Answer:
use ohms law for tabling.
Potential (V) = current (I) × R
Let R = 1 ohm.
table:-
V. (voltage difference) current (I)
1. 1 Volt 1A
2. 2 volt 2 A
3. 4 volt 4 A
4. 6 volt 6A
.
.
.
.
reduction in cerebral blood flow in areas appearing as white matter hyperintensities on magnetic resonance imaging author links open overlay paneladam m.brickman
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
a roller coaster has 800,000 joules of kinetic energy and is moving with a speed of 40 m/s. what is the mass of the roller coaster in kg?
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
In a roller coaster, how do mass and kinetic energy relate?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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four solid plastic cylinders all have radius 2.33 cm and length 5.82 cm. find the charge of each cylinder given the following additional information about each one.
the charge of each cylinder given the following additional information about each one is 10434.445 coulomb, 14906.35 Coulomb, 178876.2 coulomb, 223,595.25 coulomb.
Equation :To solve this question we are to use the equation 1 below :
Charge Q = uniform charge density p × Total area of the cylinder A
From the question, we are given radius, R to be 2.33 cm and length, L to be 5.82 cm.
Calculate for the total area of the cylinder, A. equation 2
Total area of the cylinder, A = area of the top surface + area of the bottom + area of the curved surface of the cylinder.
Hence, total area of the cylinder A is;
πR^2 + πR^2 + 2πRL.
Then, total area of the cylinder A is;
(L + R)2πR.
find the charge of each cylinder equation 3
For the first cylinder; we have the uniform charge density to be 35 nC/m^2.
Therefore, the combination of equation (1) and (3) gives equation 4
Charge Q= p × (L + R)2πR
Hence,
Charge Q = 35 × [(5.82 + 2.33) 2× 3.143 × 5.82]
Charge Q = 10434.445 coulomb.
For the cylinder B, we have a uniform charge density of 50 nC/m^2.
Using equation (4),
charge Q = p × (L + R)2πR
= 50 × [(5.82 + 2.33) 2× 3.143 × 5.82]
= 14906.35 Coulomb
For The cylinder C, the uniform charge density is 600, we use of equation (4);
Charge Q = 600 × 298.127
Charge Q = 178876.2 coulomb.
For cylinder D, the uniform charge density is 750 nC/m^2. use of equation (4)
Charge Q = 298.127 x 750
charge Q = 223,595.25 coulomb.
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Helllp
What is velocity at d,e,g,h and j
PLEASEEEEE HELP NO ONE EVER HELPS ME
suppose your bathroom scale reads your mass is 55 kg, with a 1% uncertainty. what is the uncertainty in your mass in kilograms?
Answer:
± .55 kg
Explanation:
55 * 1% = .55 kg
± .55 kg
20. A boy runs 400 m at an average speed of 4.0 m/
s. He runs the first 200 m in 40 s. How long does he
take to run the second 200 m?
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
A conducting rod is moving perpendicularly through a uniform magnetic field. Why does the motional emf drop to zero when the conductor stops moving in the field?.
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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Why is the following situation impossible? An experimenter is accelerating electrons for use in probing a material. She finds that when she accelerates them through a potential difference of 84.0 kV , the electrons have half the speed she wishes. She quadruples the potential difference to 336 kV , and the electrons accelerated through this potential difference have her desired speed.
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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What weight of the ore in lb would contain exactly 1lb of copper
The density of crushed iron ore is equivalent to 2 500 kg/m3, or 2.5 grams per cubic centimeter, which is the weight of the material. The density is equal to 156.07 pounds per cubic foot in either the Imperial or US customary measuring systems.
When weights are expressed in kilograms of air, the air gives buoyancy, and because a kilogram of copper has a larger volume than a kilogram of lead, it displaces more air and hence receives more support. As a result, on a scale in the air, a kilogram of lead will weigh more than a kilogram of copper. Copper has a density of 8.94 grams per cubic centimeter, or 8 940 kilograms per cubic meter, or 20°C (68°F or 293.15K), at ordinary atmospheric pressure.
