An electric field of magnitude 3.50 kN/C is applied along the x axis. Calculate the electric flux through a rectangular plane 0.350m wide and 0.700m long (c) if the plane contains the y axis and its normal makes an angle of 40.0° with the x axis.

Bones, Cartilage, Ligaments, Tendons, and Muscles.

Answer:

10kg

Explanation:

(I'm not super familiar with the GRESA method so apologies for any inaccuracies)

Given: We are given values for Force: 100N, and Acceleration: 10m/s2.

Required: We are trying to find Mass (m)

Equation: The best equation to use to solve this problem is F=ma,

Force = Mass x Acceleration. We can rearrange this for mass: m = F/a.

Solution: By substituting in the values we have: [tex]m = \frac{100}{10}[/tex]

Answer: Mass = 10kg

Hope this helped!

The electric charge on the rod is no longer separated.

If the magnetic field is parallel to the position of the exposed surface the magnetic flux produced will be zero if the magnetic field is non-zero. If a charged particle moving parallel to the magnetic field has a velocity parallel to the magnetic field, then the force is zero In the above case the velocity is parallel to the magnetic field lines, so the magnetic force is zero.

Induced currents can be created by changing the strength of the magnetic field changing the size of the wire loop, or changing the direction of the wire loop. The main difference between magnetic flux and flux density is that magnetic flux is a scalar quantity whereas flux density is a vector quantity. Flux is the scalar product of flux density and area vector.

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The radiation pressure is 83.3 nPa.

The mechanical pressure that is applied to any surface as a result of the exchange of momentum between an item and an electromagnetic field is known as radiation pressure.

It is the pressure that electromagnetic radiation exerts on a surface as a function of the momentum it carries; radiation pressure doubles if the radiation is reflected as opposed to absorbed.

The radiation pressure will be:

P = I/C

P = 25 / 3 / 10^8

P = 83.3 nPa.

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The repulsive force between the two pith balls is 9.6 × 10⁻⁴ N.

A pith ball, which is a tiny pith ball suspended on a thread, may detect the presence and intensity of an electric charge in an object that is near or touching it.

Two pith balls are 11 cm apart from each other and both have equal charges of -36 nC.

r = 11 cm = = 11 × 10⁻² m = 0.11 m

q = q₁ = q₂ = -36 nC = -36 × 10⁻⁹ C

According to Coulomb's law, the force between two charges is given as:

F = k [ q₁ q₂ / r² ]

The constant k = 8.99 × 10⁹ N.m² / C²

Therefore,

F = (8.99 × 10⁹ ) [ (-36 × 10⁻⁹) (-36 × 10⁻⁹) / (0.11)² ]

F = (8.99 × 10⁹ ) [ 1296 × 10⁻¹⁸ / 0.0121 ]

F = (8.99 × 10⁹ ) [ 10.7 × 10⁻¹⁴ ]

F = 9.6 × 10⁻⁴ N

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If the nails dimensions were expressed in meters, we would have to divide it by 100.

In earlier times there were 3 main types of measurement systems:

1.The CGS system(centimetre, gram and second)

2.• The FPS system( foot, pound and second)

3.• The MKS system(metre, kilogram and second)

But in the present day, we use Système Internationale d’ Unites (French for International System of Units), abbreviated as SI.

The SI Unit of length is meter, but in many cases the length is given in centimeters.

So, we have to convert the value given in centimeter to meter by using the conversion factor which is,

1 meter = 100 centimeters

1 cm = 1/100 meter

Therefore, if the nails dimensions were expressed in meters, we would have to divide it by 100.

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The speed of rocket is ,V₀ = 268.42ft/sec

A projectile is an object upon which the only force is gravity.

a rocket launched in vertical direction by a group of students.

The path of projectile will be parabola.

to find the speed of the rocket is  we use the equation of a projectile motion the data determined by them is

Initial height is.[tex]y_{0 } =[/tex] 89.6 ft

 final height is  [tex]y_{f} =[/tex] 0ft

 time taken, t = 17sec

as we know , g = 32.2 ft/s²

g is acceleration due to gravity.

 [tex]y_{f} - y_{o} = V_{o}t - \frac{1}{2} g t^{2}[/tex]

0 = 89.6 + 17V₀ - 16.1 × (17)²

V₀ = 268.42ft/sec

The speed of rocket is ,V₀ = 268.42ft/sec

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Answer: Conclusion White matter hyperintensities predict an increased risk of stroke, dementia, and death. Therefore white matter hyperintensities indicate an increased risk of cerebrovascular events when identified as part of diagnostic investigations, and support their use as an intermediate marker in a research setting

Answer:

± .55 kg

Explanation:

55  * 1% = .55 kg

± .55 kg

The magnetic field inside the solenoid is

B = μo nI

Here n = N/l

Here I be the current

N be the number of turns

l be the length of solenoid

Because the magnetic field in this instance is independent of radius, "c) it is unaffected."

