the length vector of the wire, and x is the cross product. Because the wire is bent at a right angle, the force will have two components: one for each section of the wire.

The force along the z-axis is given by F = I*LB*sin(theta), where theta is the angle between L and B. Here, L = 35.5/2 = 17.75 cm, B = 0.318 T, I = 22.5 A, and theta = 90 degrees (since B is along x-axis and L is along z-axis). Calculating gives Fz = 22.5 A * 17.75 cm * 0.318 T * sin(90) = 126.6 N.

The force along the y-axis is similar, but with the length vector along y=2x in the xy plane and B still along the x-axis, so theta = 180 degrees - arctan(2). Substituting the values we get Fy = - 22.5 A * 17.75 cm * 0.318 T * sin(180 - arctan(2)) = -73.8 N.

Therefore, the total force is given by the vector sum of Fz and Fy, which has magnitude sqrt(Fz^2 + Fy^2) = 145 N, in the direction of atan2(Fy, Fz) = -30 degrees relative to the z-axis.

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Answer:

    a = 4.8 10⁻⁵ m

Explanation:

The diffraction phenomenon is described by the expression

        a sin θ = m λ

How the pattern is observed on a distant screen

        tan θ = y / L = sin θ / cos θ

Since the angle is very small in these experiments ’we can approximate the tangent function

    tan θ = sin θ = y / L

We substitute

           a y / L = m λ

The first minimum occurs for m = 1

        a = λ L / y

        a = 680 10⁻⁹ 5.5 / 0.078

        a = 4.8 10⁻⁵ m

Answer:

5.02 m/s

Explanation:

given,

frequency of sound emitted by bat, [tex]f_1[/tex] = 30 kHz = 30000 Hz

beat frequency detected  by bat[tex]f_b[/tex] = 900 Hz

Speed of sound, v = 340 m/s

[tex]f_b = f_1 -f_2[/tex]

[tex]900 = 30000-f_2[/tex]

[tex]f_2 = 30900[/tex]

Using Doppler's effect formula

[tex]f_2=f_1(\dfrac{v+v_b}{v-v_b})[/tex]

[tex]v_b = velocity\ of\ bat[/tex]

[tex]30900=30000(\dfrac{340+v_b}{340-v_b})[/tex]

[tex]1.03 =(\dfrac{340+v_b}{340-v_b})[/tex]

[tex]v_b = 5.02 m/s[/tex]

Hence, the speed of bat is equal to 5.02 m/s

To solve this problem we will apply the concepts related to energy conservation. We know that potential energy is transformed into kinetic and vice versa energy. Since the energy accumulated in the upper part is conserved as potential energy, when the object is thrown all that energy will be converted into kinetic energy. Therefore we will have the following relation,

[tex]KE = PE[/tex]

[tex]KE = mgh[/tex]

Here,

m = mass

g = Gravitational acceleration

h = Height

Replacing,

[tex]KE = (6)(9.8)(5)[/tex]

[tex]KE = 294J[/tex]

Therefore the kinetic energy at the bottom is 294J

Answer:

225.56V

Explanation:

Q=cv

6.93 µF capacitor was charged to 94 V, therefore Q₁ =6.93 *10^-6 x 94 =651.42µC

1.76 µF capacitor charged to 81 V, therefore Q₂ =1.76 *10^-6 x 81 =142.56µC

When the capacitor are brought together, the charges on them move to maintain equilibrum, so therefore, each capacitor has 0.5(142.56µC + 651.42µC )= 396.99µC

The final potential difference across the 1.76 µF capacitor =  Q/c = 396.99µC/1.76 µF = 225.56V

To find the final potential difference across the 1.76 µF capacitor, use charge conservation by equating the total charge before and after connection. After calculating the charges on each capacitor, rearrange the equation to solve for the final potential difference across the 1.76 µF capacitor.

In order to find the final potential difference across the 1.76 µF capacitor, we can use the concept of charge conservation.

First, let's calculate the total charge on the capacitors before they are connected. The charge on a capacitor is given by Q = CV, where Q is the charge, C is the capacitance, and V is the potential difference across the capacitor.

For the 6.93 µF capacitor, the charge is Q1 = (6.93 µF)(94 V) = 650.42 µC. For the 1.76 µF capacitor, the charge is Q2 = (1.76 µF)(81 V) = 142.56 µC.

