You are pushing a crate horizontally with 100 N across a 10 m factory floor. If the force of friction on the crate is a steady 70 N, how much kinetic energy is gained by the crate

Answer:

[tex]F_G=G. \frac{m_1.m_2}{R^2}[/tex] gravitational force

[tex]F=\frac{1}{4\pi \epsilon_0} \times \frac{q_1.q_2}{R^2}[/tex] electrostatic force

Explanation:

The forces that balloons may exert on each other can be gravitational pull due to the mass of the balloon membrane and the mass of the gas contained in each. This force is inversely proportional to the square of the radial distance between their center of masses.

The Mutual force of gravitational pull that they exert on each other can be given as:

[tex]F_G=G. \frac{m_1.m_2}{R^2}[/tex]

where:

[tex]G=[/tex] gravitational constant  [tex]=6.67\times 10^{-11} m^3.kg^{-1}.s^{-2}[/tex]

[tex]m_1\ \&\ m_2[/tex] are the masses of individual balloons

[tex]R=[/tex] the radial distance between the  center of masses of the balloons.

But when  there are charges on the balloons, the electrostatic force comes into act which is governed by Coulomb's law.

Given as:

[tex]F=\frac{1}{4\pi \epsilon_0} \times \frac{q_1.q_2}{R^2}[/tex]

where:

[tex]\rm \epsilon_0= permittivity\ of\ free\ space[/tex]

[tex]q_1\ \&\ q_2[/tex] are the charges on the individual balloons

R = radial distance between the charges.

Answer:

option B

Explanation:

given,

Q = energy input = 70 J

U = increase in internal energy

W is the work done = 20 J

using first law of thermodynamics

The change in internal energy of a system is equal to the heat added to the system minus the work done.

U = Q - W

U = 70 - 20

U = 50 J

The change in internal energy is equal to 50 J

Hence, the correct answer is option B

Answer:

Explanation:

Given

speed of locomotive [tex]v=80\ km/h[/tex]

distance [tex]d=9\ km[/tex]

Relative velocity of two locomotive [tex]v_r=80-(-80)=160\ km/h[/tex]

Time taken[tex]=\frac{distance}{speed}[/tex]

[tex]t=\frac{9}{160} \hr[/tex]

[tex]t=\frac{9}{160}\times 60\ min[/tex]

[tex]t=3.375\ min[/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:

a) See attachment

b) The pattern converges towards central order

c) Very bright spot at central order and rest is dark and contrast of pattern increases.

Explanation:

b) According to Young's Double slit experiment the following relationship is given:

[tex]wavelength = \frac{x*d}{nL}[/tex]

where,

λ = wavelength of light used (m)

x = distance from central fringe (m)

d = distance between the slits (m)

n = the order of the fringe

L = length from the screen with slits to the viewing screen (m)

Using the formula if we increase the (d) i.e distance between slits we see that (x) distance between fringes decreases and the patterns of bright and dark spots is alternating more frequently.

c) When d is arbitrarily large the x is arbitrarily small.

Hence, the entire pattern converges on to the film in a small space with millions spots of bright and dark spots alternating together to forma big bright spot and contrast of pattern increases.

Adding an identical slit close to the first results in an interference pattern of light and dark fridges due to light interference. Increasing the slits' distance results in narrower fringes, while in the limit that this distance becomes arbitrarily large, interference effects vanish and we see two independent single-slit diffraction patterns.

The question discusses an experiment in Physics related to wave interference, specifically light interference in a double-slit experiment. When you add an identical slit a distance of 0.25mm away from the first, an interference pattern of light and dark fringes (bright and dark spots) is created on the screen due to constructive (light) and destructive (dark) interference of the light waves passing through the slits.

Now, as we increase the distance d between the slits, the interference fringes on the screen will become narrower, and their separation will increase. This occurs because the path difference between the light waves emerging from the two slits and reaching the screen changes.

In the limit that the distance d between the slits becomes arbitrarily large, interference effects would vanish, and one would observe two independent single-slit diffraction patterns.

