At what temperature will water begin to boil and turn to steam?
212 degrees Celsius
100 degrees Fahrenheit
212 kelvins
100 degrees Celsius

At What Temperature Will Water Begin To Boil And Turn To Steam?212 Degrees Celsius100 Degrees Fahrenheit

Answers

Answer 1

Answer:

100 degrees Celsius

Explanation:

Water starts to boil at 100 degrees celcius or 212 degrees fahrenheit.

Answer 2

At 100 degrees Celsius water begin to boil and turns to steam.

What are the boiling point and melting point of water?

The melting point for water is 0 degrees C (32 degrees F). The boiling point of water varies with atmospheric pressure. At lower pressure or higher altitudes, the boiling point is lower. At sea level, pure water boils at 212 °F (100°C).

Is boiling water always 212?

If the temperature is much above 212°F, the water will boil. That means that it won't just evaporate from the surface but will form vapor bubbles, which then grow, inside the liquid itself. If the water has very few dust flecks etc.

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Related Questions

A ball is thrown at 23.2 m/s inside a boxcar moving along the tracks at 34.9 m/s. What is the speed of the ball relative to the ground if the ball is thrown forward

Answers

Answer:

The speed of the ball relative to the ground if the ball is thrown forward is 58.1 m/s

Explanation:

Given;

speed of the ball thrown inside boxcar, [tex]V_B[/tex] = 23.2 m/s

speed of the boxcar moving along the tracks, [tex]V_T[/tex] = 34.9 m/s

Determine the speed of the ball relative to the ground if the ball is thrown forward.

If the ball is thrown forward, the speed of the ball relative to the ground will be sum of the ball's speed plus speed of the boxcar.

[tex]V_{relative \ speed} = V_B + V_T\\\\V_{relative \ speed} = 23.2 + 34.9\\\\V_{relative \ speed} = 58.1 \ m/s[/tex]

Therefore,  the speed of the ball relative to the ground if the ball is thrown forward is 58.1 m/s.

soaring birds and glider pilots can remain aloft for hours without expending power. Discuss why this is so.

Answers

Answer:

Since their wings and body develop the drag. When there is warm air then they expand their wings. Since,soaring birds and glider pilots have no engine, they always maintain their high speed to lift their weight in air for hours without expending power by convection

Explanation:

What explains why a prism separates white light into a light spectrum?
A. The white light, on encountering the prism, undergoes both reflection and refraction; some of the reflected rays re-enter the prism merging with refracted rays changing their frequencies.
B. The white light, on entering a prism, undergoes several internal reflections, forming different colors.
C. The different colors that make up a white light have different refractive indexes in glass.
D. The different colors that make up a white light are wavelengths that are invisible to the human eye until they pass through the prism.
E. The different rays of white light interfere in the prism, forming various colors.

Answers

Answer:

I think the answer probably be B

Answer :QUESTION①)

What explains why a prism separates white light into a light spectrum ?

C. The different colors that make up a white light have different refractive indexes in glass.

✔ Indeed, depending on the radiation (and therefore colors), which each have different wavelengths, the refraction index varies: the larger the wavelength (red) the less the reflection index is important and vice versa (purple).

✔ That's why purple is more deflected so is lower than red radiation.  

Two spherical objects at the same altitude move with identical velocities and experience the same drag force at a time t. If Object 1 has twice (2x) the diameter of Object 2, which object has the larger drag coefficient? Explain your answer using the drag equation.

Answers

Answer:

The object with the twice the area of the other object, will have the larger drag coefficient.

Explanation:

The equation for drag force is given as

[tex]F_{D} = \frac{1}{2}pu^{2} C_{D} A[/tex]

where [tex]F_{D}[/tex] IS the drag force on the object

p = density of the fluid through which the object moves

u = relative velocity of the object through the fluid

p = density of the fluid

[tex]C_{D}[/tex] = coefficient of drag

A = area of the object

Note that [tex]C_{D}[/tex] is a dimensionless coefficient related to the object's geometry and taking into account both skin friction and form drag. The most interesting things is that it is dependent on the linear dimension, which means that it will vary directly with the change in diameter of the fluid

The above equation can also be broken down as

[tex]F_{D}[/tex] ∝ [tex]P_{D}[/tex] A

where [tex]P_{D}[/tex] is the pressure exerted by the fluid on the area A

Also note that [tex]P_{D}[/tex] = [tex]\frac{1}{2}pu^{2}[/tex]

which also clarifies that the drag force is approximately proportional to the abject's area.