After connecting the capacitors, the total charge remains the same. So, we have Q1 + Q2 = Qf, where Qf is the final charge on the capacitors. Rearranging the equation, we can find the final potential difference across the 1.76 µF capacitor:

Vf = (Qf - Q1) / C2 = (142.56 µC - 650.42 µC) / (1.76 µF).

Calculating this expression will give us the final potential difference across the 1.76 µF capacitor.

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Explanation:

Given data:

G = 11.5×10⁶psi

d₂ = 2.0 inch

d₁ = 1.5 inch

ε[tex]_{max}[/tex] = 170 × 10⁻⁶

Y[tex]_{max}[/tex] = 2ε

T/J = τ[tex]_{max}[/tex] /R

[tex]\frac{Td_{2} }{2J}[/tex] = τ[tex]_{max}[/tex]         (1)

τ[tex]_{max}[/tex] = G Y

from 1 and 2

[tex]\frac{T d_{2} }{2J} = G Y_{max}[/tex]

T = [tex]\frac{2 G Y_{max}J }{d_{2} }[/tex]

[tex]\frac{2* 11.5*10^{6}*0.006895*10^{6}*340*10*^{-6} *\frac{\pi }{32}[2^{4}-1.5^{4}]*(0.0254)^{4} }{2* 0.0254}[/tex]

  = 474.14 Nm

Final answer:

The magnitude of the electric force on the second object remains the same when the first object is moved to a distance 2r from it.

Explanation:

To find the magnitude of the electric force on the second object (B) when the first object (A) is moved to a distance 2r from B, we can use Coulomb's law. Coulomb's law states that the magnitude of the electric force between two charged objects is directly proportional to the product of their charges and inversely proportional to the square of the distance between them.

Given that the charge of A is Q and the charge of B is 4Q, the force exerted on B when A is at distance r is F. Now, when A is moved to a distance 2r from B, the new distance between A and B is 2r. Using Coulomb's law and the proportionalities mentioned earlier, the magnitude of the new electric force on B can be calculated as:

F' = (k * Q * 4Q) / (2r)^2

Where k is the constant in Coulomb's law. Simplifying the equation gives:

F' = (4 * F) / 4 = F

Therefore, the magnitude of the electric force on the second object B remains the same when the first object A is moved to a distance 2r from B.

Answer:

The coefficient of kinetic friction between the puck and the ice is 0.05.

Explanation:

Given that,

Weight of the hockey puck, W = mg = 3.5 N

Acceleration of the hockey puck, [tex]a=0.45\ m/s^2[/tex]

The frictional force opposing the motion of the puck is given by

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

This force is balanced by the force due to motion of the hockey puck. So,

[tex]\mu N=ma\\\\\mu mg=ma\\\\\mu=\dfrac{a}{g}\\\\\mu=\dfrac{0.45}{9.8}\\\\\mu=0.045[/tex]

or

[tex]\mu=0.05[/tex]

So, the coefficient of kinetic friction between the puck and the ice is 0.05. Hence, this is the required solution.

Answer:

The negative electrons moves to the left While the positively charged current moves in the opposite direction to the right when an external field is applied.

The positive atomic nuclei remains almost motionless.

In an external field, electrons and atomic nuclei move differently. In an electric field, electrons tend towards the positive field direction, whereas nuclei tend towards the negative. In a magnetic field, their paths are perpendicular to the field lines.

The motion of negative electrons and positive atomic nuclei under the influence of an external field depends on the nature of the external field. This field could be of many types such as electric, magnetic, or electromagnetic. In an electric field, electrons move towards the positive field direction while the nuclei move towards the negative field direction. In a magnetic field, the motion of electrons and nuclei is governed by the right-hand rule, which means they move in a path perpendicular to the magnetic field lines.

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Answer:

Φ = 2.26 10⁶ N m² / C

Explanation:

The electric flow is

           Ф = E .ds = [tex]q_{int}[/tex] / ε₀

Since we have two components for this flow. The uniform electric field and the flux created by the point charge.

