The analogous formula for magnetic fields is the Ampere's law.
To find the answer, we need to know about the Ampere's law of magnetism.
What's Ampere's law of magnetism?Ampere's law states that the close line integral of magnetic field around a current carrying loop is directly proportional to the current enclosed within it.
What's is the mathematical expression of Ampere's law?Mathematically, Ampere's law is
B•dl= μ₀I
Thus, we can conclude that the analogous formula for gauss law is the Ampere's law in magnetism.
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What are the sign and magnitude in coulombs of a point charge that produces a potential of -3.50 V at a distance of 2.00 mm
The magnitude and sign of the charge are 0.8 MC and negative respectively.
To find the answer, we need to know about the electric potential of a point charge.
What's the mathematical expression of the electric potential of a point charge?Mathematically, the electric potential at a distance 'r' from a point charge 'q' is given as (Kq)/r. K is the electrical constant with value 9×10^(-9) in vaccum. What will be the magnitude and sign of a charge, if potential is -3.50V at 2mm?From the expression of electric potential, charge is
q= (potential ×r)/K
= (-3.5×0.002)/ (9×10^(-9))
= -0.8 mega coulomb.
Thus, we can conclude that the magnitude and sign of the charge are 0.8 MC and negative respectively.
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Suppose you put very large-diameter tires on your car. Then, your speedometer will show a speed that is
Answer:
The speed registered by the speedometer depends on the rotations of the axle to which it is connected -
Larger diameter tires will contribute to a larger distance traveled in one rotation of the axle and hence a larger speed and the speed shown on the speedometer will be less than the actual speed.
Mass of empty density bottle - 43.2g Mass of density bottle full of water 66.4g Mass of density bottle with some sand - 67.5g Filled up with water 82.3g he above data to determine the: Mass of the water that completely filled the bott Volume of water that completely filled the bottl Volume of the density bottle: Mass of sand Mass of water that filled the space above the sa Volume of the sand: Mass of empty density bottle - 43.2g Mass of density bottle full of water 66.4g Mass of density bottle with some sand - 67.5g Filled up with water 82.3g he above data to determine the : Mass of the water that completely filled the bott Volume of water that completely filled the bottl Volume of the density bottle : Mass of sand Mass of water that filled the space above the sand
Volume of the sand :
The values of the density, volume and masses are:
mass of water that completely filled the bottle = 23.2 gvolume of water that completely filled the bottle = 23.2 cm³volume of the density bottle = 23.2 cm³mass of sand = 24.4 gmass of water that filled the space above the sand = 14.8 gvolume of sand = 8.4 cm³density of sand = 2.9 g/cm³What is density?Density is defined as the ratio of mass and volume of a substance.
Density = Mass/volumeThe given data is as follows:
Mass of empty density bottle = 43.2 g
Mass of density bottle full of water = 66.4 g
Mass of density bottle with some sand = 67.5 g
Mass of density bottle with the sand
filled up with water = 82.3 g:
(a) mass of water that completely filled the bottle = 66.4 - 43.2
mass of water that completely filled the bottle = 23.2 g
(b) volume of water that completely filled the bottle = mass/density
density of water = 1 g/cm³
volume of water that completely filled the bottle = 23.3 g/1 g/cm³
volume of water that completely filled the bottle = 23.2 cm³
(c) volume of the density bottle = 23.2 cm³
(d) mass of sand = 67.5 g - 43.2 g = 24.4 g
(e) mass of water that filled the space above the sand = 82.3 - 67.5 = 14.8 g
(f) volume of sand = volume of bottle - volume of water
volume of water = 14.8 cm³
volume of sand = 23.2 cm³ - 14.8 cm³
volume of sand = 8.4 cm³
(g) density of sand = 24.4 g/8.4 cm³ = 2.9 g/cm³
Therefore, the density of sand is obtained by taking the ratio of the mass of the sand and the volume if the sand.
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what are the wave model and particle model of light
Answer:
Newton's corpuscular theory stated that light consisted of particles that travelled in straight lines. Huygens argued that if light were made of particles, when light beams crossed, the particles would collide and cancel each other. He proposed that light was a wave.