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M Two ships are moving along a line due east (Fig. P17.58). The trailing vessel has a speed relative to a landbased observation point of v₁ = 64.0km/h , and the leading ship has a speed of v₂ = 45.0 km/h relative to that point. The two ships are in a region of the ocean where the current is moving uniformly due west at v_current = 10.0km/h . The trailing ship transmits a sonar signal at a frequency of 1200.0Hz through the water. What frequency is monitored by the leading ship?
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 the frequency and distribution of a disease in a defined population isThe study of illness frequency and distribution in a specific community is known as epidemiology.
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a copper cable carries a current of 300 a. if the power loss is 2 w per meter, find the radius of the cable. (the resistivity of copper is 1.7x10-8 ωm.)
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:
Current (I) = 300 APower Loss (P) = 2 W / m Resistivity (ρ) = 1.7 X 10 -8 Ω mThe formula that will be used to solve for the radius of the cable is shown below.
Power (P) = [tex]I^{2} * R[/tex] R = (ρ*L) / AA = π*[tex]radius^{2}[/tex]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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what is the repulsive force between two pith balls that are 11.0 cm apart and have equal charges of −36.0 nc?
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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the value, v(m), of a comic book m months after publication has an average rate of change of –0.04 between m
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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9. Which parts of the ear have a solid medium?
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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inclined plane how much work does it take to slide a crate 20 m along a loading dock by pulling on it with a 200-n force at an angle of 30° from the horizontal?
The work it takes to slide a crate 20 m along a loading dock by pulling on it with a 200-n force at an angle of 30° from the horizontal is 346.4J.
What is work?
The energy that is transmitted to or from an object when a force is applied along a displacement is referred to as work in physics. In its simplest form, it is typically explained as the outcome of force and displacement. Positive work is the term for the component of a force that moves the point of application when it is applied. A force is said to do negative work if, at the point of application, one of its components points in the opposite direction of the displacement.
Calculations:
Work done= component of force along displacement x displacement
= F cosΘ x d
= 200cos 30 x2
= 346.4 J
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A proton having an initial velvocity of 20.0i Mm/s enters a uniform magnetic field of magnitude 0.300 T with a direction perpendicular to the proton's velocity. It leaves the field-filled region with velocity -20.0j Mm/s. Determine(c) the distance the proton traveled in the field.
The distance the proton traveled in the field is 2.09m.
What is a proton?A stable subatomic particle known as a proton with the symbols p, H+, or 1H+ and an elementary electric charge of +1e. It has a mass that is somewhat lower than that of a neutron, and because of the proton to electron mass ratio, it has an 1836-fold greater mass than an electron. The term "nucleons" refers to protons and neutrons, both of which have masses of roughly one atomic mass unit (particles present in atomic nuclei).
In the nucleus of every atom, there are one or more protons. They offer the central force of electrostatic attraction that holds the atomic electrons together.
Explanation:
Now as the initial and final velocity makes 180degree angle, So, the proton must have travelled only half a revolution. So, the distance ,d will be :
d=2πr/2=πR
d=2.09m
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What is the mass of an object that requires 100N (kg-m/s2) of force in order to accelerate it at 10m/s2 (Please use G-R-E-S-A)
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!
A ball is thrown straight up into the air with a velocity of 63. 7 m/s. What is the velocity after 3s?
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:
v₀ = initial velocityg = gravityt = timevf= final velocityInformation about the problem:
g = 9.8 m/s²v₀ = 63. 7 m/st = 3 svf =?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
What is vertical launch upwards?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 micrometer 1 is often called the micron , a) how many microns make up 1.8km ? b) how many centimeters equal 1.8 ? c) how many microns are in 1.8 yd ?
By unit conversion, the converted unit is
a. 1.8 km = 1.8 x 10⁹ μm
b. 1.8 μm = 1.8 x 10¯⁴ cm
c. 1.8 yard = 1645920 μm
We need to know about unit conversion to solve this problem. The unit conversion can be used to convert a unit to another unit. It can be defined as
a = xb
where a is unit a, b is unit b and x is the constant of conversion.