Electric charges in motion create magnetic fields. Everything is constructed of atoms, and each atom contains a nucleus that is composed of protons and neutrons, with orbiting electrons. Each atom creates a tiny magnetic field because the orbiting electrons are tiny moving charges.

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The rest energy of the proton is 900 MeV.

We need to know about relativistic energy to solve this problem. The rest energy of the object can be determined by

Eo = m₀ . c²

where Eo is rest energy, m₀ is rest mass and c is speed of light (3 x 10⁸ m/s).

From the question above, we know that :

m₀ = 1.6 x 10¯²⁷ kg

c = 3 x 10⁸ m/s

By substituting the following parameter, we get

Eo = m₀ . c²

Eo = 1.6 x 10¯²⁷ . (3 x 10⁸)²

Eo = 1.44 x 10¯¹⁰ joule

Eo = 1.44 x 10¯¹⁰ / (1.6 x 10¯¹⁹) eV

Eo = 900 x 10⁶ eV

Eo = 900 MeV

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Speed and velocity are similar in term of their unit. the speed of the jet is 42 m/s

What is Speed ?

Speed can simply be defined as how fast an object moves. It is the distance covered per time taken.

Given that acceleration = 2.8 m/s² and If after take-off, the jet continues to accelerate at the same rate for another 15 s, to know how fast it will be going at that time, we will use acceleration formula.

Acceleration = Velocity / time

speed = acceleration × time

speed = 2.8 × 15

Speed = 42 m/s

Therefore, the speed or the velocity of the jet is 42 m/s

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The sound level from 1 km away is 46.4 dB and the sound energy radiated through the windows and doors in 20 min is 332.6 J

Multiple plate modes participate in the forced vibration of a baffled plate that produces sound. The emitted sound power depends on the activation of each plate mode by the external pressures as well as the radiation characteristics of the individual plate modes and their mutual interaction via their radiated sound, as shown in eq (20) and (24). The net contribution of the mutual interaction between plate modes to the overall sound power may be disregarded if a plate is activated in a manner that results in a plate vibration field that can be defined as reverberant with an equal distribution of energy over all modes.

(Lw) = 10·log (W/Wo) dB

Given:

sound level,  [tex]\beta= 101 dB[/tex]

Area, A = [tex]22\;m^{2}[/tex]

Time, [tex]\triangle t = 20\;min=1200\;s[/tex]

Intensity, [tex]I=1\times 10^{-12}\;W/m^{2}[/tex]

[tex]r=1\;km=1000\;m[/tex]

(a)

We know that, Sound level is,

[tex]\beta=10\times log(\frac{I}{I_{o} } )[/tex]

Solving the above equation for sound intensity,

[tex]I=I_{o} \times 10^{\frac{\beta}{10} }[/tex]

[tex]I=1 \times 10^{-12} \times 10^{\frac{101}{10} }[/tex]

[tex]I=0.0126\;W/m^{2}[/tex]

Therefore, The sound energy is,

[tex]E=P\times \triangle t[/tex]

Substitute [tex]P=I \times A[/tex] in the above equation,

[tex]E=I \times A \times \triangle t[/tex]

[tex]E=0.0126 \times 22 \times 1200[/tex]

[tex]E=332.6\;J[/tex]

(b)

Half of a sphere area with 1 km radius is equal to Area of the sound at 1 km distance,

[tex]A_{hemisphere} = \frac{1}{2} \times 4 r^{2} \pi[/tex]

Substitute the known value in the above equation ,

[tex]A_{hemisphere} = \frac{1}{2} \times 4 (1000)^{2} \pi[/tex]

[tex]A_{hemisphere} = 6283185\;m^{2}[/tex]

Sound Intensity is,

[tex]I = \frac{P}{A_{hemisphere}}[/tex]

Substitute [tex]P=I \times A[/tex] in the above equation,

[tex]I = \frac{I \times A}{A_{hemisphere}}[/tex]

Substitute the known value in the above equation,

[tex]I = \frac{0.0126 \times 22}{6283185}[/tex]

[tex]I = 4.4 \times 10^{-8}\;W/m^{2}[/tex]

Sound level is,

[tex]\beta=10\times log(\frac{I}{I_{o} } )[/tex]