If the mass of the earth and all objects on it were suddenly doubled, but the size remained the same, the acceleration due to gravity at the surface would become.
When the mass of Earth doubles, acceleration due to gravity doubles as well.
Acceleration due to gravity
Apply Newton's second law of motion;
F = mg --- (1)
where;
m is mass of the objectg is acceleration due to gravityF = GmM/R² --- (2)
where;
M is mass of EarthR is radius of EarthSolve (1) and (2)
mg = GmM/R²
g = GM/R²
when the mass of Earth doubles, acceleration due to gravity becomes;
g' = G(2M)/R²
g' = 2(GM/R²)
g' = 2g
Thus, when the mass of Earth doubles, acceleration due to gravity doubles as well.
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A resistor dissipates 2.00 W when the rms voltage of the emf is 10.0 V. At what voltage will the resistor dissipate 10.0 W
50 V to dissipate a power of 10 W
Given
A resistor dissipates 2.00 W when the rms voltage of the emf is 10.0 V.
Resistor and Voltage The difference in charge between two places is known as voltage. The passage of charge is measured as current. A material's propensity to oppose the flow of charge is known as resistance (current).Ohm's Law describes the relationship between current, voltage, and resistance. This asserts that, given the temperature is constant, the current flowing in a circuit is directly proportional to the applied voltage and inversely proportional to the circuit resistance.Explanation[tex]$\mathrm{V}_{r m s}=\mathrm{P}$[/tex]
10 v provides 2 W of power whereas V provides 10 W of power.
10 V = 2 W
x V=10 W
By using cross multiplication
2 x = 100
x = 50 V
So, 50 V to dissipate a power of 10 W
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The work that is done when twice the load is lifted twice the distance is _______. View Available Hint(s)for Part A The work that is done when twice the load is lifted twice the distance is _______. the same three times as much twice as much four times as much
The work that is done when twice the load is lifted twice the distance is
four times as much
The net work performed by forces acting on an object equals the change in kinetic energy, according to the work-energy theorem.
when an item slows down, the net work applied to it decreases, its change in kinetic energy is negative, and its ultimate kinetic energy is less than its starting kinetic energy. When an item accelerates, positive net work is done on it. All the forces acting on an item must be taken into consideration when determining the net work. You will obtain an incorrect result if you exclude any forces that affect an item or if you add any forces that do not affect it.
Hence The work that is done when twice the load is lifted twice the distance is four times as much
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A dog running at 10 m/s is 30m behind a rabbit moving at 5 m/s. when will the dog catch up with the rabbit assuming both their velocities remain constant during the chase?
The will dog catch up with the rabbit in 6 minutes assuming both their velocities remain constant during the chase.
What time will the dog catch the rabbit?The time that the dog will catch up with the rabbit is given as follows:
Let the distance covered by the rabbit be x.
Distance covered by dog = x + 30
Time taken = distance/speedThe time taken will be the same T
Time taken by dog, T = (x + 30)/10Time taken by rabbit, T = x/5Equating both times.
(x + 30)/10 = x/5
x = 30 m
Solving for T in equation (ii);
T = 30/5 = 6 minutes
In conclusion, time is obtained as a ratio of distance and speed.
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A container of PS5s with a mass of 451kg is loaded onto a Walmart truck using a ramp. The ramp is 6.12m long and
the bed of the truck is 1.53m above the ground. A force of 2025N applied parallel to the ramp moves the precious
consoles at a constant speed up the ramp. Find the efficiency of the ramp.
a) 32%
Ob) 54.65 %
O c) 46.4%
d) 54.6 %
D. The efficiency of the ramp is 54.6 %.
Velocity ratio of the rampV.R = distance moved by effort/distance moved by load = L/h
V.R = (6.12)/(1.53) = 4
Mechanical advantage of the rampM.A = Load/Effort
M.A = (451 x 9.8)/(2025)
M.A = 2.183
Efficiency of the rampE = (M.A / V.R) x 100%
E = (2.183 / 4) x 100%
E = 54.6 %
Thus, the efficiency of the ramp is 54.6 %.