The flux of the uniform electric field is zero since the flux entering is equal to the flux leaving the Gaussian surface

The flow created by the point load at the origin is

              Φ = q_{int} /ε₀

Let's calculate

              Φ = 2.00 10⁻⁶ /8.85 10¹²

              Φ = 2.26 10⁶ N m² / C

Answer:

Explanation:

Since , resistances are joined in parallel , potential difference across either of R₁ or R₂ will be same , and it will be equal to emf of the cell provided no internal resistance exists in the cell.

Since potential difference across R₁ is equal to emf and resistance is R₁ , current through it too will be same as that in the previous case.

The potential difference across resistor R1 remains the same when R2 is added in parallel, and the current through R1 also remains unchanged because the voltage across it does not change.

When R2 is added in parallel to R1, the potential difference (voltage) across R1 remains the same because in a parallel circuit, all components have the same potential difference across them. However, the current i1 through R1 remains the same as previously as well since the potential difference across it hasn't changed and Ohm's Law (I = V/R) dictates that the current through a resistor will depend only on the potential difference across it and its resistance, which remains unchanged.

Adding R2 in parallel with R1 has the effect of decreasing the overall resistance of the circuit as found using the equation 1/Rp = 1/R1 + 1/R2. This means that the total current drawn from the battery will increase, but it will be divided among R1 and R2. Since the voltage across R1 remains constant, the current through R1 is unaffected by the addition of R2.

Answer:

The centripetal acceleration at highway speed is greater.

Explanation:

We assume the motion of the car is uniformly accelerated. Let the highway speed be v.

By the equation of motion,

[tex]v=u+at[/tex]

[tex]a=\dfrac{v-u}{t}[/tex]

u is the initial velocity, a is acceleration and t is time

Because the car starts from rest, u = 0.

[tex]a_T=\dfrac{v}{t}[/tex]

This is the tangential acceleration of the thread of the tire.

The centripetal acceleration is given by

[tex]a_C=\dfrac{v^2}{r}[/tex]

r is the radius of the tire.

Comparing both accelerations and applying commonly expected values to r and t, the centripetal acceleration is seen to be greater. The radius of a tyre is, on the average, less than 0.4 m. Then the centripetal acceleration is about

[tex]a_C=\dfrac{v^2}{0.3}=2.5v^2[/tex]

The tangential acceleration can only be greater in the near impossible condition that the time to attain the speed is on the order of microseconds.

Answer:

Explanation:

Given that,

Sphere of radius R

At radial distance ¼R it has an electric field of Eo

Magnitude of electric field E at r=2R

Solution

electric field is given as

E=KQ/r²

Now at R,

E=KQ/R²

When, at r=R/4 inside the sphere is given as

Eo=kQ/R³ • R/4

Eo=kQ/4R²

When, r =2R

Then,

E=kQ/(2R)²

E=kQ/4R²

Comparing this to Eo

Then, E=Eo

Then, the electric field at r=2R is equal to the electric field at r=R/4

The angular velocity of the merry-go-round increases as a child walks towards the center, due to the conservation of angular momentum and a decrease in the system's moment of inertia.

When a child walks from the outer edge of a rotating merry-go-round towards the center, the system's angular momentum remains constant due to the conservation of angular momentum. However, the moment of inertia of the system decreases as the child moves closer to the center. Since angular momentum is the product of the moment of inertia and angular velocity (L = I∙ω), and it is conserved, the angular velocity of the merry-go-round must increase to compensate for the decrease in moment of inertia.

This concept can be compared to a figure skater pulling in their arms to spin faster. As the skater's arms come closer to the body, their moment of inertia decreases, causing an increase in their spinning speed, or angular velocity, with no external torques acting on the system.

The same principle applies to the merry-go-round: as the child moves inward, the system's angular velocity increases to conserve angular momentum. This can be observed in playgrounds and is a practical example of conservation laws in rotational motion.

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As the child walks towards the center of the rotating merry-go-round, the system's angular momentum remains constant due to the conservation of angular momentum. The moment of inertia decreases, causing the angular velocity to increase.

A child is standing on the edge of a merry-go-round that is rotating with frequency f.

As the child walks towards the center of the merry-go-round, the angular momentum of the system (child plus merry-go-round) remains constant.

This is due to the conservation of angular momentum, which states that if no external torque acts on the system, the total angular momentum remains unchanged.

