A ball is tossed so that it bounces off the ground, rises to a height of 2.50 m, and then hits the ground again 0.90 m away from the first bounce.
1.How long is the ball in the air between the two bounces?2.What is the ball's velocity in the x-direction?3.What is the ball's speed just before the second bounce?4.What is the angle of the velocity vector with respect to the ground right after the first bounce

Answer:

iv) It is 9x bigger than before

Explanation:

As the amplitudes of the new speakers add directly with the original one, taking into account the phase that they have, the composed amplitude of the sound wave is as follows:

At = A + 4A -2A = 3 A

The intensity of the wave, assuming it propagates evenly in all directions, is constant at a given distance from the source, and can be expressed as follows:

I = P/A

where P= Power of the wave source, A= Area (for a point source, is equal to the surface area of a sphere of radius r, where is r is the distance to the source along a straight line)

For a sinusoidal wave, the power is proportional to the square of the amplitude, so the intensity is proportional to the square of the amplitude also.

If the amplitude changes increasing three times, the change in intensity will be proportional to the square of the change in amplitude, i.e., it will be 9 times bigger.

So, the statement iv) is the right one.

A light bulb acts as a non-ohmic resistor due to its changing resistance with temperature changes, primarily caused by increased atomic vibrations impeding electron flow. This contrasts with Ohm's Law's assumption of constant resistance, illustrating the complex atomic level interactions that govern a light bulb's resistance and thereby its non-ohmic behavior.

The reason a light bulb is considered a non-ohmic resistor primarily lies in its changing resistance with variations in temperature, a characteristic that defies the principle of Ohm's Law which presumes constant resistance. In a light bulb, particularly an incandescent one, the filament's resistance increases significantly as it heats up from room temperature to its operating temperature. This increase in temperature, and consequently resistance, is not simply due to thermal expansion but is rooted in atomic level interactions.

At the atomic level, as the filament's temperature rises, the atoms inside the metal filament vibrate more vigorously. This enhanced vibration creates more impediments for the free flow of electrons, which is the principal cause of electrical current. Hence, with more obstacles in their path, electrons face increased resistance. This change in resistance with temperature illustrates the non-ohmic behavior as it shows that the resistance isn't constant but varies with temperature. It's clear that thermal expansion plays a role in this scenario, but the key factor is the increased atomic vibrations that hinder electron flow.

The power dissipation in resistors, and by extension in light bulbs, can be described by the equations P = V^2/R and P = I^2R. These seemingly contradictory formulae actually complement each other in explaining how power dissipation can either increase or decrease with rising resistance, depending on whether the scenario is considered from the perspective of voltage or current, further illustrating the complex relationship between these variables in a non-ohmic conductor like a light bulb.

Answer:

The sound level is 138.97 dB.

Explanation:

Given that,

Sound level L= 145 dB

We need to calculate the intensity

Using formula of sound intensity

[tex]L=10\log(\dfrac{I}{I_{0}})[/tex]

Put the value into the formula

[tex]145=10\log(\dfrac{I}{I_{0}})[/tex]

[tex]\log(\dfrac{I}{I_{0}})=\dfrac{145}{10}[/tex]

[tex]\log(\dfrac{I}{I_{0}})=14.5[/tex]

[tex]\dfrac{I}{I_{0}}=10^{14.5}[/tex]

[tex]I=10^{14.5}\times I_{0}[/tex]

[tex]I=10^{14.5}\times10^{-12}[/tex]

[tex]I=316.2[/tex]

We need to calculate the noise of one engine

Using formula of intensity

[tex]I'=\dfrac{I}{4}[/tex]

Put the value into the formula

[tex]I'=\dfrac{316.2}{4}[/tex]

[tex]I'=79.05[/tex]

We need to calculate the intensity level due to one engine

Using formula of sound intensity

[tex]L'=10\log(\dfrac{I'}{I_{0}})[/tex]

[tex]L'=10\log(\dfrac{79.05}{10^{-12}})[/tex]

[tex]L'=138.97\ dB[/tex]

Hence, The sound level is 138.97 dB

In a simplified situation, if the pilot shuts down all engines but one, the person would experience an estimated sound level of approximately 141.5 dB. This is because decibels combine logarithmically, not linearly.

The sound of an airplane engine is a result of four equally noisy jet engines operating together, hence the combined sound level is 145 dB. The sound level is measured in decibels (dB), which is a logarithmic unit. The unit being logarithmic implies that when you decrease the number of sound sources, in this case the number of engines, the sound level does not decrease linearly.

Let's consider each engine contributes equally to the total sound; thus, each engine would approximately produce a sound level of 141.5 dB (since dB is a logarithmic unit, they combine logarithmically, not linearly). Therefore, if the pilot shuts down all engines but one, the resulting sound level experienced would be approximately 141.5 dB.

It's important to comprehend that actual schoolwork problems involving sound levels are more complicated. This simplified explanation does not consider other factors like interference, which can enhance or diminish the overall sound level.

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Answer:

The velocity of the skateboard is 0.774 m/s.

Explanation:

Given that,

The spring constant of the spring, k = 3086 N/m

The spring is stretched 18 cm or 0.18 m

Mass of the student, m = 100 kg

Potential energy of the spring, [tex]P_f=20\ J[/tex]

To find,

The velocity of the car.