In this case, the object with the twice the area of the other object, will have the larger drag coefficient.

A sailor strikes the side of his ship just below the surface of the sea. He hears the echo of the wave reflected from the ocean floor directly below 2.5 ss later.
How deep is the ocean at this point? (Note: Use the bulk modulus method to determine the speed of sound in this fluid, rather than using a tabluated value.)
_____ m

Answers

Answer:

1248m

The time that wave moves from the wave source to the ocean floor is half the total travel time: t = 2.5/2 = 1.25s

The speed of sound in seawater is 1560 m/s

Therefore, s = vt = (1560 m/s)(1.25s) =1248 m = 1.2km

A physics professor wants to perform a lecture demonstration of Young's double-slit experiment for her class using the 633-nm light from a He-Ne laser. Because the lecture hall is very large, the interference pattern will be projected on a wall that is 4.0 m from the slits. For easy viewing by all students in the class, the professor wants the distance between the m
What slit separation is required in order to produce the desired interference pattern?
d=________m

Answers

Note: if the professor wants the distance between the m = 0 and m = 1 maxima to be 25 cm

Answer:

d = 1.0128×10⁻⁵m

Explanation:

given:

length L = 4.0m

maximum distance between m = 0 and m = 1 , y = 25cm = 0.25m

wavelength λ = 633nm = 633×10⁻⁹m

note:

dsinθ = mλ (constructive interference)

where d is slit seperation, θ is angle of seperation , m is order of interference , and λ is wavelength

for small angle

sinθ ≈ tanθ

[tex]d (\frac{y}{L}) =[/tex] mλ

[tex]d (\frac{y}{L}) = (1)(633nm)[/tex]

[tex]d(\frac{0.25}{4} ) = (1)(633nm)[/tex]

d = 1.0128×10⁻⁵m

A segment of wire of total length 3.0 m carries a 15-A current and is formed into a semicircle. Determine the magnitude of the magnetic field at the center of the circle along which the wire is placed.

Answers

Answer:

4.9x10^-6T

Explanation:

See attached file

. A 24-V battery is attached to a 3.0-mF capacitor and a 100-ohm resistor. If the capacitor is initially uncharged, what is the voltage across the capacitor 0.16 seconds after the circuit is connected to the battery

Answers

Answer:

The  voltage is [tex]V_c = 9.92 \ V[/tex]

Explanation:

From the question we are told that

     The voltage of the battery is  [tex]V_b = 24 \ V[/tex]

     The capacitance of the capacitor is  [tex]C = 3.0 mF = 3.0 *10^{-3} \ F[/tex]

     The  resistance of the resistor is [tex]R = 100\ \Omega[/tex]

     The time taken is  [tex]t = 0.16 \ s[/tex]  

Generally the voltage of a charging charging capacitor after time t is mathematically represented as

       [tex]V_c = V_o (1 - e^{- \frac{t}{RC} })[/tex]

Here [tex]V_o[/tex] is the voltage of the capacitor when it is fully charged which in the case of this question is equivalent to the voltage of the battery so  

      [tex]V_c = 24 (1 - e^{- \frac{0.16}{100 * 3.0 *10^{-1}} })[/tex]

      [tex]V_c = 9.92 \ V[/tex]

A wheel rotating about a fixed axis has a constant angular acceleration of 4.0 rad/s2. In a 4.0-s interval the wheel turns through an angle of 80 radians. Assuming the wheel started from rest, how long had it been in motion at the start of the 4.0-s interval

Answers

Answer:

The  time interval is  [tex]t = 3 \ s[/tex]

Explanation:

From the question we are told that

    The angular acceleration is  [tex]\alpha = 4.0 \ rad/s^2[/tex]

     The  time taken is  [tex]t = 4.0 \ s[/tex]

      The angular displacement is  [tex]\theta = 80 \ radians[/tex]

     

The angular displacement can be represented by the second equation of motion as shown below

          [tex]\theta = w_i t + \frac{1}{2} \alpha t^2[/tex]

where  [tex]w_i[/tex] is the initial velocity at the start of the 4 second interval

So substituting values

        [tex]80 = w_i * 4 + 0.5 * 4.0 * (4^2)[/tex]

=>    [tex]w_i = 12 \ rad/s[/tex]

Now considering this motion starting from the start point (that is rest ) we have