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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: 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.

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:

Volume.

Explanation:

Boyle’s law experiment :

This is also known as Mariotte's law or in the other words it is also known as Boyle–Mariotte law.

This law tell us about the variation of gas pressure when the volume of the gas changes at the constant temperature.According to this law the abslute pressure is inversely proportional to the volume .

We can say that

[tex]P\alpha \dfrac{1}{V}[/tex]

Where P=Pressure

V=Volume

PV = K

K=Constant

When volume decrease then the pressure of the gas will increase.

That is why the answer is "Volume".

Answer:

[tex]<3068.2352, 800, 0>\ m/s[/tex]

Explanation:

F = Force = [tex]<-1.12\times 10^{-11}, 0, 0>[/tex]

m = Mass of proton = [tex]1.7\times 10^{-27\ kg[/tex]

t = Time taken = [tex]2\times 10^{-14}\ s[/tex]

Acceleration is given by

[tex]a=\dfrac{F}{m}\\\Rightarrow a=\dfrac{<-1.12\times 10^{-11}, 0, 0>}{1.7\times 10^{-27}}\\\Rightarrow a=<-6.58824\times 10^{15}, 0, 0>\ m/s^2[/tex]

[tex]v=u+at\\\Rightarrow v=<3200, 800, 0>+<-6.58824\times 10^{15}, 0, 0>\times 2\times 10^{-14}\\\Rightarrow v=<3200, 800, 0>+<-6.58824\times 10^{15}, 0, 0>\times 2\times 10^{-14}\\\Rightarrow v=<3200, 800, 0>+<-131.7648, 0, 0>\\\Rightarrow v=<3068.2352, 800, 0>\ m/s[/tex]

The velocity of the proton is [tex]<3068.2352, 800, 0>\ m/s[/tex]

Answer:

option a

Explanation:

Size of an atom (diameter) = 10⁻¹⁰ m

There are approximately 10²² atoms in a single drop of water. If they are put in  a straight line, the length would be

l = diameter of an atom × number of atoms

l = 10²²×  10⁻¹⁰ m = 10¹² m

Distance between the Sun and the Earth is 1.47 × 10¹¹ m. The calculated length is greater than the distance between the Sun and the Earth.

Thus, option a is correct.

Answer:

Van will stop after a distance of 183.216 ft

Explanation:

We have given initial speed of the van = 36 mi/hr

As the van finally stops so final velocity v = 0 mi/hr

Distance after which van stop = 50 ft

As 1 ft = 0.000189 mi

So 50 ft [tex]=50\times 0.000189=0.00945mi[/tex]

From third equation of motion [tex]v^2=u^2+2as[/tex]

[tex]0^2=36^2+2\times a\times 0.00945[/tex]

[tex]a=-68571.42mi/hr^2[/tex]

In second case u = 69 mi/hr

And acceleration is same

So [tex]0^2=69^2-2\times 68571.42\times s[/tex]

[tex]s=0.03471mi[/tex]

As 1 mi = 5280 ft

So [tex]0.0347mi=0.0347\times 5280=183.216ft[/tex]

Answer: E = 0.85

Therefore the efficiency is: E = 0.85 or 85%

Explanation:

The efficiency (e) of a Carnot engine is defined as the ratio of the work (W) done by the engine to the input heat QH

E = W/QH.

W=QH – QC,

Where Qc is the output heat.

That is,

E=1 - Qc/QH

E =1 - Tc/TH

where Tc for a temperature of the cold reservoir and TH for a temperature of the hot reservoir.

Note: The unit of temperature must be in Kelvin.

Tc = 300K

TH = 2000K

Substituting the values of E, we have;

E = 1 - 300K/2000K

E = 1 - 0.15

E = 0.85

Therefore the efficiency is: E = 0.85 or 85%

To calculate the ideal efficiency of an engine with specified temperatures, convert to Kelvin and use the Carnot efficiency formula.