Solution,

It is a case of conservation of energy. The total energy of the system remains conserved. So,

[tex]P_i=K_f+P_f[/tex]

[tex]\dfrac{1}{2}kx^2=\dfrac{1}{2}mv^2+20[/tex]

[tex]\dfrac{1}{2}\times 3086\times (0.18)^2=\dfrac{1}{2}mv^2+20[/tex]

[tex]50-20=\dfrac{1}{2}mv^2[/tex]

[tex]30=\dfrac{1}{2}mv^2[/tex]

[tex]v=\sqrt{\dfrac{60}{100}}[/tex]

v = 0.774 m/s

So, the velocity of the skateboard is 0.774 m/s.

The minimum intensity of light required to release 1.40 x 10^11 electrons with a kinetic energy of 3.61 x 10^-19 J in a 0.20 second pulse of light is 25.25 Watts.

The question requires us to calculate the minimum intensity of light needed to release a certain number of electrons with a given kinetic energy in a set amount of time. This can be done using the formula for intensity (I), which is total energy (E) divided by time (t).

The total energy can be found by multiplying the kinetic energy of each electron (Ke) by the total number of electrons (N). Given, Ke = 3.61 x 10^-19 J, and N = 1.40 x 10^11, we get E = Ke x N = 3.61 x 10^-19 J x 1.40 x 10^11 = 5.05 J. Next, we divide this total energy by the pulse duration to find the intensity. Given t = 0.20 s, we have I = E/t = 5.05 J / 0.20 s = 25.25 Watts.

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Answer:

[tex]\vec{A}\times \vec{B}=-51.12\hat{k}[/tex]

[tex]\theta=83.2^{\circ}[/tex]

Explanation:

We are given that

[tex]\vec{A}=4.2\hat{i}+7.2\hat{j}[/tex]

[tex]\vec{B}=5.70\hat{i}-2.40\hat{j}[/tex]

We have to find the scalar product and the  angle between these two vectors

[tex]\vec{A}\times \vec{B}=\begin{vmatrix}i&j&k\\4.2&7.2&0\\5.7&-2.4&0\end{vmatrix}[/tex]

[tex]\vec{A}\times \vec{B}=\hat{k}(-10.08-41.04)=-51.12\hatk}[/tex][tex]\hat{k}[/tex]

Angle between two vectors is given by

[tex]sin\theta=\frac{\mid a\times b\mid}{\mid a\mid \mi b\mid}[/tex]

Where [tex]\theta[/tex] in degrees

[tex]\mid{\vec{A}}\mid=\sqrt{(4.2)^2+(7.2)^2}=8.3[/tex]

Using formula[tex]\mid a\mid=\sqrt{x^2+y^2}[/tex]

Where x= Coefficient of unit vector i

y=Coefficient of unit vector j

[tex]\mid{\vec{B}}\mid=\sqrt{5.7)^2+(-2.4)^2}=6.2[/tex]

[tex]\mid{\vec{A}\times \vec{B}}\mid=\sqrt{(-51.12)^2}=51.12[/tex]

Using the formula

[tex]sin\theta=\frac{51.12}{8.3\times 6.2}=0.993[/tex]

[tex]\theta=sin^{-1}(0.993)=83.2[/tex]degrees

Hence, the angle between given two vectors=[tex]83.2^{\circ}[/tex]

The scalar product of vectors A = 4.20 i + 7.20 j and B = 5.70 i - 2.40 j is 6.66. To find the angle between A and B, calculate the magnitudes of each vector and use the cosine formula involving the scalar product and magnitudes.

Finding Scalar Product and Angle Between Two Vectors

To find the scalar product (also known as the dot product) of two vectors A and B, you multiply the corresponding components of the vectors and add them up. For vectors A = 4.20 i + 7.20 j and B = 5.70 i - 2.40 j, the scalar product A · B is:

Scalar Product = (Ax * Bx) + (Ay * By) = (4.20 * 5.70) + (7.20 * -2.40) = 23.94 - 17.28 = 6.66.

To find the angle between the two vectors, we use the formula that involves the scalar product and the magnitudes of the two vectors:

cos(θ) = (A · B) / (|A| * |B|).

First, calculate the magnitudes of A and B:

Then, calculate cos(θ) and finally θ using the arccos function on a calculator. Plug in the scalar product and the magnitudes into the formula to get the angle.

Answer:

Explanation:

Given

Two masses M and 2 M with velocity v in opposite direction

After collision they stick together

Initial momentum

[tex]P_i=Mv-2Mv[/tex]

final momentum

[tex]P_f=3Mv'[/tex]

Conserving momentum

[tex]P_i=P_f[/tex]

[tex]-Mv=3Mv'[/tex]

[tex]v'=\frac{v}{3}[/tex]

i.e. system moves towards the direction of 2M mass

Initial kinetic Energy [tex]K_1=\frac{1}{2}Mv^2+\frac{1}{2}2Mv^2[/tex]

[tex]K_1=\frac{3}{2}Mv^2[/tex]

Final Kinetic Energy [tex]K_2=\frac{1}{2}\cdot (3M)\cdot (\frac{v}{3})^2=\frac{3}{18}Mv^2[/tex]

loss of Energy[tex]=K_1-K_2[/tex]

[tex]=\frac{3}{2}Mv^2-\frac{3}{18}Mv^2[/tex]

[tex]=\frac{4}{3}Mv^2[/tex]

The mechanical energy is lost to other forms of energy during the collision is [tex]\bold { \dfrac 43 mv^2}[/tex].