       [tex]w__{4.0 }} = w__{0}} + \alpha * t[/tex]

Where  [tex]w__{0}}[/tex] is the angular velocity at rest which is zero  and  [tex]w__{4}}[/tex] is the angular velocity after 4.0 second which is calculated as 12 rad/s s

        [tex]12 = 0 + 4 t[/tex]

=>       [tex]t = 3 \ s[/tex]

Following are the response to the given question:

Given:

[tex]\to \alpha = 4.0 \ \frac{rad}{s^2}\\\\[/tex]

[tex]\to \theta= 80\ radians\\\\\to t= 4.0 \ s\\\\ \to \theta_0=0\\[/tex]

To find:

[tex]\to \omega=?\\\\\to t=?\\\\[/tex]

Solution:

Using formula:

[tex]\to \theta- \theta_0 = w_{0} t+ \frac{1}{2} \alpha t^2\\\\ \to 80-0= \omega_{0}(4) + \frac{1}{2} (4)(4^2)\\\\ \to 80= \omega_{0}(4) + \frac{1}{2} (4)(16)\\\\\\to 80= \omega_{0}(4) + (4)(8)\\\\\to 80= \omega_{0}(4) + 32\\\\\to 80-32 = \omega_{0}(4) \\\\\to \omega_{0}(4)= 48 \\\\\to \omega_{0}= \frac{48}{4} \\\\ \to \omega_{0} = 12 \frac{rad}{ s} \\\\[/tex]  

It would be the angle for rotation at the start of the 4-second interval.

This duration can be estimated by leveraging the fact that the wheel begins from rest.  

[tex]\to \omega = \omega_{0} + \alpha t\\\\\to 12 = 0 +4(t) \\\\\to 12 = 4(t) \\\\ \to t=\frac{12}{4}\\\\\to t= 3\ s[/tex]

Therefore, the answer is "[tex]12\ \frac{rad}{s}[/tex] and [tex]3 \ s[/tex]".

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. If you live in a region that has a particular TV station, you can sometimes pick up some of its audio portion on your FM radio receiver. Explain how this is possible. Does it imply that TV audio is broadcast as FM

Answers

Answer:

Please see below as the answer is self-explanatory.

Explanation:

The low band of the VHF TV Spectrum, spans channels 2-6, from 54 to 88 Mhz.

In the analog TV, in the Americas, the total bandwidth of any channel is 6 Mhz, with the visual carrier modulated in VSS (Vestigial Side Band) at 1.25 Mhz from the lowest frequency of the channel.

The aural carrier is located at 4.5 Mhz from the visual carrier, and is FM modulated.

For Channel 6, which spans between 82 and  88 Mhz, the visual carrier is at 83.25 Mhz, so the aural carrier is at 87.75 Mhz, which falls within the FM Band, so it is possible to listen the audio part of this channel in a FM radio receiver, even at a lower volume, due to the FM radio has a greater deviation than TV aural carrier.

The reason why it is possible for TV station to sometimes pick up some of the audio portion on your FM radio receiver is because; TV waves can sometimes deviate into the FM radio frequency range.

Let us start with explaining the waves of TV and radio.

The frequency range utilized by TV stations is either the range 54 MHz to 88 MHz or 174 MHz to 222 MHz. In contrast, the frequency range utilized by FM Radio band is between 88 MHz and 174 MHz.

Now, in some cases, it is possible that the TV signal may deviate into the range of the FM Radio and as such in that case, the TV signal will pick the audio portion of an FM Radio. These TV waves are very high frequency waves.

Finally, it does not imply that the TV wave is broadcasting as an FM because it only deviated a bit from the TV range and not like that is where it is made to operate.

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If the current flowing through a circuit of constant resistance is doubled, the power dissipated by that circuit will

Answers

Assuming that the voltage is constant, and the resistance was doubled, since I=V/R, the current through the circuit will be halved. As P=IV, with the same voltage and halved current, the power dissipated by the circuit will be halved

Find the work done in pumping gasoline that weighs 6600 newtons per cubic meter. A cylindrical gasoline tank 3 meters in diameter and 6 meters long is carried on the back of a truck and is used to fuel tractors. The axis of the tank is horizontal. The opening on the tractor tank is 5 meters above the top of the tank in the truck. Find the work done in pumping the entire contents of the fuel tank into the tractor.