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What is the gravitational potential energy stored in a rock of 100 kg
mass, placed on top of a hill, which is 100 m high?
O 45 kJ
O 98 kJ
O 50 kJ
O 100 kJ
A positively charged insulating rod is brought close to an object that is suspended by a string. If the object is repelled away from the rod we can conclude:
Answer: A positively charged insulating rod is brought close to an object that is suspended by a string. If the object is repelled away from the rod, we can conclude that the object is positively charged.
Explanation: To find the answer, we need to know more about the Electric charges and its properties.
What is meant by electric charge?It can be defined as the basic fundamental property of matter causes to experience a force, when it is placed in an external electric field.The charge of the particle can be positive, negative, and zero.What are the fundamental properties of electric charge?Additivity of chargesConservation of chargesQuantization of charges.Like charges repels and unlike charges attract.It's a scalar quantity.Thus, from the above, we can conclude that, when a positively charged insulating rod is brought close to an object that is suspended by a string. If the object is repelled away from the rod, we can conclude that the object is positively charged.
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Answer:
positively charged
Explanation:
when a positively charged object is being repelled, that means the charge should be the same for both.
A mass is placed on a smooth inclined plane with an angle of 37e to the horizontal. If theinclined plane is 5.0-m long, how long does it take for the mass to reach the bottom of theinclined plane after it is released from rest?
The time taken for the mass to reach the bottom of the inclined plane after it is released from rest is 0.78 s.
Height of the inclined plane
The height of the inclined plane is calculated as follows;
h = L(sin 37)
where;
h is height of the planeL is length of the planeh = 5 x sin(37)
h = 3.01 m
Time of motion of the masst = √(2h/g)
t = √(2 x 3.01 / 9.8)
t = 0.78 s
Thus, the time taken for the mass to reach the bottom of the inclined plane after it is released from rest is 0.78 s.
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PLEASE HELP! An airplane is moving at a velocity of 90.1 m/s and requires 1435 m to reach that velocity. Determine the acceleration of the plane and the time required to reach this velocity.
The time required to reach the velocity is 15.9s.
The acceleration of the plane is 5.67m/s²
Explanation:Acceleration is solved by the given formula.
Acceleration = velocity ÷ Time
Finding Time
Time = distance ÷ speed
Time = 1435m ÷ 90.1m/s = 15.9s
Finding Acceleration
Acceleration = velocity ÷ time
Acceleration = 90.1m/s ÷ 15.9s = 5.67m/s²
The law of combining volumes only applies under these conditions EXCEPT :_________.
i. the pressure must be the same. ii. the temperature must be the same. iii. the reactants must be liquids. iv. all reactants must be gases.
Answer:
measured at constant temperature and pressure
Hope this helps :)
The table shows the percentage of carbon dioxide in the Earth’s atmosphere in the years 1800 and 2013. Calculate the difference in the mass of carbon dioxide in 500 kg of air in 2013 compared to 1800. Select the correct answer.
The difference in the mass of carbon dioxide in 500 kg of air in 2013 compared to 1800 is 0.06 Kg
Data obtained from the questionYear 1800 percent = 0.028%Year 2013 percent = 0.040%Mass of air = 500 KgDifference =?How to determine the mass of CO₂ in 500 Kg in year 1800Year 1800 percent = 0.028% Mass of air = 500 KgMass of CO₂ =?Mass = percent × mass of air
Mass of CO₂ = 0.028% × 500
Mass of CO₂ = 0.14 Kg
How to determine the mass of CO₂ in 500 Kg in year 2013Year 1800 percent = 0.040% Mass of air = 500 KgMass of CO₂ =?Mass = percent × mass of air
Mass of CO₂ = 0.040% × 500
Mass of CO₂ = 0.2 Kg
How to determine the differenceMass of CO₂ in year 1800 = 0.14 KgMass of CO₂ in year 2013 = 0.2 KgDifference =?Difference = mass in 2013 - mass in 1800
Difference = 0.2 - 0.14
Difference = 0.06 Kg
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A duck flies in a straight line for 500m (60° N of W)
a.) What angle is this in polar coordinates?
b.) What is the North/South and East/West components of its displacement?