Given here,

Mass of the first object = M

Mass of the second  object = 2M

Thy are moving at same speed = v,

The initial momentum,

[tex]\bold {Pi = Mv - 2mv}[/tex]

The Final momentum,

Pf = 3 Mv'

From the conservation of momentum,

Pi = pf

Mv - 2Mv = 3mv'

[tex]\bold {v' = \dfrac v3}[/tex]

The final velocity is one-third of the initial speed.

Since, the system moves towards the direction second object,

The loss of energy  = Ki - Kf

Where,

Ki - Initial kinetic energy,

[tex]\bold {Ki = \dfrac 12 mv^2 + \dfrac 12 mv^2 = \dfrac 32 mv^2}[/tex]

Kf = final kinetic energy

[tex]\bold {Kf = \dfrac 12 (3m)(\dfrac v3)^2 = \dfrac {3}{18 } mv^2}[/tex]

Thus,

[tex]\bold {\Delta E = \dfrac 32 mv^2 - \dfrac 3{18} mv^2 }\\\\\bold {\Delta E = \dfrac 43 mv^2}[/tex]

Therefore, the mechanical energy is lost to other forms of energy during the collision is [tex]\bold { \dfrac 43 mv^2}[/tex].

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Answer:

Explanation:

When  a positively charged glass rod is brought near the uncharged sphere (which contains equal amount of positive and negative charge) then it started attracting metal sphere. This occurs because glass rod polarize metal sphere such that negative charge is induced at end near to the rod.

But when rod is touched to sphere it started to repel because some of the rod positive charge goes into metal sphere and thus similar charges repel each other.

Answer: Yes, it matters.

Explanation:

Torque is a measure of the ability of an applied force to cause an object to turn and is the rotational analogue to force. In a static system, torque should be measured due to a force and one needs adjust the magnitude of one or more forces.

Answer:

[tex]\theta = 947.7\ rad[/tex]

Explanation:

given,

initial rotating speed = 33.3 rad/s

angular acceleration = 2.15 rad/s²

final angular speed = 72 rad/s

using equation of rotating wheel

[tex]\omega_f = \omega_i + \alpha t[/tex]

[tex]72 = 33.3+ 2.15 t[/tex]

    2.15 t = 38.7

         t = 18 s

now, Again using equation of motion for the calculation of angle

[tex]\theta = \omega_o t +\dfrac{1}{2}\alpha t^2[/tex]

[tex]\theta = 33.3 \times 18 +\dfrac{1}{2}\times 2.15 \times 18^2[/tex]

[tex]\theta = 947.7\ rad[/tex]

the angle record turned is equal to 947.7 radians.

Final answer:

The LP record has turned approximately 1066.51 radians when its angular speed reaches 72.0 rad/s.

Explanation:

To find the angle turned by the LP record when its angular speed reaches 72.0 rad/s, we can use the kinematic equation:

ω^2 = ω0^2 + 2αθ

Where:

ω = final angular velocity = 72.0 rad/s

ω0 = initial angular velocity = 33.3 rad/s

α = angular acceleration = 2.15 rad/s²

θ = angle turned

Plugging in the values:

72.0^2 = 33.3^2 + 2(2.15)θ

Simplifying:

θ = (72.0^2 - 33.3^2) / (2(2.15))

θ ≈ 1066.51 radians

Therefore, the LP record has turned approximately 1066.51 radians when its angular speed reaches 72.0 rad/s.

Answer:

 r = 22.78 m

Explanation:

For this exercise let's use Newton's second law

Axis y

                 N- W = 0

                 N = W

X axis

                F = m a

Where the acceleration is centripetal

               a = v² / r

The force is the friction that the formula has

               fr = μ N

               fr = μ mg

Let's replace

            μ m g = m v² / r

           r = v² / μ g

Let's reduce the speed to the SI system

         v = 45 km / h (1000 m / 1 km) (1h / 3600 s) = 12.5 m / s

          r = 12.5 2 / (0.7 9.8)

          r = 22.78 m

Answer:

d. 2T

Explanation:

Period of a simple pendulum: This can be defined as the time taken for a simple pendulum to complete an oscillation.

The S.I unit of the period of a simple pendulum is second (s).

Mathematically the period of a simple pendulum can be represented as

T = 2π√(L/g)................................................ Equation 1

Where T = period of the pendulum, L = length of the pendulum, g = acceleration due to gravity,  π = pie.

Note: If the length of a simple pendulum is increased, the period will also increase, while acceleration due to gravity is constant.

Hence,

T' = 2π√(4L/g) ........................... Equation 2

Where T' is the new period when the length of the pendulum is increased by 4 time its original length.

Dividing equation 1 by equation 2

T/T' = 2π√(L/g)/2π√(4L/g)

T/T' = √L/2√L

T/T' = 1/2

T' = 2T

Thus, the right option is d. 2T

Final answer:

At t = 6 seconds, two wave pulses on a string moving toward each other at 1 m/s would have just passed through each other, provided they began more than 6 meters apart. If starting exactly 12 meters apart, they would be overlapping completely at this moment.

Explanation:

Understanding Wave Pulse Interaction on a String

When two wave pulses approach each other on a string, each moving at a speed of 1 m/s, their interaction can be understood through the principles of superposition. At any given moment before and during the overlap, the shape of the string is the sum of the individual displacements of the pulses. So, at t = 6 seconds, since each pulse has traveled a distance of 6 meters toward each other, they would have just passed through each other, assuming that they started more than 6 meters apart. If they started exactly 12 meters apart, at 6 seconds, we would observe them just meeting and overlapping at the midpoint. The string's shape would display features of both pulses combined at the instant of full overlap; thereafter, they would continue on their paths as though they'd passed through each other, effectively unchanged in shape apart from any damping effects.