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a) Total power output: [tex]3.845\cdot 10^{26} W[/tex]

b) The relative percentage change of power output is 1.67%

c) The intensity of the radiation on Mars is [tex]540 W/m^2[/tex]

Explanation:

a)

The intensity of electromagnetic radiation is given by

[tex]I=\frac{P}{A}[/tex]

where

P is the power output

A is the surface area considered

In this problem, we have

[tex]I=1360 W/m^2[/tex] is the intensity of the solar radiation at the Earth

The area to be considered is area of a sphere of radius

[tex]r=1.5\cdot 10^{11} m[/tex] (distance Earth-Sun)

Therefore

[tex]A=4\pi r^2 = 4 \pi (1.5\cdot 10^{11})^2=2.8\cdot 10^{23}m^2[/tex]

And now, using the first equation, we can find the total power output of the Sun:

[tex]P=IA=(1360)(2.8\cdot 10^{23})=3.845\cdot 10^{26} W[/tex]

b)

The energy of the solar radiation is directly proportional to its frequency, given the relationship

[tex]E=hf[/tex]

where E is the energy, h is the Planck's constant, f is the frequency.

Also, the power output of the Sun is directly proportional to the energy,

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

where t is the time.

This means that the power output is proportional to the frequency:

[tex]P\propto f[/tex]

Here the frequency increases by 1 MHz: the original frequency was

[tex]f_0 = 60 MHz[/tex]

so the relative percentage change in frequency is

[tex]\frac{\Delta f}{f_0}\cdot 100 = \frac{1}{60}\cdot 100 =1.67\%[/tex]

And therefore, the power also increases by 1.67 %.

c)

In this second  case, we have to calculate the new power output of the Sun:

[tex]P' = P + \frac{1.67}{100}P =1.167P=1.0167(3.845\cdot 10^{26})=3.910\cdot 10^{26} W[/tex]

Now we want to calculate the intensity of the radiation measured on Mars. Mars is 60% farther from the Sun than the Earth, so its distance from the Sun is

[tex]r'=(1+0.60)r=1.60r=1.60(1.5\cdot 10^{11})=2.4\cdot 10^{11}m[/tex]

Now we can find the radiation intensity with the equation

[tex]I=\frac{P}{A}[/tex]

Where the area is

[tex]A=4\pi r'^2 = 4\pi(2.4\cdot 10^{11})^2=7.24\cdot 10^{23} m^2[/tex]

And substituting,

[tex]I=\frac{3.910\cdot 10^{26}}{7.24\cdot 10^{23}}=540 W/m^2[/tex]

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The total power output of the Sun is 3.8 x 10^26 W and does not change with frequency variations. Therefore, increasing frequency by 1MHz won't change the power output. Using a distance that takes into account Mars being farther from the Sun, the radiation intensity at Mars can be calculated.

The total power output of the sun can be calculated using the equation for the intensity of radiant power (P = I * 4πd^2). So, the total power output of the sun is P = 1360 W/m^2 * 4π * (1.5 x 10^11 m)^2 ≈ 3.8 x 10^26 W.

As for the relative change in power output with frequency change, it is important to note that the power output of the Sun is independent of frequency. Therefore, an increase in frequency by 1 MHz would not result in any relative change in power output. The power output remains 3.8 x 10^26 W regardless of frequency.

Lastly, to find the intensity at Mars which is 60% farther from the Sun than Earth, we use the same intensity equation (I = P / 4πd²) but adjust the distance accordingly. If r is the distance to Earth, then the distance to Mars is 1.6r. Substituting these values, we get I = 3.8 x 10^26 W / 4π * (1.6 * 1.5 x 10^11 m)². From this, you can calculate the radiation intensity at Mars.