Please help me t-t
(A) The angle of this in polar coordinates is 30⁰.
(B) The North/South and East/West components of its displacement is 250 m and 433 m respectively.
The angle is this in polar coordinates60° N of W lies in the fourth quadrant;
θ = 90 - 60 = 30⁰
in polar coordinate = (r,θ) = (500 m, 30⁰)
Vertical component of the displacement (North/South)Dy = D sinθ
Dy = 500 m x sin(30) = 250 m
Horizontal component of the displacement (East/West)Dx = D cosθ
Dx = 500 m x cos(30) = 433 m
Thus, the angle of this in polar coordinates is 30⁰.
The North/South and East/West components of its displacement is 250 m and 433 m respectively.
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A vehicle comes to a stop 10.0 s after the brakes are applied. While the brakes were applied, the vehicle travelled a distance of
75.0 m. If the vehicle's acceleration remained constant during the braking, what was the vehicle's initial speed?
Since acceleration is constant, the average and instantaneous accelerations are the same, so that
[tex]a = a_{\rm ave} = \dfrac{\Delta v}{\Delta t} = -\dfrac{v_i}{10.0\,\rm s}[/tex]
By the same token, we have the kinematic relation
[tex]v^2 - {v_i}^2 = 2a\Delta x[/tex]
where [tex]v[/tex] is final speed, [tex]v_i[/tex] is initial speed, [tex]a[/tex] is acceleration, and [tex]\Delta x[/tex] is displacement.
Substitute everything you know and solve for [tex]v_i[/tex] :
[tex]0^2 - {v_i}^2 = 2\left(-\dfrac{v_i}{10.0\,\rm s}\right)(75.0\,\mathrm m)[/tex]
[tex]\implies {v_i}^2 - \left(15.0\dfrac{\rm m}{\rm s}\right) v_i = 0[/tex]
[tex]\implies v_i \left(v_i - 15.0\dfrac{\rm m}{\rm s}\right) = 0[/tex]
[tex]\implies v_i = \boxed{15.0\dfrac{\rm m}{\rm s}}[/tex]
Calculate the speed of sound in air when the temperature is 20°C. 
A footballer kicks a ball at angle teta to horizontal with some initial velocity a) what are the two types of independent motion takes place at the same time? b.how long is the ball in the air?
a) The motion along the vertical direction and the motion along the horizontal direction.
b) The object remains in the air for a time period of 2usin(θ)/g.
Any object that is thrown in the air when gravity is acting on it is called a projectile. The motion of this projectile is called projectile motion.
When the projectile is thrown in the air at some angle θ, then there are two independent motions taking place at the same time. First is the component of motion along the vertical direction along which gravity acts. Second is the component of motion along the horizontal direction along which the object moves with a constant velocity. No force acts along the horizontal direction. The horizontal motion does not affect the vertical motion and the converse is also true. So these are independent of each other.
The time of flight is the time during which a projectile remains in the air. This time of flight is calculated using the formula,
T=2usin(θ)/g
where T is the time of flight, u is the initial velocity and g is the acceleration due to gravity.
Hence, the object remains in the air for a time period of 2usin(θ)/g.
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The figure below shows a ball on a curved surface. The ball is released at point A. At which point does the ball have maximum gravitational potential energy?
Answer:
Point A
Explanation:
Because it reaches maximum height
1. An object moving with velocity 15 ms accelerates at 2 m/s for 5 seconds. Calculate the final velocity.
Answer:
final velocity=u+at 15+2×5=25
You hear three beats per second when two sounds tones are generated. The frequency of one tone is known to be 610 Hz. The frequency of the other is:
The frequency of the other wave is 613 Hz or 607 Hz.
The difference between the frequencies of two waves is called the beat frequency.
Here, one wave has a frequency 610 Hz and the beat frequency is 3 beats per second.