Answer:

[tex]v_{y_0} = \frac{\frac{g}{2}t(t - 2\sqrt{\frac{2h}{g}})}{\sqrt{\frac{2h}{g}} - t}[/tex]

Explanation:

We will apply the equations of kinematics to both stones separately.

First stone:

Let us denote the time spent after the second stone is thrown as 'T'.

[tex]y - y_0 = v_{y_0}(t+T) + \frac{1}{2}a(t+T)^2\\0 - h = 0 + \frac{1}{2}(-g)(t+T)^2\\(t+T)^2 = \frac{2h}{g}\\T = \sqrt{\frac{2h}{g}}-t[/tex]

Second stone:

[tex]y - y_0 = v_{y_0}T + \frac{1}{2}aT^2\\0 - h = v_{y_0}T -\frac{1}{2}gT^2\\-h = v_{y_0}(\sqrt{\frac{2h}{g}} - t) - \frac{g}{2}(\sqrt{\frac{2h}{g}} - t)^2\\-h = v_{y_0}(\sqrt{\frac{2h}{g}} - t) - \frac{g}{2}(\frac{2h}{g} + t^2 - 2t\sqrt{\frac{2h}{g}})\\-h = v_{y_0}\sqrt{\frac{2h}{g}} - v_{y_0}t - h -\frac{g}{2}t^2 + gt\sqrt{\frac{2h}{g}}\\v_{y_0}(\sqrt{\frac{2h}{g}} - t) = \frac{g}{2}t^2 - gt\sqrt{\frac{2h}{g}}\\v_{y_0} = \frac{\frac{g}{2}t(t - 2\sqrt{\frac{2h}{g}})}{\sqrt{\frac{2h}{g}} - t}[/tex]

The initial speed of the second stone thrown vertically down is the same as its final velocity when it reaches the water, which is given by gt.

To find the initial speed of the second stone, we can use the equation of motion for an object in free fall. The equation is given by: h = ½gt^2, where h is the height of the bridge, g is the acceleration due to gravity, and t is the time taken by the second stone to reach the water.

Rearranging the equation, we can solve for t: t = √(2h/g). Since the second stone is thrown vertically down, its initial velocity is zero. The final velocity of the second stone when it reaches the water is given by: v = gt. Therefore, the initial speed of the second stone is simply the same as its final velocity, which is gt.

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Answer:

a) On the way to Grandmother´s the average velocity is 48 mi/h.

b) In the return trip, the average velocity is 50 mi/h

Explanation:

Hi there!

The average velocity (AV) is calculated as follows:

AV = Δx / Δt

Where:

Δx = traveled distance = final position - initial position

Δt = elapsed time

a) Let´s find the time it takes Julie to travel 50 mi (half the total distance) at 60 and 40 mph:

At 60 mph:

60 mi/h = 50 mi / Δt

Δt = 50 mi / 60 mi/h = 5/6 h

At 40 mph:

40 mi/h = 50 mi / Δt

Δt = 50 mi / 40 mi/h = 1.25 h

The total time of travel is 1.25 h + 5/6 h = 25/12 h

Then, the AV will be:

AV = 100 mi / 25/12 h = 48 mi/h

On the way to Grandmother´s the average velocity is 48 mi/h.

b)Now let´s calculate the AV on the return trip. During the first half of the trip, the elapsed time is half of the total time (Δt/2) and the traveled distance is unknown, Δx1.

Then, the average velocity during the first half of the trip can be written as follows. :

40 mi/h = Δx1 / (1/2 Δt)

Solving for Δx1:

20 mi/h · Δt = Δx1

During the second half of the trip the average velocity will be:

60 mi/h = Δx2 / (1/2 Δt)

Solving for Δx2:

30 mi/h · Δt = Δx2

The average velocity for the entire trip can be expressed as follows:

AV = Δx / Δt

Δx = Δx1 + Δx2 = 20 mi/h · Δt + 30 mi/h · Δt = 50 mi/h · Δt

Then:

AV = 50 mi/h · Δt / Δt = 50 mi/h

In the return trip, the average velocity is 50 mi/h

The energy released by the body is approximately 1500 times more than that of a 1 megaton hydrogen bomb.

The energy released from converting mass to energy using the formula [tex]\rm \(E=mc^2\)[/tex] can be calculated by substituting the mass (m) into the equation, where c is the speed of light [tex]\rm (\(3.00 \times 10^8\) m/s)[/tex].

Given:

Mass (m) = 70.0 kg

Speed of light (c) = [tex]\rm \(3.00 \times 10^8\) m/s[/tex]

Using the formula [tex]\rm \(E = mc^2\)[/tex]:

[tex]\rm \[ E = (70.0 \, \text{kg}) \times (3.00 \times 10^8 \, \text{m/s})^2 \][/tex]

Now, calculate E:

[tex]\rm \[ E = 70.0 \, \text{kg} \times 9.00 \times 10^{16} \, \text{J/kg} \]\\\ E = 6.3 \times 10^{18} \, \text{Joules} \][/tex]

So, the energy released from converting the mass of your body to energy is [tex]\rm \(6.3 \times 10^{18}\)[/tex] Joules.

The energy released by a 1 megaton hydrogen bomb can be calculated as [tex]\(1 \, \text{megaton} = 4.18 \times 10^{15}\)[/tex] Joules.