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Answer:

[tex]\mathbf{1\ L=61.02384\ in^3}[/tex]

[tex]\mathbf{0.135\ kW=99.5709185406\ ft-lb/s}[/tex]

[tex]\mathbf{304\ kPa=44.091435811\ psi}[/tex]

[tex]\mathbf{122\ ft^3=3.45465495258\ m^3}[/tex]

[tex]\mathbf{100\ hp=74.6\ kW}[/tex]

[tex]\mathbf{1000\ lbm=453.592\ kg}[/tex]

Explanation:

[tex]1\ L=10^{-3}\ m^3[/tex]

[tex]1\ m=39.3701\ in[/tex]

[tex]10^{-3}\ m^3=10^{-3}\times 39.3701^3=61.02384\ in^3[/tex]

[tex]\mathbf{1\ L=61.02384\ in^3}[/tex]

[tex]0.135\ kW=135\ W[/tex]

[tex]135\ W=135\ Nm/s[/tex]

[tex]1\ N=0.224809\ lbf[/tex]

[tex]1\ m=3.28084\ ft[/tex]

[tex]135\times 0.224809\times 3.28084=99.5709185406\ ft-lb/s[/tex]

[tex]\mathbf{0.135\ kW=99.5709185406\ ft-lb/s}[/tex]

[tex]304\ kPa=304000\ N/m^2[/tex]

[tex]1\ m=39.3701\ in[/tex]

[tex]304000\ N/m^2=304000\times 0.224809\times \dfrac{1}{39.3701^2}=44.091435811\ psi[/tex]

[tex]\mathbf{304\ kPa=44.091435811\ psi}[/tex]

[tex]122\ ft^3=\dfrac{122}{3.28084^3}=3.45465495258\ m^3[/tex]

[tex]\mathbf{122\ ft^3=3.45465495258\ m^3}[/tex]

[tex]1\ hp=746\ W[/tex]

[tex]100\ hp=100\times 746\ W=74.6\ kW[/tex]

[tex]\mathbf{100\ hp=74.6\ kW}[/tex]

[tex]1\ lbm=0.224809\ kg[/tex]

[tex]1000\ lbm=1000\times 0.453592=453.592\ kg[/tex]

[tex]\mathbf{1000\ lbm=453.592\ kg}[/tex]

Answer:

[tex]\bar x=4679496.086\ m=4679.496086\ km[/tex] from the center of the earth.

Explanation:

We have a system of Earth & Moon:

Now we assume the origin of the system to be at the center of the earth.

Now for the center of mass of this system:

[tex]\bar x=\frac{m_e.x_e+m_m.x_m}{m_e+m_m}[/tex]

here:

[tex]x_e\ \&\ x_m[/tex] are the distance of the centers (center of masses) of the Earth and the Moon from the origin of the system.

[tex]x_e=0[/tex] ∵ since we have taken the point as the origin of the system.

[tex]x_m=d[/tex]

now putting the values in the above equation:

[tex]\bar x=\frac{(5.972\times 10^{24}\times 0)+(7.348\times 10^{22}\times 385000\times 1000)}{5.972\times 10^{24}+7.348\times 10^{22}}[/tex]

[tex]\bar x=4679496.086\ m=4679.496086\ km[/tex] from the center of the earth.

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:

No, they are not able to make this calculation

Explanation:

Acceleration Equation of satellite:

                                g = (G • Mcentral)/R2 ..............(1)

Acceleration equation of a satellite in circular motion:

                                a=(G • Mcentral)/R2...................(2)

Newton's form of Kepler's third law .

The period of a satellite (T) and the mean distance from the central body (R) are related by the following equation:

                                     [tex]\frac{T^{2} }{R^{3} }[/tex]=4×[tex]\pi ^{2}[/tex]/G×[tex]M_{central}[/tex].............(3)

T= the period of the satellite

R= The average radius of orbit for the satellite

G=6.673 x 10-11 N•m2/kg2.

According to all these three equations(1)(2)(3)

The period, speed and the acceleration of an orbiting satellite are independent upon the mass of the satellite.

Answer: Maximum distance

= {s²/g} * sine(2*theta)unit

Explanation: This is a projectile motion problem. The horizontal distance between the tennis player and where the tennis reaches over the net is given by the horizontal Range.