Answers

Answer:

work done in pumping the entire fuel is 1399761 J

Explanation:

weight per volume of the gasoline = 6600 N/m^3

diameter of the tank = 3 m

length of the tank = 6 m

The height of the tractor tank above the top of the tank = 5 m

The total volume of the fuel is gotten below

we know that the tank is cylindrical.

we assume that the fuel completely fills the tank.

therefore, the volume of a cylinder =  

where r = radius = diameter ÷ 2 = 3/2 = 1.5 m

volume of the cylinder = 3.142 x  x 6 = 42.417 m^3

we then proceed to find the total weight of the fuel in Newton

total weight = (weight per volume) x volume

total weight = 6600 x 42.417 = 279952.2 N

therefore,

the work done to pump the fuel through to the 5 m height = (total weight of the fuel) x (height through which the fuel is pumped)

work done in pumping = 279952.2 x 5 = 1399761 J

Two identical wooden barrels are fitted with long pipes extending out their tops. The pipe on the first barrel is 1 foot in diameter, and the pipe on the second barrel is only 1/2 inch in diameter. When the larger pipe is filled with water to a height of 20 feet, the barrel bursts. To burst the second barrel, will water have to be added to a height less than, equal to, or greater than 20 feet? Explain.

Answers

Answer:

The 1/2 inch barrel will burst at the same height of 20 ft

Explanation:

The pressure on a column of fluid increases with depth, and decreases with height. This means that if you increase the height of the fluid in the column, the pressure at the bottom will increase.

From the equation of fluid pressure,

P = ρgh

where

P is the pressure at the bottom of the fluid due to its height

ρ is the density of the fluid in question

h is the height to which the water stand.

You notice how apart from the height 'h' in the equation, all the other parts of the right hand side of the equation cannot be varied; they are a fixed property of the fluid and gravity. And there is no consideration for the horizontal diameter of the water's cross section area.

We can also think of the pressure at the bottom of the fluid to be as a result of an incremental weight of an infinitesimally small vertical section of the water down.

That been said, we can then say that if the barrel with the 1 ft diameter dimension bursts when filled with water up to 20 ft, then, the barrel with the reduced diameter will still burst at the same height as the former pipe.

NB: The only way to stop the pipe from bursting is to increase the thickness of the barrel wall to counteract the pressure forces due to the height.

A charge is placed on a spherical conductor of radius r1. This sphere is then connected to a distant sphere of radius r2 (not equal to r1) by a conducting wire. After the charges on the spheres are in equilibrium:__________.
1. the electric fields at the surfaces of the two spheres are equal.
2. the amount of charge on each sphere is q/2.
3. both spheres are at the same potential. the potentials are in the ratio V2/V1 = q2/q1.
4. the potentials are in the ratio V2/V1 = r2/r1 .

Answers

Answer:

Option 3 = both spheres are at the same potential.

Explanation:

So, let us complete or fill the missing gap in the question above;

" A charge is placed on a spherical conductor of radius r1. This sphere is then connected to a distant sphere of radius r2 (not equal to r1) by a conducting wire. After the charges on the spheres are in equilibrium BOTH SPHERES ARE AT THE SAME POTENTIAL"

The reason both spheres are at the same potential after the charges on the spheres are in equilibrium is given below:

=> So, if we take a look at the Question again, the kind of connection described in the question above (that is a charged sphere, say X is connected another charged sphere, say Y by a conducting wire) will eventually cause the movement of charges(which initially are not of the same potential) from X to Y and from Y to X and this will continue until both spheres are at the same potential.

An electron moves to the left along the plane of the page, while a uniform magnetic field points into the page. What direction does the force act on the moving electron

Answers

Answer:

acting force is the answer

The direction of the magnetic force on the moving electron is upward.

The direction of the magnetic force on the electron can be determined by applying right hand rule.

This rule states that when the thumb is held perpendicular to the fingers, the thumb will point in the direction of the speed while the fingers will point in the direction of the field and the magnetic force will be perpendicular to the field.

Thus, we can conclude that, the direction of the magnetic force on the moving electron is upward.