Comparing the two energies:

[tex]\[ \text{Energy released by body} = 6.3 \times 10^{18} \, \text{Joules} \]\\\ \text{Energy released by 1 megaton hydrogen bomb} = 4.18 \times 10^{15} \, \text{Joules} \][/tex]

The energy released by the body is approximately 1500 times more than that of a 1-megaton hydrogen bomb.

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The energy obtained from completely converting a 70kg mass into energy using the E=mc^2 formula is approximately 6.3 x 10^18 joules. This is about 1500 times greater than the energy released by a 1 megaton hydrogen bomb, which is about 4.18 x 10^15 joules.

The mass-energy equivalence principle, stated by the formula E=mc^2, allows us to calculate the energy resulting from the conversion of mass to energy. Here, 'E' is energy, 'm' is mass, and 'c' is the speed of light in a vacuum (about 3.00x10^8 meters per second). If we substitute your mass (70.0 kilograms) into the formula, we find that the energy released would amount to approximately 6.3 x 10^18 joules of energy.

For comparison, a 1 megaton hydrogen bomb releases about 4.18 x 10^15 joules. Therefore, the energy from a complete mass to energy conversion of a 70-kg person would be approximately 1500 times more than a 1 megaton hydrogen bomb.

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Answer:

the direction that should be walked by Ricardo to go directly to Jane is 23.52 m, 24° east of south

Explanation:

given information:

Ricardo walks 27.0 m in a direction 60.0 ∘ west of north, thus

A= 27

Ax =  27 sin 60 = - 23.4

Ay = 27 cos 60 = 13.5

Jane walks 16.0 m in a direction 30.0 ∘ south of west, so

B = 16

Bx = 16 cos 30 = -13.9

By = 16  sin 30 = -8

the direction that should be walked by Ricardo to go directly to Jane

R = √A²+B² - (2ABcos60)

   = √27²+16² - (2(27)(16)(cos 60))

   = 23.52 m

now we can use the sines law to find the angle

tan θ = [tex]\frac{R_{y} }{R_{x} }[/tex]

         = By - Ay/Bx -Ax

         = (-8 - 13.5)/(-13.9 - (-23.4))

     θ  = 90 - (-8 - 13.5)/(-13.9 - (-23.4))

         = 24° east of south

Answer:

a. [tex]\bar{d}=4.34 cm[/tex]

b. [tex]\sigma=0.023 cm[/tex]

c. [tex]\rho=(0.0089\pm 0.00058) kg/cm^{3}[/tex]

Explanation:

a) The average of this values is the sum each number divided by the total number of values.

[tex]\bar{d}=\frac{\Sigma_{i=1}^(N)x_{i}}{N}[/tex]

[tex]\bar{d}=\frac{4.32+4.35+4.31+4.36+4.37+4.34}{6}[/tex]

[tex]\bar{d}=4.34 cm[/tex]

b) The standard deviation equations is:

[tex]\sigma=\sqrt{\frac{1}{N}\Sigma^{N}_{i=1}(x_{i}-\bar{d})^{2}}[/tex]

If we put all this values in that equation we will get:

[tex]\sigma=0.023 cm[/tex]

Then the mean diameter will be:

[tex]\bar{d}=(4.34\pm 0.023)cm[/tex]

c) We know that the density is the mass divided by the volume (ρ = m/V)

and we know that the volume of a cylinder is: [tex]V=\pi R^{2}h[/tex]

Then:

[tex]\rho=\frac{m}{\pi R^{2}h}[/tex]

Using the values that we have, we can calculate the value of density:

[tex]\rho=\frac{1.66}{3.14*(4.34/2)^{2}*12.6}=0.0089 kg/cm^{3}[/tex]

We need to use propagation of error to find the error of the density.

[tex]\delta\rho=\sqrt{\left(\frac{\partial\rho}{\partial m}\right)^{2}\delta m^{2}+\left(\frac{\partial\rho}{\partial d}\right)^{2}\delta d^{2}+\left(\frac{\partial\rho}{\partial h}\right)^{2}\delta h^{2}}[/tex]  

Let's find each partial derivative:

1. [tex]\frac{\partial\rho}{\partial m}=\frac{4m}{\pi d^{2}h}=\frac{4*1.66}{\pi 4.34^{2}*12.6}=0.0089[/tex]

2.  [tex]\frac{\partial\rho}{\partial d}=-\frac{8m}{\pi d^{3}h}=-\frac{8*1.66}{\pi 4.34^{3}*12.6}=-0.004[/tex]

3. [tex]\frac{\partial\rho}{\partial h}=-\frac{4m}{\pi d^{2}h^{2}}=-\frac{4*1.66}{\pi 4.34^{2}*12.6^{2}}=-0.00071[/tex]

Therefore:

[tex]\delta\rho=\sqrt{\left(0.0089)^{2}*0.05^{2}+\left(-0.004)^{2}*0.023^{2}+\left(-0.00071)^{2}*0.5^{2}}[/tex]

[tex]\delta\rho=0.00058[/tex]

So the density is:

[tex]\rho=(0.0089\pm 0.00058) kg/cm^{3}[/tex]

I hope it helps you!

When a glass rod is rubbed with a silk cloth, electrons are removed from the rod, resulting in a positive charge.

When an object, such as a glass rod, is rubbed with a silk cloth, it can acquire a positive charge. In this case, electrons are removed from the object, leaving behind a net positive charge. The process of rubbing causes the cloth to transfer some of its excess electrons to the glass rod, resulting in the glass rod gaining a positive charge.