Which has a higher frequency is not mentioned. Therefore, there are two possibilities.
Δf = | 610 - 613 | = 3
or
Δf = | 610 - 607 | = 3
Therefore, the frequency of the other wave is 613 Hz or 607 Hz.
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A car approaches a stationary police car at 36 m/s. The frequency of the siren (relative to the police car) is 500 Hz. What is the frequency (in Hz) heard by an observer in the moving car as he approaches the police car
The frequency (in Hz) heard by an observer in the moving car as he approaches the police car is 552 Hz
The Doppler Effect equation says
f = f0 c+v / c
Given
f0 = 500 Hz
c = 343 m/s
v = 36 m/s
by plugging the given data, we get
500 (343+36) / 343 = 552 Hz
The frequency (in Hz) heard by an observer in the moving car as he approaches the police car is 552 Hz
Doppler effect is the apparent difference between the frequency at which sound or light waves leave a source and that at which they reach an observer, caused by the relative motion of the observer and the wave source.
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The positron created in pair production travels a certain distance and loses all of its kinetic energy. It finally annihilates with an electron and releases two photons. What is the angular separation between the two photons
Answer:
In pair production, after the loss of Kinetic energy, the angular separation between the two photons is 180°.
Explanation:
Pair production is the process of formation of two electrons, one negative and the other positive (positron), from a pulse of electromagnetic energy traveling through matter.It is a process of direct conversion of radiant energy to matter.The sum of the Kinetic energies of the formed particles amounts to a value of 4 MeV.When the kinetic energy is lost, emission of two photons, each with an energy of approximately 1 MeV in the form of gamma rays takes place ( in opposite direction).
Therefore, the angle of separation between the two photons is 180°.
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Why is the potential difference higher between any two points on a single E-field line than between two points the same distance apart on different E-field lines
While moving over a separate electric field line, however, has a non-zero value. If the test charge is transferred in a direction perpendicular to E, no work is done and the electric potential does not change. A smooth equipotential surface characterizes every such path.
Why is the potential difference higher between any two points on a single E-field line than between two points the same distance apart on different E-field lines?Generally, The equation of the potential difference is mathematically given as
[tex]V = Ed cos \theta.[/tex]
Where
[tex]\theta=Angle[/tex]
E=electric field
d= the varing distance
When the charge to be tested is transferred via an electric field line. Since =0, the maximum possible velocity (Vmax) is calculated.
In conclusion, Since no work is done when the test charge is shifted in a direction perpendicular to E, the electric potential does not shift in this case. Any such route is a smooth equipotential surface.
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Which hrase describes what atoms are made of? OA. A cloud of protons and neutrons surrounded by a nucleus of electrons OB. A nucleus of neutrons and electrons surrounded by a cloud of protons OC. A nucleus of protons and neutrons surrounded by a cloud of electrons OD. A nucleus of protons and electrons surrounded by a cloud of neutrons
Answer:
(OC) a nucleus of protons and neutrons surrounded by a cloud of electrons - modern description of an atom - where the nucleus may be hundreds of times smaller in diameter than the cloud of electrons
Which class of hazards is characterized by thermal and mechanical hazards in the form of blast pressure waves, shrapnel and fragmentation, and incendiary thermal effects
Class 1 (Explosives) is the class of hazards that is characterized by thermal and mechanical hazards in the form of blast pressure waves, shrapnel and fragmentation, and incendiary thermal effects.
There are different classes of Hazards
Class 1 - Explosives
Class 2 - Gases
Class 3 - Flammable liquids
Class 4 - Flammable solids
Class 5 - Oxidizers
Class 6 - Toxic materials
Class 7 - Radioactive materials
Class 8 - Corrosive materials
Class 9 - Miscellaneous dangerous goods
Any substance or item, including a gadget, that is intended to function by explosion or that, through a chemical reaction inside itself, is capable of functioning similarly even if not intended to function by explosion, falls within the category of explosive materials (class 1).
Hence, Class 1 (Explosives) is the class of hazards that is characterized by thermal and mechanical hazards.
