explain why cooking food with pressure pot is faster than cooking with an ordinary pot​

Using Stefan's equation, which states that the power of the emitted radiation will be proportional to the fourth power of the absolute temperature, we treat the filament of the evacuated light bulb as a black body.

P = σ AeT⁴

Where A is the surface area of the emitter, σ is Stefan's constant, and e is the emissivity of the surface.

Now, from Stefan's Law, the radiated power is proportional to the fourth power of temperature.

P =IV

IV ∝ T⁴

Therefore,

I₁ V₁ / (I₂ V₂) = (T₁ / T₂)⁴

T₁ / T₂ = (I₁ V₁ / I₂ V₂)^{1/4}

T₂ = T₂ [ (I₁ V₁ / I₂ V₂)^{1/4} ]

Now we have,

I₂ = 2I₁ and

V₂ = 8V₁

Therefore,

T₂ = T₁ [ (2I₁ )(8V₁)/(I₁ V₁) ]^{1/4}

T₂ = T₁ (16)^{1/4}

T₂ = 2T₁

The correct option is (d).

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Light source will pass through and form a shadow, when I hold the largest cardboard between the light source and the wall. Since, the cardboard is not so thick it will allow rays to pass through it.

Light rays always travel in a straight line. A shadow is formed when the light rays hit an opaque object. Sun is a natural source of light. The rays from the sun fall on objects on earth and forms shadows. Light rays do not interfere with each other if they come cross each other. Light rays travel from the direction of source to the direction of eye. They show property of reflection wherein after colliding a surface they trace their path back.

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The efficiency of moving the box is 60%.

Given in the question

Force exerted by peter = 500 N

Force due to friction on box = 200 N

Distance Traveled by box = 3 meters

Now work done by the force is given by

Work Done = (Magnitude of Force) × (Distance traveled)

Let us find work done by each force,

Work done by peter

Work Done = (Magnitude of Force) × (Distance traveled)

Put in the value, we get

Work Done =500 × 3

Work Done by peter = 1500 J

Similarly, Work done by friction

Work Done = (Magnitude of Force) × (Distance traveled)

Put in the value, we get

Work Done =200 × 3

Work Done by Friction = 600 J

As we know the work done by friction force is always negative,

Work Done by Friction = -600 J

So, net Work done = Work Done by peter + Work Done by Friction

net Work done = 1500 - 600

Net Work done = 900 J

Now Efficiency = (Net work done/work done by peter) × 100

Put in the values, we get

Efficiency = (900/1500) × 100

Efficiency = (3/5) × 100

Efficiency = 300/5

Efficiency = 60 %

So, the efficiency of moving the box is 60%.

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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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Plane used to remove the tip of your nose is known to be called option C. Frontal (front/back).

In regards to biology, If a person were to cut their nose off, it is one that is known to be called coronal.

The coronal plane is said to be a term that is often referred to as the frontal plane and it is seen as any vertical plane that shares the body into ventral as well as dorsal (belly and back) parts.

Therefore one can say that the coronal Plane (Frontal plane) used to remove the tip of your nose.

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The ideal selection is C. The spheres will each grow evenly.

The coefficient of expansion of any material is the property of the material which defines the expansion or contraction of the material with the variation of the temperature.

The ratio of a body's length, area, or volume at a given temperature, usually 0° C, to its length, area, or volume at that temperature, respectively, while maintaining constant pressure.

Given that two spheres with the same radius and formed of the same metal are hollow and solid, respectively. The same temperature rise is applied to the spheres.

Since both bodies in this instance are formed of the same substance, their orientation has no bearing on how much it expands. Because the materials used to create both spheres were the same, we can thus conclude that they will expand equally.

Therefore, both materials will expand equally.

The complete options are given below:-

A Hollow sphere will expand more than the solid one.

B Solid sphere will expand more than the hollow one.

C Both spheres will expand equally.

D None of these

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[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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The gauge pressure at the bottom of the second cylinder is 0.10 atm.

The pressure of a system above the atmospheric pressure is known as gauge pressure, commonly referred to as overpressure. The pressure from the weight of the atmosphere is included in gauge pressure readings because gauge pressure is zero-referenced against ambient air (or atmospheric) pressure. This indicates that gauge pressure varies in response to both weather and height above sea level. Gauge pressure measurement is adequate for the majority of industrial applications because every operation in a refinery or manufacturing facility operates at the same air pressure. Absolute pressure is a measurement of pressure that is based on a reference pressure of zero or absolutely no pressure. The only place where this happens naturally is in a perfect vacuum, which can only be found in space.

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The force per unit length exerted by I₁ on I₂ is [tex]8\times10^-^5 \ N/m[/tex] .

It is given that two long, parallel conductors, separated by 10.0cm having current  I₁ = 5.00A and  I₂ = 8.00 A.