Final answer:

When two conducting spheres come into contact, they share their total charge evenly. If sphere A has a charge of -5 nC and sphere B has -3 nC, each sphere will have -4 nC after contact and separation.

Explanation:

When two identical conducting spheres are brought into contact, charge redistribution occurs. The charge will spread evenly across both spheres, because both have equal capacity to hold charge being identical conductors. To find the final charge on each sphere, one must add the initial charges of both spheres and then divide by two because the total charge will be shared equally.

In the scenario given where sphere A has a charge of −5 nC and sphere B has a charge of −3 nC:

Total charge before contact: -5 nC + (-3 nC) = -8 nC

After contact and separation, each sphere will have half the total charge: -8 nC / 2 = -4 nC

Therefore, both sphere A and sphere B will end up with a charge of −4 nC after being brought into contact and separated. The equivalent number of electrons is determined by the charge on each sphere divided by the elementary charge (approximately 1.6 x 10^-19 Coulombs).

Answer:A

Explanation:

A positively charged glass rod attracts object x. So, object x must be negatively charged or uncharged.

This occurs because opposite charges attract each other or either object x is uncharged and a negative charge is induced in it as glass rod approach the object x.

So option A is correct

Answer:

The speed of the boat is equal to 13.50 ft/s.

Explanation:

given,

1  nautical mile = 6076 ft

1 knot = 1 nautical mile /hour

1 knot = 6076 ft/hr

speed of boat = 8 knots

 8 knots = 8 nautical mile /hour

               =[tex]8 \times \dfrac{6076\ ft}{1\nautical\ mile}\times \dfrac{1\ hour}{60\times 60\ s}[/tex]

               = 13.50 ft/s

The speed of the boat is equal to 13.50 ft/s.

The speed of the boat will be    [tex]V=13.50\dfrac{ft}{sec}[/tex]

It is given that

nautical mile   6076 feet

1 knot= 1 nautical mile per hour

[tex]\rm 1\ knot = 6076\dfrac{ft}{hr}[/tex]

Speed of the boat is = 8 knots

[tex]\rm 8 \ knot = 8\ nautical \dfrac{Mile }{Hour}[/tex]

[tex]=\dfrac{8 \times 6076\tines 1}{60\times 60}[/tex]

[tex]\rm 8 \ k not= 13.50\dfrac{ft}{s}[/tex]

Thus  The speed of the boat will be    [tex]V=13.50\dfrac{ft}{sec}[/tex]

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Answer: For the given process it is true that the balloon has a net charge on it, but the wall may be neutral (have no net charge).

Explanation:

When we tend to rub a balloon against a woolen sweater then it is possible that the balloon might stick to the wall. This might be due to the increase in number of electrons on the balloon and the wall gets positively charged.

Since, opposite charges tend to attract each other hence, the balloon will stick to the wall.

A neutral object also tends to acquire a certain charge when it is rubbed by a charged object.

Therefore, we can conclude that the balloon has a net charge on it, but the wall may be neutral (have no net charge).

The balloon acquires a net negative charge after being rubbed against the wool sweater, then it attracts to a possibly neutral wall due to polarization, not because of new charges being created or requiring a net transfer of charge to the wall.

When a balloon is rubbed against a wool sweater, electrons are transferred from the sweater to the balloon, giving the balloon a net negative charge. The correct statement regarding the interaction of the balloon with the wood wall is that the balloon has a net charge on it, but the wall may be neutral (have no net charge). This is because the negatively charged balloon can induce a polarization in the neutral wall, where the positive charges in the wall are slightly attracted towards the balloon, causing an attractive force. No new charges are created during this process; instead, electrons are simply transferred from one material to another, and the overall charge in the system remains constant.

To solve this problem we will define the two frequencies given. From there and in international units, we will proceed to clear the variable concerning the spring constant from both equations. We will match the two equations and find the mass. With the mass found we will replace in any of the two system equations and find the spring constant

[tex]f_1 = \frac{1}{2\pi} (\sqrt{\frac{k}{m}} ) = 94\frac{cycles}{minute} \times \frac{1 minute}{60s} = \frac{47}{30} cycles/s[/tex]

Now

[tex]f_2 = \frac{1}{2\pi} (\sqrt{\frac{k}{m+0.453}}) = 76.7\frac{cycles}{minute} \times \frac{1 minute}{60s} = \frac{767}{600} cycles/s[/tex]

The two equations could be described as:

1)

[tex]\frac{1}{2\pi} \sqrt{\frac{k}{m}} = \frac{47}{30}[/tex]

[tex]\sqrt{\frac{k}{m}} = \frac{47\pi}{15}[/tex]

[tex]\sqrt{k} = \frac{47\pi}{15}\sqrt{m}[/tex]

2)

[tex]\frac{1}{2\pi} \sqrt{\frac{k}{m+0.453}}) = \frac{767}{600}[/tex]

[tex]\sqrt{\frac{k}{m+0.453}} = \frac{767\pi}{300}[/tex]

[tex]\sqrt{k} = \frac{767\pi}{300}\sqrt{m+0.453}[/tex]

Equation both expression we have that,

[tex]\frac{47\pi}{15}\sqrt{m} = \frac{767\pi}{300}\sqrt{m+0.453}[/tex]

[tex]\sqrt{m}= \frac{767}{940} \sqrt{m+0.453}[/tex]