Range = {s² * sine2*theta}/g

(s)is the initial speed of projection

Theta is the angle of projection

g is acceleration due to gravity 10m/s².

Answer: according to the Avagadro's law, volume is directly propotional to no of moles: VXn

according to the Charles law, volume is directly propotional to  temperatue: VXT

according to the Boyle's law, volume is inversely propotional to P: VX1/P

when we combine them we get:

VXnT1/P

V=knT/P

k= R(universal gas constant)

V=RnT/P

PV=nRT  

Answer:

Biological system is one of the major causes of oscillation due to sensitive negative feedback loops. For instance, imagine a father teaching his son how to drive, the teen is trying to keep the car in the centre lane and his father tell him to go right or go left as the case may be. This is a example of a negative feedback loop of a biological system. If the father's sensitivity to the car's position on the road is reasonable, the car will travel in a fairly straight line down the centre of the road. On the other hand, what happens if the father raise his voice at the son "go right" or when the car drifts a bit to the left? The startled the son will over correct, taking the car too far to the right. The father will then starts yelling "go left" then the boy will over correct again and the car will definitely oscillate back and forth. A scenario that indicates the behavior of a car driver under a very steep feedback control mechanism. Since the driver over corrects in each direction. Therefore causes oscillations.

Explanation:

In physics, a system's equilibrium point is considered stable if the system returns to that point after being slightly disturbed. For an n value less than 8, the system remains in stable equilibrium; once n exceeds 8, stable oscillations indicate an unstable equilibrium. This concept can be visualized by the marble in a bowl analogy, where the bowl's orientation determines the stability of the marble's equilibrium.

The stability of an equilibrium point is determined by the response of a system to a disturbance. If an object at a stable equilibrium point is slightly disturbed, it will oscillate around that point. The stable equilibrium point is characterized by forces that are directed toward it on either side. On the contrary, an unstable equilibrium point will not allow the object to return to its initial position after a slight disturbance, since the forces are directed away from that point. For values of n less than 8, the system finds a stable equilibrium. However, when n surpasses 8, the system exhibits unstable equilibrium, leading to stable oscillations.

As an example, consider a marble in a bowl. When the bowl is right-side up, the marble represents a stable equilibrium; when disturbed, it returns to the center. If the bowl were inverted, the marble on top would be at an unstable equilibrium point; any disturbance would cause it to roll off, as the forces on either side would be directed outwards. Extending this concept to potential energy, n in a potential energy function acts as an adjustable parameter. For n=<8, the system remains in stable equilibrium, as with NaCl, which has an n value close to 8. Beyond this value, the equilibrium becomes unstable, giving rise to oscillatory behavior.

Stability also depends on the nature of damping in the system. An overdamped system moves slowly towards equilibrium without oscillations, an underdamped system quickly returns but oscillates, and a critically damped system reaches equilibrium as quickly as possible without any oscillations.

We will make the comparison between each of the sizes against the known wavelengths.

In the case of the hydrogen atom, we know that this is equivalent to [tex]10^{-10}[/tex] m on average, which corresponds to the wavelength corresponding to X-rays.

In the case of the Virus we know that it is oscillating in a size of 30nm to 200 nm, so the size of the virus is equivalent to the range of the wavelength of an ultraviolet ray.

In the case of height, it fluctuates in a person around [tex]10 ^ 0[/tex] to [tex]10 ^ 1[/tex] m, which falls to the wavelength of a radio wave.

The region of the spectrum that best corresponds to light with a wavelength equal to the diameter of a hydrogen atom is the X-ray region. The visible light spectrum corresponds to objects around the same size as the wavelength. The size of a virus is much smaller than the wavelength of visible light

The region of the spectrum that best corresponds to light with a wavelength equal to the diameter of a hydrogen atom is the X-ray region of the electromagnetic spectrum.