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A flat loop of wire consisting of a single turn of cross-sectional area 8.20 cm2 is perpendicular to a magnetic field that increases uniformly in magnitude from 0.500 T to 2.60 T in 1.02 s. What is the resulting induced current if the loop has a resistance of 2.70

Answers

Answer:

The  induced current is [tex]I = 6.25*10^{-4} \ A[/tex]

Explanation:

From the question we are told that  

    The number of turns is  [tex]N = 1[/tex]

     The  cross-sectional area is  [tex]A = 8.20 cm^2 = 8.20 * 10^{-4} \ m^2[/tex]

    The  initial magnetic field is  [tex]B_i = 0.500 \ T[/tex]

     The  magnetic field at time =  1.02 s  is  [tex]B_t = 2.60 \ T[/tex]

     The  resistance is  [tex]R = 2.70\ \Omega[/tex]

The  induced emf is mathematically represented as

       [tex]\epsilon = - N * \frac{ d\phi }{dt}[/tex]

The  negative sign tells us that the induced emf is moving opposite to the change in magnetic flux

      Here  [tex]d\phi[/tex] is the change in magnetic flux which is mathematically represented as

        [tex]d \phi = dB * A[/tex]

Where  dB  is the change in magnetic field which is mathematically represented as

        [tex]dB = B_t - B_i[/tex]

substituting values

        [tex]dB = 2.60 - 0.500[/tex]

        [tex]dB = 2.1 \ T[/tex]

Thus  

      [tex]d \phi = 2.1 * 8.20 *10^{-4}[/tex]

     [tex]d \phi = 1.722*10^{-3} \ weber[/tex]

So  

     [tex]|\epsilon| = 1 * \frac{ 1.722*10^{-3}}{1.02}[/tex]

     [tex]|\epsilon| = 1.69 *10^{-3} \ V[/tex]

The  induced current i mathematically represented as

      [tex]I = \frac{\epsilon}{ R }[/tex]

  substituting values

       [tex]I = \frac{1.69*10^{-3}}{ 2.70 }[/tex]

       [tex]I = 6.25*10^{-4} \ A[/tex]

If a bicycle starts from rest and is pedaled normally until the bike is moving at 6 meters per second across level ground, what kinds of energy have its tires been given? (Select all that apply) g

Answers

Answer: Translational Kinetic Energy

Rotational Kinetic Energy

Explanation:

An object has translational kinetic energy when it is undergoing through a linear displacement.

Rotational energy is kinetic energy due to the rotation of an object .

Here the wheel of bicycle undergoes both translational and rotational kinetic energy has it moves with linear displacement with rotation in it.

Hence, the tires have been two kinds of energy : translational and rotational kinetic energy

a uniform rod of 30cm is pivoted at its center.a 40N weight is hung 5cm from left.from where 50N weight be hung to maintain equilibrium?

Answers

Answer:

The 50N weight be hung  at 23 cm to maintain equilibrium

Explanation:

Given;

length of the uniform rod = 30 cm

center of the uniform rod = 15 cm

weight of 40N is hung at 5 cm mark

weight of 50 N will be hung at ?

     0------5cm-----------------15cm-------------P---------30cm

              ↓             10cm      Δ       xcm    ↓

             40N                                             50N

Take moment about the pivot point and apply the principle of moment

50N (x cm) = 40N (10 cm)

x = (400) / 50

x = 8cm

P = x cm + 15 cm

P = 8 cm + 15 cm

P = 23 cm

Therefore, the 50N weight be hung  at 23 cm to maintain equilibrium

An RC circuit is connected across an ideal DC voltage source through an open switch. The switch is closed at time t = 0 s. Which of the following statements regarding the circuit are correct?
a) The capacitor charges to its maximum value in one time constant and the current is zero at that time.
b) The potential difference across the resistor and the potential difference across the capacitor are always equal.
c) The potential difference across the resistor is always greater than the potential difference across the capacitor.
d) The potential difference across the capacitor is always greater than the potential difference across the resistor
e) Once the capacitor is essentially fully charged, There is no appreciable current in the circuit.

Answers

Answer:

e)

Explanation:

In an RC series circuit, at any time, the sum of the voltages through the resistor and the capacitor must be constant and equal to the voltage of the DC voltage source, in order to be compliant with KVL.

At= 0, as the voltage through the capacitor can't change instantaneously, all the voltage appears through the resistor, which means that a current flows, that begins to charge the capacitor, up to a point that the voltage through the capacitor is exactly equal to the DC voltage, so no current flows in the circuit anymore, and the charge in the capacitor reaches to its maximum value.

A device called an insolation meter is used to measure the intensity of sunlight. It has an area of 100 cm2 and registers 6.50 W. What is the intensity in W/m2

Answers

Answer:

650W/m²

Explanation:

Intensity of the sunlight is expressed as I  = Power/cross sectional area. It is measured in W/m²

Given parameters

Power rating = 6.50Watts

Cross sectional area = 100cm²

Before we calculate the intensity, we need to convert the area to m² first.