[tex]m(1-\frac{588289}{883600}) = \frac{588289}{883600} \times 0.453[/tex]

[tex]m = \frac{\frac{588289}{883600} \times 0.453}{(1-\frac{588289}{883600})}[/tex]

[tex]m \approx 0.902kg[/tex]

Use one of the formulas from the system

[tex]\sqrt{k} = \frac{47\pi}{15}\sqrt{m}[/tex]

[tex]\sqrt{k} = \frac{47\pi}{15}\sqrt{0.902}[/tex]

[tex]k = 87.4N/m[/tex]

Answer:

The diameters of the disk is 1.23 cm

Explanation:

Given information:

the space of two disks, d =0.5 mm = 0.0005 m =  5 x [tex]10^{-4}[/tex] m

the transferred electron, n = 1.70 x [tex]10^{9}[/tex]

electric field strength, E = 2.6 x [tex]10^{5}[/tex] N/C

to find the diameter of the disk we can use the following equation

A = πD²/4 ...........................................(1)

where

D = the distance of the disk

A = the area of the disk

first, we have to find the are of the disk using the capacitance equation

C = ε₀A/d...........................................(2)

A = Cd/ε₀ where C = Q/V (Q is total charge and V is potential difference)

thus

A = Qd/Vε₀.........................................(3)

now substitute V = Ed and Q = ne, so

A = (ned)/(Edε₀)

   = ne/Eε₀..........................................(4)

e = 1.6 x [tex]10^{-19}[/tex] C

now can substitute equation (4) to the first equation

A = πD²/4

D² = 4A/π

D  = √4A/π

    = √(4ne)/(πEε₀) , ε₀ = 9.85 x [tex]10^{-12}[/tex] C²/Nm²

    = √4(1.70 x [tex]10^{9}[/tex])(1.6 x [tex]10^{-19}[/tex])/π(2.6 x [tex]10^{5}[/tex] )(8.85 x [tex]10^{-12}[/tex] )

    = 0.0123 m

    = 1.23 cm

Answer:

Explanation:

mass, m = 4.1 x 10^-6 g

Electric field, E = 466 N/C

Electric force = weight

(a) Electric force = q x E = m x g

where, q be the charge on the dust particle.

q x 466 = 4.1 x 10^-6 x 9.8

q = 8.62 x 10^-8 C

(b) Number of electron  = charge / charge of one electron

n = 8.62 x 10^-8 / (1.6 x 10^-19)

n = 5.4 x 10^11

Answer:

-0.912 m/s

Explanation:

When the package is thrown out, momentum is conserved. The total momentum after is the same as the total momentum before, which is 0, since the boat was initially at rest.

[tex] (m_c + m_b)v_b + m_pv_p = 0[/tex]

where [tex]m_c = 24.6 kg, m_b = 39 kg, m_p = 5.8 kg[/tex] are the mass of the child, the boat and the package, respectively. [tex], v_p = 10m/s, v_b[/tex] are the velocity of the package and the boat after throwing.

[tex] (24.6 + 39)v_b + 5.8*10 = 0[/tex]

[tex]63.6v_b + 58 = 0[/tex]

[tex]v_b = -58/63.6 = -0.912 m/s[/tex]

Final answer:

The velocity of the ball 0.3 seconds after being kicked is <-8, 16.06, -3> m/s. The average velocity of the ball over this time interval is <0, 53.53, 13.33> m/s. The location of the ball 0.5 seconds after being kicked is <8, 26.765, -0.335> m.

Explanation:

(a) To find the velocity of the ball 0.3 seconds after being kicked, we can use the Momentum Principle. The Momentum Principle states that the change in momentum of an object is equal to the force exerted on it multiplied by the time over which the force was exerted. In this case, we can assume that the only force acting on the ball is gravity. Thus, the change in momentum is equal to the weight of the ball, which is equal to its mass multiplied by the acceleration due to gravity. The time over which the force was exerted is 0.3 seconds. Using the formula for momentum (momentum = mass * velocity), we can calculate the velocity of the ball after 0.3 seconds.

Given initial velocity: <-8, 19, -3> m/s

Acceleration due to gravity (in the vertical direction): -9.8 m/s^2

Time: 0.3 seconds

Using the formula for velocity (velocity = initial velocity + acceleration * time), we can calculate the velocity of the ball after 0.3 seconds:

Velocity = (-8, 19, -3) + (0, -9.8, 0) * 0.3

Velocity = (-8, 19, -3) + (0, -2.94, 0)

Velocity = (-8, 16.06, -3)

Therefore, the velocity of the ball 0.3 seconds after being kicked is <-8, 16.06, -3> m/s.

(b) To find the average velocity of the ball over this time interval, we can divide the displacement of the ball by the time interval. The displacement of the ball is the difference between its final position and its initial position. The time interval is 0.3 seconds. Using the formula for average velocity (average velocity = displacement / time interval), we can calculate the average velocity of the ball over this time interval:

Initial position: <8, 0, -7> m

Final position: <8, 16.06, -3> m

Displacement = Final position - Initial position

Displacement = <8, 16.06, -3> - <8, 0, -7>

Displacement = <0, 16.06, 4> m

Average velocity = Displacement / Time interval

Average velocity = <0, 16.06, 4> / 0.3

Average velocity = <0, 53.53, 13.33> m/s

Therefore, the average velocity of the ball over this time interval is <0, 53.53, 13.33> m/s.