The wavelength of visible light corresponds to the size of objects that are around the same size as the wavelength. For example, if you want to use light to see a human, you would need to use a wavelength at or below 1 meter, since humans are about 1 meter in size.

The size of a virus is much smaller than the wavelength of visible light, so the wavelength is very small compared to the object's size.

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:

a) [tex]\lambda=1\ m[/tex]

b) [tex]f=122.47\ Hz[/tex]

c) [tex]\lambda_s=2.8\ m[/tex]

Explanation:

Given:

distance between the fixed end of strings, [tex]l=1.5\ m[/tex]

mass of string, [tex]m=5\ g=0.005\ kg[/tex]

tension in the string, [tex]F_T=50\ N[/tex]

a)

Since the wave vibrating in the string is in third harmonic:

Therefore wavelength λ of the string:

[tex]l=1.5\lambda[/tex]

[tex]\lambda=\frac{1.5}{1.5}[/tex]

[tex]\lambda=1\ m[/tex]

b)

We know that the velocity of the wave in this case is given by:

[tex]v=\sqrt{\frac{F_T}{\mu} }[/tex]

where:

[tex]\mu=[/tex] linear mass density

[tex]v=\sqrt{\frac{50}{(\frac{m}{l}) } }[/tex]

[tex]v=\sqrt{\frac{50}{(\frac{0.005}{1.5}) } }[/tex]

[tex]v=122.47\ m.s^{-1}[/tex]

Now, frequency:

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

[tex]f=\frac{122.47}{1}[/tex]

[tex]f=122.47\ Hz[/tex]

c)

When the vibrations produce the sound of the same frequency:

[tex]f_s=122.47\ Hz[/tex]

Velocity of sound in air:

[tex]v_s=343\ m.s^{-1}[/tex]

Wavelength of the sound waves in air:

[tex]\lambda_s=\frac{v_s}{f_s}[/tex]

[tex]\lambda_s=2.8\ m[/tex]

Answer:

A

Explanation:

i took the test pimp

Answer:

a)time t = 47.4s

b)speed v = 38.7cm/s

Explanation:

Given:

Total volume of blood = 4.93L × 1000cm^3/L = 4930cm^3

Volumetric rate of flow = 104.1cm^3/s

a) Time taken for all the blood to pass through the aorta is:

Time t = total volume/ volumetric rate

t = 4930/104.1

t = 47.4s

b) Given that the diameter of the aorta is 1.85cm.

V = Av

Where V = Volumetric rate

A = area of aorta

v = speed of blood

v = V/A ...1

Area of a circular aorta = πr^2 = (πd^2)/4

d = 1.85cm

A = (π×1.85^2)/4

A = 2.69cm^2

From equation 1.

v = V/A = 104.1/2.69

v = 38.7 cm/s

To solve this problem we will apply the physical equations of the angular kinematic movement, for which it defines the square of the final angular velocity as the sum between the square of the initial angular velocity and the product between 2 times the angular acceleration and angular displacement. We will clear said angular displacement to find the correct response

Using,

[tex]\omega^2 = \omega_0^2 +2\alpha \theta[/tex]

Here,

[tex]\omega[/tex] = Final angular velocity

[tex]\omega_0[/tex] = Initial angular velocity

[tex]\alpha =[/tex] Angular acceleration

[tex]\theta =[/tex] Angular displacement

Replacing,

[tex]59^2 = 0+2*58\theta[/tex]

[tex]\theta = 30rad[/tex]

Therefore the correct answer is C.

The angle at which the wheel turns while accelerating is 30 radians and this can be determined by using the kinematics equation.

Given :

A wheel accelerates from rest to 59 rad/[tex]\rm s^2[/tex] at a uniform rate of 58 rad/[tex]\rm s^2[/tex].

The equation of kinematics is used in order to determine the angle at which the wheel turn while accelerating.

[tex]\omega^2 = \omega^2_0+2\alpha \theta[/tex]

where [tex]\omega[/tex] is the final angular velocity, [tex]\omega_0[/tex] is the initial angular velocity, [tex]\alpha[/tex] is the angular acceleration, and [tex]\theta[/tex] is the angular displacement.