Substitute the known value in the above equation,

[tex]\beta=10\times log(\frac{4.4 \times 10^{-8} }{1 \times 10^{-12} } )[/tex]

[tex]\beta=46.4\;dB[/tex]

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The ball that is thrown straight up into the air with a velocity of 63. 7 m/s after 3s will have a velocity of: 34.3 m/s

The formula for the vertical launch upward and the procedure we will use is:

vf = v₀ - g * t

Where:

Information about the problem:

Applying the final velocity formula we get:

vf = v₀ - g * t

vf = 63. 7 m/s - 9.8 m/s² * 3 s

vf = 63. 7 m/s - 29.4 m/s

vf = 34.3 m/s

In physics vertical launch upwards is the motion described by an object that has been launched vertically upwards in which the height and the effect of the earth's gravitational force on the launched object are taken into account.

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The time when the object is not moving on the velocity time graph is  at the point in which the line crosses the x - axis.

The velocity time graph can be used to observe the movement of a particle. The graph consist of three main parts;

We can say that the time when the object is not moving on the velocity time graph is  at the point in which the line crosses the x - axis.

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Answer:

Car velocity when crossing checkpoint = 60 m/s

Explanation:

For an object traveling at an acceleration of a m/s², with an initial velocity of u m/s the displacement at time = t secs is given in meters by the equation
[tex]\displaystyle d=ut+\frac{1}{2}at^{2}[/tex]

Here we are given displacement, and time but not acceleration and initial velocity : d = 225 meters, t = 5 seconds

Let's find an equation relating u and a in terms of d and using data given

Switching sides we get
[tex]ut+\frac{1}{2}at^{2}=d[/tex]

Substituting values for t = 5, d = 225 we get
[tex]5u+\frac{1}{2}a.25=225[/tex]

Multiplying both sides by 2 yields
[tex]10u+25a=450\;\;\; ...... (1)[/tex]

We also have the formula:
[tex]\displaystyle a=\frac{{v-u}}{t}[/tex]
where v is the current velocity and u the initial velocity

So
[tex]\displaystyle a=\frac{30-u}{5}[/tex]
[tex]u+5a=30\;\;...... (2)[/tex]

Multiply equation (2) by 5 and subtract from (1) to eliminate the a terms and solve for u
[tex]\displaystyle 10u+25a-5u-25a=450-150\\\\5u=300\\\\u=60m/s\\\\\textsf{which is the speed at which the car passes the checkpoint}\\[/tex]

The value of the comic book decreased by an average of $0.04 each month between m = 36 and m = 60.

The average rate of change for the function f(x) from [tex]x_{1} =a[/tex] and [tex]x_{2}=b[/tex] is [tex]\frac{f(b)-f(a)}{b-a}[/tex]

It is the average amount by which the function changed per unit throughout that time period.

Actually, this is the angle of the line on the function's graph that passes through two points.

The value of the comic book fell by an average of $0.04 per month (a unit in this case is a month) between m = 36 and m = 60 if the average rate of change of the function f(x) between m = 36 and m = 60 is -0,04.

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Middle part of the ear have a solid medium. The ossicles, which are bones that amplify sound waves, are located in the middle ear, therefore the medium is solid.

The outer ear is the first place that sound waves pass through, followed by the middle ear and lastly the inner ear. The first medium is gas because the external auditory canal, which is part of the outer ear, is filled with ambient air.  The ossicles, which are bones that amplify sound waves, are located in the middle ear; therefore the second medium is solid. The last medium is liquid because the inner ear is loaded with fluids like endolymph and perilymph that vibrate when sound waves pass through them.

The middle ear amplifies sound after being directed toward it by the outer ear, which is in charge of catching it.  Amplified sound waves are sent from the middle ear to the inner ear, where hair cells detect the sound waves and transmit the data to the brain.

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When the parallel-plate capacitors are disconnected from a battery and plates are pulled a small distance farther apart then Q remains the same, Change in potential increases, C decreases, E stays the same and Energy stored increases.

Solution:

Let's say the capacitance of parallel plates is C

Charge stored = Q

Distance between a plate of capacitor = d

The capacitance of a parallel plate capacitor equals when a parallel- plate capacitor is disconnected.

Then, charge Q remains the same i.e.

Q = εA / d = Q / Ed = Q/V

Here Q=Charge stored in a capacitor

E = Electric field, V is the potential difference, and 'd' is the separation between plate  

Both potential difference and distance are inversely proportional to the capacitance between a plate of the capacitor.

Similarly, the potential difference increases when capacitance decreases.