(c) To find the location of the ball 0.5 seconds after being kicked, we can multiply the average velocity by the time interval and add the result to the initial position of the ball. The average velocity is <0, 53.53, 13.33> m/s and the time interval is 0.5 seconds. Using the formula for displacement (displacement = average velocity * time interval), we can calculate the displacement of the ball:

Displacement = <0, 53.53, 13.33> m/s * 0.5 seconds

Displacement = <0, 26.765, 6.665> m

Therefore, the location of the ball 0.5 seconds after being kicked is <8, 0, -7> m + <0, 26.765, 6.665> m = <8, 26.765, -0.335> m.

(a) Velocity after 0.3s: <-8, 16.057, -3> m/s.

(b) Average velocity: <-8, 15.0595, -1.4715> m/s.

(c) Location after 0.5s: <4, 7.53, -7.74> meters.

To solve this problem, we can use the Momentum Principle, which relates the change in momentum of an object to the net force acting on it. The formula for the Momentum Principle is:

Δp = FΔt

Where:

Δp = change in momentum

F = net force

Δt = change in time

First, let's calculate the change in momentum of the ball over the given time intervals.

(a) Velocity of the ball 0.3 seconds after being kicked:

We want to find the velocity of the ball at t = 0.3 seconds. We can use the following formula:

Δp = mΔv

Where:

Δp = change in momentum

m = mass of the ball (which is not given, but we can ignore it since it cancels out in this case)

Δv = change in velocity

Given initial velocity (u) = <-8, 19, -3> m/s

Time interval (Δt) = 0.3 seconds

Δv = a * Δt

Where:

a = acceleration

To find acceleration, we need to use the kinematic equation:

Δv = u + at

For each component (x, y, z), we can calculate the change in velocity:

Δvx = -8 + ax * 0.3

Δvy = 19 + ay * 0.3

Δvz = -3 + az * 0.3

Now, we need to find the acceleration (ax, ay, az) for each component. Since there's no air resistance, the only force acting on the ball is gravity (assuming the acceleration due to gravity is approximately -9.81 m/s² in the negative y-direction).

ax = 0 (no force in the x-direction)

ay = -9.81 m/s²

az = 0 (no force in the z-direction)

Now we can calculate Δv for each component:

Δvx = -8 + 0 * 0.3 = -8 m/s

Δvy = 19 + (-9.81) * 0.3 = 19 - 2.943 = 16.057 m/s

Δvz = -3 + 0 * 0.3 = -3 m/s

So, the velocity of the ball 0.3 seconds after being kicked is:

V = <Δvx, Δvy, Δvz> = <-8, 16.057, -3> m/s

(b) Average velocity of the ball over the time interval:

Average velocity is calculated by taking the total displacement and dividing it by the total time. In this case, the total displacement is the change in position from the initial point to the final point (0.3 seconds after being kicked):

Δx = ut + (1/2)at²

Δx = <-8 * 0.3, 19 * 0.3 - (1/2) * 9.81 * (0.3)², -3 * 0.3> = <-2.4, 4.51785, -0.44145> meters

So, the average velocity is:

Average velocity = Δx / Δt = <-2.4/0.3, 4.51785/0.3, -0.44145/0.3> = <-8, 15.0595, -1.4715> m/s

(c) Location of the ball 0.5 seconds after being kicked using average velocity:

We want to find the location of the ball 0.5 seconds after being kicked. We can use the formula:

Δx = v_avg * Δt

Where:

Δx = change in position

v_avg = average velocity calculated in part (b)

Δt = time interval (0.5 seconds)

Δx = <-8 * 0.5, 15.0595 * 0.5, -1.4715 * 0.5> = <-4, 7.52975, -0.73575> meters

Now, we need to add this displacement to the initial position to find the final position:

Final position = Initial position + Δx = <8, 0, -7> + <-4, 7.52975, -0.73575> = <4, 7.52975, -7.73575> meters

So, the location of the ball 0.5 seconds after being kicked is approximately <4, 7.53, -7.74> meters.

Answer:

Explanation:

To stop a ball with high momentum in a small-time imparts a high amount of impact on hands. This is the reason for the stinging of hands.

The momentum of the ball is due to the mass and velocity. To prevent stinging in the hand one needs to lower his hands to increase the time of contact. In this way, the momentum transfer to the hands will be lesser.        

Answer:

The heat is 115478.4 J.

Explanation:

Given that,

Mass of water = 0.400 kg

Power = 200 W

Suppose, we determine how much heat must be added to the water to raise its temperature from 20.0°C to 89.0°C?

We need to calculate the heat

Using formula of heat

[tex]Q=mc\Delta T[/tex]

Where, m = mass of water

c = specific heat

Put the value into the formula

[tex]Q=400\times4.184\times(89-20)[/tex]

[tex]Q=115478.4\ J[/tex]

Hence, The heat is 115478.4 J.

The subject of this question is Physics, and the student is asking about using an electric immersion heater to heat water for an all-night study session. To calculate the resistance of the heater, the formula Resistance = (Voltage)^2 / Power is used. The current flowing through the heater can be determined using Ohm's Law: Current = Voltage / Resistance.

The subject of this question is Physics. The student is asking about using an electric immersion heater to heat water for an all-night study session.

To calculate the resistance of the heater, we need to use the formula: Resistance = (Voltage)^2 / Power. The power can be calculated using the formula: Power = Current x Voltage.

We can determine the current flowing through the heater using Ohm's Law: Current = Voltage / Resistance.