Now, substitute the values of the known terms in the above formula.

[tex]59^2 =0+2\times 58 \times \theta[/tex]

Simplify the above expression.

[tex]\rm \theta = 30\; rad[/tex]

Therefore, the correct option is C).

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To solve this problem we will apply the geometric concepts of the Volume based on the consideration made of the radius measurement. The Volume must be written in differential terms of the radius and from the formula of the margin of error the respective response will be obtained.

The error in radius of sphere is not exceeding 2%

[tex]\frac{dr}{r} = \pm 0.02[/tex]

The objective is to find the percentage error in the volume.

The volume can be defined as

[tex]V = \frac{4}{3} \pi r^3[/tex]

Differentiate with respect the radius we have,

[tex]\frac{dV}{dr} = 4\pi r^2[/tex]

[tex]dV = 4\pi r^2 \times dr[/tex]

[tex]dV = 4\pi r^2 (\pm 0.02r)[/tex]

[tex]dV = \pm 4\times 0.02 \times \pi r^3[/tex]

The percentage change in the volume is as follows

[tex]\% change = \frac{dV}{V} \times 100[/tex]

[tex]\% change = \frac{\pm 4 \times 0.02 \times \pi r^3 \times 3}{4\pi r^3}\times 100[/tex]

[tex]\% change = \pm 6\%[/tex]

Therefore the percentage change in volume is [tex]\pm 6\%[/tex]

Final answer:

The wavelength of the 6th harmonic of a stretched string fixed at both ends with a length of 160 cm is 53.33 cm.

Explanation:

To calculate the wavelength of the 6th harmonic of a stretched string fixed at both ends, we need to use the formula for the wavelength of standing waves on a string. The formula for the nth harmonic on a string of length L is given by:

\[\lambda = \frac{2L}{n}\]

In this case, the length L is 160 cm, and the harmonic number n is 6.

So, the wavelength for the 6th harmonic is:

\[\lambda_6 = \frac{2 \times 160}{6}\]

\[\lambda_6 = \frac{320}{6}\]

\[\lambda_6 = 53.33 \text{ cm}\]

Therefore, the wavelength of the 6th harmonic is 53.33 cm.

Answer:

1. True

2. True

3. False

4. False

5. True

6. False

Explanation:

Acceleration: It refers to the change in velocity/speed of an object with respect to time. When the speed increases with time we call it acceleration and when its decreases it is called as deceleration.  Let us analyze each instance individually:

1. When roller coaster starts to roll down the track its speed will increase with time. That means it is accelerating.

2. When the ball reaches at the peak of its trajectory, it comes to a stop for a fraction of a second that means it decelerates.

3. Since the velocity remains constant there is no acceleration.

4. Since the speed is no changing with time, there is no acceleration.

5. Since the moving plane comes to a stop, it is a case of deceleration.

6. Since the truck is moving at a constant speed so the acceleration is zero.

A roller coaster as it starts to roll down a track and an airplane skidding to a stop are examples of acceleration, while a ball tossed straight up, a planet tracing a circular orbit, a block sliding down a ramp, and a dump truck carrying a load at a constant speed are not examples of acceleration.

1) True. When a roller coaster starts to roll down the track, it experiences a change in velocity, which means it is accelerating.

2) False. At the peak of its trajectory, a ball tossed straight up has zero velocity and is momentarily at rest, so it is not accelerating.

3) False. A planet tracing out a circular orbit at a constant speed is not experiencing a change in velocity, so it is not accelerating.

4) False. The block sliding down a ramp at a constant speed is not experiencing a change in velocity, so it is not accelerating.

5) True. An airplane skidding to a stop on a runway experiences a change in velocity, so it is accelerating.

6) False. A dump truck carrying a load straight forward at a constant speed is not experiencing a change in velocity, so it is not accelerating.