100cm² = 10cm * 10cm

SInce 100cm = 1m

10cm = (10/100)m

10cm = 0.1m

100cm² = 0.1m * 0.1m = 0.01m²

Area (in m²) = 0.01m²

Required

Intensity of the sunlight I

I = P/A

I = 6.5/0.01

I = 650W/m²

Hence, the intensity of the sunlight in W/m² is 650W/m²

An electron initially at rest is accelerated over a distance of 0.210 m in 33.3 ns. Assuming its acceleration is constant, what voltage was used to accelerate it

Answers

Answer:

V = 451.47 volts

Explanation:

Given that,

Distance, d = 0.21 m

Initial speed, u = 0

Time, t = 33.3 ns

Let v is the final velocity. Using second equation of motion as :

[tex]d=ut+\dfrac{1}{2}at^2[/tex]

a is acceleration, [tex]a=\dfrac{v-u}{t}[/tex] and u = 0

So,

[tex]d=\dfrac{1}{2}(v-u)t[/tex]

[tex]v=\dfrac{2d}{t}\\\\v=\dfrac{2\times 0.21}{33.3\times 10^{-9}}\\\\v=1.26\times 10^7\ m/s[/tex]

Now applying the conservation of energy i.e.

[tex]\dfrac{1}{2}mv^2=qV[/tex]

V is voltage

[tex]V=\dfrac{mv^2}{2q}\\\\V=\dfrac{9.1\times 10^{-31}\times (1.26\times 10^7)^2}{2\times 1.6\times 10^{-19}}\\\\V=451.47\ V[/tex]

So, the voltage is 451.47 V.

What is the distance in m between lines on a diffraction grating that produces a second-order maximum for 775-nm red light at an angle of 62.5°?

Answers

Answer:

The distance is [tex]d = 1.747 *10^{-6} \ m[/tex]  

Explanation:

From the question we are told that  

       The order of maximum diffraction is  m =  2

         The wavelength is   [tex]\lambda = 775 nm = 775 * 10^{-9} \ m[/tex]

         The angle is  [tex]\theta = 62.5^o[/tex]

Generally the   condition for  constructive  interference for diffraction grating  is mathematically represented as

          [tex]dsin \theta = m * \lambda[/tex]

where  d is  the distance between the lines on a  diffraction grating

     So  

            [tex]d = \frac{m * \lambda }{sin (\theta )}[/tex]

substituting values  

           [tex]d = \frac{2 * 775 *1^{-9} }{sin ( 62.5 )}[/tex]

          [tex]d = 1.747 *10^{-6} \ m[/tex]

   

A double-slit experiment uses coherent light of wavelength 633 nm with a slit separation of 0.100 mm and a screen placed 2.0 m away. (a)How wide on the screen is the central bright fringe

Answers

Answer:

0.0127m

Explanation:

Using

Ym= (1)(633x10^-9m)(2m) / (0.1x10^-3m) = 0.0127m

6. How would the measurements for potential difference and current change if a 200 Ω resistor was used in Circuit 1 instead of the 100 Ω resistor? Explain your answer.

Answers

Answer:

Explanation:

Resistance is defined as the opposition to the flow of an electric current in a circuit. This means that a higher amount of resistance tends to reduce the amount of current flowing through the resistance. The lower the current, the greater the possibility for the resistor to allow current to pass through it.  if a 200 Ω resistor was used in Circuit 1 instead of the 100 Ω resistor, then the current in the circuit will tends to increase since we are replacing the load with a lesser resistor and a smaller resistance tends to allow more current to flow through it

For the potential difference, a decrease in the resistance value will onl decrease the potential difference flowing in the circuit according to ohm's law. According to the law the pd in a circuit is directly proportional to the current which means an increase in the resistance value will cause an increase in the corresponding pd and vice versa.

An apple falls from a tree and hits your head with a force of 9J. The apple weighs 0.22kg. How far did the apple fall?

Answers

Answer:

The apple fell at a distance of 4.17 m.

Explanation:

Work is defined as the force that is applied on a body to move it from one point to another. When a force is applied, an energy transfer occurs. Then it can be said that work is energy in motion.

When a net force is applied to the body or a system and this produces displacement, then that force is said to perform mechanical work.