If the current flowing through the wire are in same direction then they the wire attract each other and if the current flowing through the wire are opposite in direction then they the wire repel each other

Putting r = 10.0cm [tex]=10\times10^{-2}m[/tex] ,  I₁ = 5.00A and  I₂ = 8.00 A in equation (1) , we get

[tex]f=\frac{4\pi \times10^-^7\times5.00\times8.00}{2\times\pi\times10.00\times10^{-2}}\\\\f=\frac{2\times40\times10^-^5}{10} \\\\f=8\times10^{-5}N/m[/tex]

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The distance the proton traveled in the field is 2.09m.

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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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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Radio waves allows the charges in the receiving antenna to respond to the electric field portion of the carrier wave.

This refers to a type of disturbance which transports energy from one place to another without the actual movement of the particles and of various types such as  microwaves, radio waves etc.

The radio wave allows the electrons in the rod to vibrate with amplitudes which emulates those of the carrier wave and in this case, it is the antenna thereby resulting in the signals in the car radio being provided as a result of these type of interactions.

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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 velocity in meters per second of the object at t = 2 seconds is 2.5 m/s

Velocity is simply defined as the rate of change of displacement with time. Mathematically, it can be expressed as:

Velocity = displacement / time

From the question given above, the following data were obtained:

Velocity = displacement / time

Velocity = 5 / 2

Velocity = 2.5 m/s

Thus, the velocity is 2.5 m/s

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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:

The independent variable is the cause. Its value is independent of other variables in your study. The dependent variable is the effect. Its value depends on changes in the independent variable.

Explanation:

Hope this helps!

The total capacitance of the capacitor circuit is 40 μF.

We need to know about capacitor circuits to solve this problem. The capacitance will be summed up when the capacitor is combined in parallel. The total capacitance can be determined as

Ct = C1 + C2 + C3

From the question above, we know that:

C1 = 25 μF

C2 = 3 μF

C3 = 12 μF

By substituting the parameters, we can get

Ct = C1 + C2 + C3

Ct = 25 μF + 3 μF + 12 μF

Ct = 40 μF

Hence, the total capacitance of the circuit is 40 μF.

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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.

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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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

Kinetic energy is K = 1.82 × 10⁻³ eV.

⁵⁷Fe nucleus emits 14 keV photons at rest.

We have, Mc² = 8.6 × 10⁻⁹ J in the final state of the nucleus.

Let p(nucleus) be the moment of the nucleus and p(photon) be the momentum of the photon.

Then, by conservation of momentum since the total moment is zero.

p(nucleus) = p(photon) = E(γ) / c = 140 keV / c

The energy for the recoiling nucleus is given as:

E² = p²c² + (mc²)²

Now, Mc² = 8.60 × 10⁻⁹J =5.38 × 10¹⁰ eV = 5.38  × 10⁷ keV

Therefore,

(Mc² + K²) ² = (14 keV)² + (Mc²)²

(1 + K²/Mc²)² =(14/Mc²)² +1

The term (1 + K²/Mc²)² is less than 1  therefore, (14/Mc²)² + 1 is equal to 1 by clacualtor.

Now using binomial theorem,

(1 + K²/Mc²)² = √[1 + ((14/Mc²)²] = 1 + (1/2)(14/Mc²)²

K = 1.82 × 10⁻³ eV

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The magnitude of the minimum deceleration to avoid an accident is 2,186.5 m/s².

The minimum acceleration to avoid accident is calculated as follows;

s = ut + ¹/₂at²

where;

Substitute the given parameters and solve for the acceleration as follows;

350 = 12(0.56) +  ¹/₂(0.56²)a

350 = 6.72 + 0.157a

0.157a = 350 - 6.72

0.157a = 343.28

a = 343.28/0.157

a = 2,186.5 m/s²

Thus, the magnitude of the minimum deceleration to avoid an accident is 2,186.5 m/s².

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Answer: 5.4 hours.

Explanation:

So, this would simply be divison. So, we can take 540 km, and seperate it into 500 km, and 40 km.

That being said, we can divide 500 km by 100, so we can see how many hours it takes for 500 km, which is 5.

Now, we divide 40/100, which gives us 0.4.

Add the two together, and it takes 5.4 hours to reach 540 km at a speed of 100 km/hr

Answer:

Touch the open ends of the two wires to each other to form a circuit and test the bulb. Touch the two open ends of the wire to each material you are testing, one at a time. If the bulb lights up, it is a good conductor. If it doesn't, it is a good insulator.How do you know if an object is a good conductor and poor conductor of electricity?

Image result for Suppose you have found a piece of material. You are not sure what the material is. You want to find out if it is a good conductor or a good insulator. Describe in words what you would do to determine if the material is an electrical conductor.​

Difference between Good Conductor and Bad Conductor of Electricity. Good conductors are those materials which allow electricity to pass through them easily. Bad conductors are those materials which do not allow electricity to pass through them easily.

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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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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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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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

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.

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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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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