From the question above, we know that
a. 1.8 km = .... μm
b. 1.8 μm = .... cm
c. 1.8 yard = .... μm
Find the microns (1 km = 10⁹ μm)
1.8 km = 1.8 x 10⁹ μm
Find the centimeter (1 cm = 10⁴ μm)
1cm = 10⁴ μm
1/10⁴ cm = 1 μm
Hence,
1.8 μm = 1.8/10⁴ cm
1.8 μm = 1.8 x 10¯⁴ cm
Find the microns (1 yd = 914400 μm)
1.8 yard = 1.8 x 914400 μm
1.8 yard = 1645920 μm
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hurriquake nails help homes better withstand the damages of both earthquakes and hurricanes. the cause is largely because the larger heads they have enable them to resist 120 kilograms of uplift force. the nails dimensions are 6.4 cm x 0.3 cm. if the nail dimensions were expressed in meters, you would have to divide each number by .
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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Give a physical argument that shows it is impossible to accelerate an object of mass m to the speed of light, even with a continuous force acting on it.
It is not possible to accelerate an object to the speed of light.
What is 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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Review. For a certain type of steel, stress is always proportional to strain with Young's modulus 20 × 10¹⁰ N/m² . The steel has density 7.86× 10³kg / m³. It will fail by bending permanently if subjected to compressive stress greater than its yield strength бy = 400MPa. A .rod 80.0cm long, made
of this steel, is fired at 12.0 m/s straight at a very hard wall.(e) the stress in the rod.
[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.
What is Stress?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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As the people sing in church, the sound level everywhere inside is 101 dB . No sound is transmitted through the massive walls, but all the windows and doors are open on a summer morning. Their total area is 22.0 m² . (b) Suppose the ground is a good reflector and sound radiates from the church uniformly in all horizontal and upward directions. Find the sound level 1.00 km away.
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
What do you mean by sound radiates?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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(i) What happens to the magnitude of the magnetic field inside a long solenoid if the current is doubled? (a) It becomes four times larger. (b) It becomes twice as large. (c) It is unchanged. (d) It becomes one-half as large. (e) It becomes one-fourth as large.
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."
What is magnetic field made of?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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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
The position of the wire is at a distance of 12 cm, in the direction left of the wire.
The force per unit length between two parallel thin current-carrying [tex]I_1[/tex] and [tex]I_2[/tex] wires at distance ' r ' is given by [tex]f=\frac{u_0I_1I_2}{2\pi r}[/tex] ....(1) . If the current is flowing in both wires in the same direction, and the force between them will be the attractive force and if the current is flowing in opposite direction in wires then the force between them will be the repulsive force.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
A speeding car passes a highway patrol checkpoint, then decelerates at a constant rate. After 5 s, the car is 225 m from the checkpoint, and its speed is then 30 m/s. What was the car’s velocity when it passed the checkpoint?.
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]
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. (a) Is this situation possible? Is it possible in more than one way? Describe
Yes, this situation is possible. No, It is not possible in more than one way.
The force per unit length between two parallel thin current-carrying [tex]I_1[/tex] and [tex]I_2[/tex] wires at distance ' r ' is given by [tex]f=\frac{u_0I_1I_2}{2\pi r}[/tex] ....(1) If the current is flowing in both wires in the same direction, and the force between them will be the attractive force and if the current is flowing in opposite direction in wires then the force between them will be the repulsive force.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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a group of students launches a model rocket in the vertical direction. based on tracking data, they determine that the altitude of the rocket was 89.6 ft at the end of the powered portion of the flight and that the rocket landed 17 s later. the descent parachute failed to deploy so that the rocket fell freely to the ground after reaching its maximum altitude. assume that g
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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Acceleration = 2.8 m/s^2
If after take-off, the jet continues to accelerate at the same rate for another
15 s, how fast will it be going at that time?
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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On a velocity-time graph, when is the object not moving?
when the slope is a straight line rising to the right
when the slope is a straight line rising to the right
at the point in which the line crosses the x-axis
at the point in which the line crosses the x -axis
when the slope is a straight line falling to the right
when the slope is a straight line falling to the right
when the slope is a line curving upward to the right
when the slope is a line curving upward to the right
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.
What is the velocity time graph?The velocity time graph can be used to observe the movement of a particle. The graph consist of three main parts;
Uniform accelerationConstant velocityUniform decelerationWe 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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