E = V/d

While the value of V and d decrease.

E remains constant, therefore energy stored in the capacitor will be E = 1/2CV^2

The decrease in C causes an increase in V, as a result of which energy stored in the capacitor increases.

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When the electrons accelerated through a potential difference of 84 kV, the electrons have half the speed that she wishes.

When the potential is quadrupled to 336 kV, the electrons acquire the desired speed.

Let u be the speed of the electrons after accelerating through a potential difference ΔV.

Now,

K = eΔV = (γ - 1)mc² where K is the kinetic energy.

K = (γ - 1)mc² =[ 1/ ( √1 - (u/c)² ) - 1]mc²

Therefore,

[ 1/ ( √1 - (u/c)² ) - 1] = (eΔv / mc² ) + 1 = (eΔV + mc² ) / (mc²)

1 - (u/c)² = [ (mc²) / (eΔV + mc² ) ]²

u/c = √[ 1 - ( m / {(eΔV/c²) + m})² ]

u / c = √[ 1 - ( 9.11 × 10⁻¹³ / {(1.6 × 10⁻¹⁹ × 8.4 × 10⁴ /(3 × 10⁸)² +  9.11 × 10⁻¹³ })² ]

u = 0.512c

Despite the increased accelerating voltage, this speed cannot be doubled because it is greater than half that of light. The electrons move at u = 0.798c faster if the accelerating voltage is quadrupled to 336 kV.

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Yes, this situation is possible. No, It is not possible in more than one way.

A schematic of the information provided in the question can be seen in the image attached below.

 From the picture: Assuming that the forces exerted on the wire by the other two  wires are equal and opposite,  the third wire exerts no net force. So you can say: Yes the situation is possible.

The forces acting on the other two wires will be in opposite directions, so they cannot have multiple passes, but they will not be of the same  magnitude.

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Answer:

40s to run the second 200 m

Explanation:

the average speed is 4.0 m/s

if he is moving at a constant speed he should be able to run the first and second have at the same time

[tex]4.78\times 10^8\ \mathrm{Pa}$[/tex]  is the stress in the rod.

Given:

Young's modulus, Y = [tex]20 \times 10^{10}\;N/m^{2}[/tex]

Steel density, [tex]\rho = 7.86 \times 10^{3}\;kg/m^{3}[/tex]

Length of the rod, L = 80 cm = 0.800m

The speed of the wave in the rod is,

[tex]$t=\frac{L}{v} = \frac{0.800\ \mathrm{m}}{5044\ \mathrm{m/s}} = 1.58\times 10^{-4}\ \mathrm{s}$[/tex]

Hence the time taken by the wave to travel the end of the rod is,

[tex]$t=\frac{L}{v} = \frac{0.800\ \mathrm{m}}{5044\ \mathrm{m/s}} = 1.58\times 10^{-4}\ \mathrm{s}$[/tex]

The velocity of the rod is, [tex]$v_r = 12.0\ \mathrm{m/s}$[/tex]

The other end will keep moving after the front end hits the wall until the wave reaches the other end, given the period determined by the above problem.

As a result, before the wave reaches the end of the rod, it has traveled a distance of

[tex]$\Delta L = v_r t = (12.0\ \mathrm{m/s}) (1.58\times\ \mathrm{s}) = 1.90\times 10^{-3}\ \mathrm{m} = 1.90\ \mathrm{mm}$[/tex]

The strain in the rod is given by,

[tex]$\frac{\Delta L}{L} = \frac{1.90\times 10^{-3}\ \mathrm{m}}{0.800\ \mathrm{m}} = 2.37\times 10^{-3}$[/tex]

The Young's modulus is given by,

[tex]$Y = \frac{F/A}{\Delta L/L}$[/tex]

Now, we may write that the stress is given by using the equation above,

[tex]$\frac{F}{A} = Y\frac{\Delta L}{L} = \left( 20\times 10^{10}\ \mathrm{N/m^2} \right) \left ( 2.73\times 10^{-3} \right ) = 4.78\times 10^8\ \mathrm{Pa}$[/tex]

Hence, [tex]4.78\times 10^8\ \mathrm{Pa}$[/tex] is the stress in the rod.

Stress is a tangible property in continuum mechanics. It happens as a result of a body being subjected to tension or compression forces. Stress is defined as a force per unit area within a material that results from externally applied forces, unequal heating, or persistent deformation and that enables an accurate description and prediction of elastic, plastic, and fluid behavior in physical sciences and engineering. By dividing a force by an area, a stress is expressed.