Angular momentum (L) is given by the formula:

where I is the moment of inertia and ω is the angular velocity.

complete question:

A child is standing on the edge of a merry-goround that is rotating with frequency f. The child then walks towards the center of the merry-go-round. For the system consisting of the child plus the merry-go-round, what remains constant as the child walks towards the center of the rotating merry-go-round?

To solve this problem we will apply the concepts related to Power, such as the product of voltage and current. Likewise, Power is defined as the amount of energy per unit of time, therefore, using this relationship we will solve the second part.

PART A)

We know that Power is,

[tex]P = VI[/tex]

Substitute [tex]25.0V[/tex] for V and 12.0 A for I in the expression,

[tex]P=(25.0V)(12.0A)[/tex]

[tex]P = 300W[/tex]

Therefore the power deliver to the body is 300W

PART B) Converting the time to SI,

[tex]t = 3ms = 0.003s[/tex]

Therefore if we have that the expression for energy is,

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

Here,

E = Energy,

P = Power,

t = Time,

Replacing,

[tex]E = (300W)(0.003s)[/tex]

[tex]E = 0.9J[/tex]

Therefore the energy transferred is equal to 0.9J

This question involves the concepts of electrical power and energy.

(a) The power delivered by the defibrillator is "300 W".

(b) The energy trasferred is "0.9 J".

The electrical power delivered by the defibrillator can be given by the following formula:

[tex]P=IV[/tex]

where,

Therefore,

[tex]P=(12\ A)(25\ V)[/tex]

P = 300 W

The transferred energy can be given by the following formula:

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

where,

Therefore,

[tex]E=(300\ W)(3\ x\ 10^{-3}\ s)\\[/tex]

E = 0.9 J

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Answer: The intensity level of sound in the bedroom is 80dB

Explanation:

Intensity of lawn mower at r=1m is 100dB

Beta1= 10dBlog(I1/Io)

100dB= 10dB log(I1/Io)

10^10= I1/Io

I1= Io(10^10)

10^12)×(10^10)= I1

I1=10^-2w/m^2

Intensity of lawn mower at r=20m

I2/I1=(r1/r2)^2 =(1/20)^2

I2= I1(1/400)

I2=2.5×10^-3W_m^2

Intensity of 4 lown mowers at 20m fro. Window

= 10dBlog(4I2/Io)

= 10^-4/10^-12

=80dB

The minimum coefficient of static friction between the pavement and the tires is 0.156.

The minimum coefficient of static friction is calculated by applying Newton's second law of motion in determining the net force.

[tex]F_c = Wsin(\theta) + \mu_s W cos(\theta)\\\\\frac{mv^2}{r} = mg sin(\theta) + \mu_s mg cos(\theta)\\\\\mu_s mg cos(\theta)\ = \frac{mv^2}{r} - mg sin(\theta) \\\\\mu _ s = \frac{mv^2 \ - \ mgr sin(\theta)}{mg rcos(\theta)}[/tex]

where;

[tex]\mu _ s = \frac{1200 \times 18.1^2 \ - \ 1200 \times 9.8 \times 100 sin(10.4)}{1200 \times 9.8 \times 100 \times cos(10.4)}\\\\\mu_s = 0.156[/tex]

Thus, the minimum coefficient of static friction between the pavement and the tires is 0.156.

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Answer:

210,600πft-lb

Explanation:

Force is a function F(x) of position x then in moving from

x = a to x= b

Work done = [tex]\int\limits^b_a {Fx} \, dx[/tex]

Consider a water tank conical in shape

we will make small horizontal section of the water at depth h and thickness dh and also assume radius at depth h is w

we will have ,

[tex]\frac{w}{12} = \frac{(18-h)}{18} \\w = \frac{2}{3} (18-h)a[/tex]

weight of slice under construction

weight = volume × density × gravitational constant

[tex]weight = \pi \times w^2 \times dh \times 62.4\\= (62.4\pi w^2dh)lb[/tex]

Now we can find work done

[tex]W = \int\limits \, dw\\[/tex]

[tex]\int\limits^{18}_{3} {(62.4\pi } \, dx w^2dh)h\\= 62\pi \frac{4}{9} \int\limits^{18}_{3} {(18-h)^2hdh} \,[/tex]

= [tex]62.4\pi \times\frac{4}{9} (\frac{324}{2} h^2-\frac{36}{3} + \frac{h^4}{4})^1^8_3[/tex]

= 210,600πft-lb