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Dinesh takes 20 minutes to cover a distance of 5 km on a bicycle. Calculate his average speed in.
A) 1m/s
B) 2km/h
Please explain in detail.
Answer:
Soln:Here
Given,
Distance(s)=5m
Time taken(t)=20min=2400seconds
Speed(s)=?
We know that,
Speed(s)=Distance/timetaken
=5m/2400sec
580m/s
Thus the average speed of his bicycle is 580 m/s..
Thank you...
The speed that a tsunami can travel is modeled by the equation , where s is the speed in kilometers per hour and d is the average depth of the water in kilometers. What is the approximate depth of water for a tsunami traveling at 200 kilometers per hour?.
The speed of tsunami is a.0.32 km.
Steps involved :
The equation s = 356d models the maximum speed that a tsunami can move at. It reads as follows: s = 200 km/h d =?
Let's now change s to s in the equation to determine d: s = 356√d 200 = 356√d √d = 200 ÷ 356 √d = 0.562 Let's square the equation now by squaring both sides: (√d)² = (0.562) ² d = (0.562)² = 0.316 ≈ 0.32
As a result, 0.32 km is roughly the depth (d) of water for a tsunami moving at 200 km/h.
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If an electron is released at PP , what is the magnitude of the net force that these rods exert on it?
The magnitude of the net force that the rods exert after an electron is released at point P is 2.885 × 10⁻¹⁵ N.
Given values:
Length of non-conducting rod, l = 1.20 m
Charge on positive rod, +Q = +2.50 μC = +2.50 × 10⁻⁶ C
Charge on negative rod, -Q = -2.50 μC = -2.50 × 10⁻⁶ C
Distance from point P of each rod, x = 60 cm = 0.60 m
Calculation of Net electric force exerted on point P:
Consider an electron released at point P, then the net electric force exerted will be given as:
F = e. E_net - ( 1 )
Step 1:
The net electric field value is given as:
E_net = E₁ cos Φ + E₂ cos Φ
= 2E₁ cos Φ -( 2 )
where, E₁ & E₂ are electric fields due to positive and negative rod
respectively.
Φ is phase angle
Step 2:
The electric field due to positive rod is given as:
E₁ = k (λ/r) - ( 3 )
where, k is Coulomb's force constant
λ is linear charge density
r is distance between point P and half of the rod.
Now, the linear charge density is given as:
λ = Charge/length = Q/x
The value of r is given as:
r = √x²-a²
where, x is length of rod
a is half length of rod
Applying values in above equation, we get:
r = √x²-(x/2)²
r = √(1.20 m)²-(1.20/2)²
= √1.08
= 1.04 m
Substituting all the determined values in equation 3 we get:
E₁ = k (λ/r)
= k [(Q/x)/r]
= k [ Q/xr ]
= (9×10⁹ Nm²/C²) [ |+2.50×10⁻⁶ C|/(1.20 m)(1.04 m)]
= 1.803×10⁴ N/C
Step 3:
Similarly, the electric field due to negative rod is given as:
E₂ = k [ Q/xr ]
= (9×10⁹ Nm²/C²) [ |-2.50×10⁻⁶ C|/(1.20 m)(1.04 m)]
= 1.803×10⁴ N/C
Step 4:
Consider equation 2:
E_net = 2E₁ cos Φ
From the figure we get the phase angle as:
Φ = tan⁻¹ (0.60 m/0.60 m)
= tan⁻¹ ( 1 )
= π/4
Now, the electric field produced due to each rod is equal and mutually perpendicular. Thus, the net electric field after applying values can be calculated as:
E_net = 2(1.803×10⁴ N/C) cos π/4
= 2(1.803×10⁴ N/C) (0.5)
= 18030 N/C
Step 5:
Consider equation 1 :
F = e. E_net
where, e is charge on an electron
Applying values in above equation we get:
F = (1.6 × 10⁻¹⁹ C)(18030 N/C)
= 2.885 × 10⁻¹⁵ N
Therefore, the magnitude of the net force that the rods exert after an electron is released at point P is 2.885 × 10⁻¹⁵ N.
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