Various forms of stress exist. Shear stress results from forces that are parallel to and reside in the plane of the material's cross-section, whereas normal stress results from forces that are perpendicular to the area.

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Frequency 1200 Hz  is monitored by the leading ship.

Here, the leading ship serves as the observer while the trailing ship serves as the source.

Therefore, the source's velocity is v' = 64 km/h = 17.777 m/s.

Observer's speed v "= 45 km / h = 12.5 m / s

The leading ship measures frequency using the formula f'= [(v -v ') / (v - v "()] f where v = source speed = 1520 m/s

Values obtained by substitution are f' = 1502.23/1507.5*1200 = 1195.8

The study of illness frequency and distribution in a specific community is known as epidemiology.

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Yes, if the two 50 kg objects are significantly farther apart, the gravitational force between them may be less than the gravitational force between a 50 kg and a 5 kg object.

The gravitational force increases as the mass of an object increases (also called the gravity force). Since gravitational force is inversely proportional to the square of the distance between the two interacting objects, greater separation distance will result in less gravitational forces.

Therefore, the gravitational attraction between two things weakens as they become farther apart. The size of this force is determined by the mass of each object and the separation of their centers.

According to mathematics, the force of gravity is proportional to the square of the distance between the objects and directly related to the masses of the items.

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The radius of the copper cable, carrying 300A current with 2 W/ m power loss, is 0.0156 m or 15.6 mm.

First, let us list the given:

The formula that will be used to solve for the radius of the cable is shown below.

Wherein: R = resistance, L= length of cable, and A = cross-section area of cable.

In solving for R, the length of the cable is assumed to be 1 m since the unit of power should be in W.

(2 W / m)(1 m) = [tex](300 A)^{2}*R[/tex]

R= 2.22 x 10 -5 Ω

Solve for the cross-section area of the cable.

R = 2.22 x 10 -5 Ω = [(1.7 X 10 -8 Ω m)*(1 m)] / A

A = 7.65 x 10 -4 square meters

Solve for the radius of the cable.

A = 7.65 x 10 -4 square meters = *[tex]radius^{2}[/tex]

radius = [tex]\sqrt{\frac{7.65x10^{-4} }{pi} }[/tex] = 0.0156 m or 15.6 mm

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The position of the wire is at a distance of 12 cm, in the direction left of the wire.

It is given that I₁ = 1.50 A and I₂ = 4.00 A and the distance between wire 1 and wire 2 = 20cm = 0.2m

Let the distance of wire 3 from wire 2 be " d " .

We have to place the 3 wire such that each wire experiences no net force which means force on wire 2 due to wire 3 = force on wire 1 due to wire 3

   [tex]F_2=F_1[/tex]

Using equation (1) , we get

[tex]\frac{u_0I_2I_3}{0.2+d}=\frac{u_0I_1I_3}{d} \\\\\frac{I_2}{0.2+d} =\frac{I_1}{d} \\\\\frac{4.00}{0.2+d} =\frac{1.50}{d} \\\\4.00d=1.50(0.2+d)\\\\4.00d=0.3+1.50d\\\\\2.5d=0.3\\\\d=0.12m[/tex]

d= 12 cm in the direction left of the wire.

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The complete question is given below .

A schematic of the information provided in the question can be seen in the image attached below.Two long wires hang vertically. Wire 1 carries an upward current of 1.50 A . Wire 2,20.0cm to the right of wire 1, carries a downward current of 4.00 A . A third wire, wire 3 , is to be hung vertically and located such that when it carries a certain current, each wire experiences no net force. (b) the position of wire 3 and

1000kg  is the mass of the roller coaster in kg

The speed of the roller coaster is, v=20ms−1

Ek=1/2⋅1000⋅20∧2=2⋅10∧5J

The mass and speed of an item affect its kinetic energy, which is the energy of motion. As the coaster cars lose height, the train accelerates. As a result, their initial potential energy, which was caused by their height, is converted into kinetic energy (revealed by their high speeds).

Kinetic energy has the following formula where m is the object's mass and v is its square velocity. The kinetic energy is measured in kilograms-meters squared per second squared if the mass is measured in kilogrammes and the velocity is measured in metres per second.

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It is not possible to accelerate an object to the speed of light.

The speed of light is generally regarded as the fastest speed on earth. We can tell from the Newton's law that the force that acts on a body is the product of the mass and the acceleration of the body.

It then follows that the force that is required to accelerate an object to the speed of light is a very high force that s somewhat unattainable and as the object approaches the speed of light, the mass of the object has to become infinitely small.

Hence, it is not possible to accelerate an object to the speed of light.

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