In the International System of Units, work is measured in Joule. Joule is equivalent to Newton per meter.  

The work is equal to the product of the force by the distance and by the cosine of the angle that exists between the direction of the force and the direction that travels the point or the object that moves.  

Work=Force*distance* cosine(angle)

On the other hand, Newton's second law says that the acceleration of a body is proportional to the resultant of forces on it acting and inversely proportional to its mass. This is represented by:

F=m*a

where F is Force [N], m is Mass [kg] and a Acceleration [m / s²]

In this case, the acceleration corresponds to the acceleration of gravity, whose value is 9.81 m / s². So you have:

Work= 9 JF=m*a=0.22 kg*9.81 m/s²= 2.1582 Ndistance= ?angle=0 → cosine(angle)= 1

Replacing:

9 J= 2.1582 N* distante* 1

Solving:

[tex]distance=\frac{9J}{2.1582 N*1}[/tex]

distance= 4.17 m

The apple fell at a distance of 4.17 m.

In a polar coordinate system, the velocity vector can be written as . The term theta with dot on top is called _______________________ angular velocity transverse velocity radial velocity angular acceleration

Answers

Answer:

I believe it's called rapid growth

Explanation:

that is my answer no matter what

Two point charges of +2.0 μC and -6.0 μC are located on the x-axis at x = -1.0 cm and x 12) = +2.0 cm respectively. Where should a third charge of +3.0-μC be placed on the +x-axis so that the potential at the origin is equal to zero?

Answers

Answer:

 x = 0.006 m

Explanation:

The potential at one point is given by

          V = k ∑ [tex]q_{i} / r_{i}[/tex]

remember that the potential is to scale, let's apply to our case

          V = k (q₁ / x₁ + q₂ / x₂ + q₃ / x)

in this case they indicate that the potential is zero

          0 = k (2 10⁻⁶ / (- 1 10⁻²) + (-6 10⁻⁶) / 2 10⁻² + ​​3 10⁻⁶ / x)

         3 / x = + 2 / 10⁻² + ​​3 / 10⁻²

         3 / x = 500

          x = 3/500

          x = 0.006 m

An electron traveling with a speed v enters a uniform magnetic field directed perpendicular to its path. The electron travels for a time t0 along a half-circle of radius R before leaving the magnetic field traveling opposite the direction it initially entered the field. Which of the following quantities would change if the electron had entered the field with a speed 2v? (There may be more than one correct answer.)
a. The radius of the circular path the electron travels
b. The magnitude of the electron's acceleration inside the field
c. The time the electron is in the magnetic field
d. The magnitude of the net force acting on the electron inside the field

Answers

Answer:

Explanation:

For circular path in magnetic field

mv² / R = Bqv ,

m is mass , v is velocity , R is radius of circular path , B is magnetic field , q is charge on the particle .

a )

R = mv / Bq

If v is changed  to 2v , keeping other factors unchanged , R will be doubled

b )

magnitude of acceleration inside field

= v² / R

= Bqv / m

As v is doubled , acceleration will also be doubled

c )

If T be the time inside the magnetic field

T = π R / v

=  π  / v x  mv / Bq

= π m / Bq

As is does not contain v that means T  remains unchanged .

d )

Net force acting on electron

= m v² / R = Bqv

Net force = Bqv

As v becomes twice force too becomes twice .

So a . b , d are correct answer.

Two cars are moving towards each other and sound emitted by first car with real frequency of 3000 hertz is detected by a person in second with apparent frequency of 3400 Hertz what was the speed of cars

Answers

Answer:

 v ’= 21.44 m / s

Explanation:

This is a doppler effect exercise that changes the frequency of the sound due to the relative movement of the source and the observer, the expression that describes the phenomenon for body approaching s

           f ’= f (v + v₀) / (v-[tex]v_{s}[/tex])

where it goes is the speed of sound 343 m / s, v_{s} the speed of the source v or the speed of the observer

in this exercise both the source and the observer are moving, we will assume that both have the same speed,

                v₀ = v_{s} = v ’

we substitute

               f ’= f (v + v’) / (v - v ’)

               f ’/ f (v-v’) = v + v ’

               v (f ’/ f -1) = v’ (1 + f ’/ f)

               v ’= (f’ / f-1) / (1 + f ’/ f) v

               v ’= (f’-f) / (f + f’) v

let's calculate

                v ’= (3400 -3000) / (3000 +3400) 343

                v ’= 400/6400 343

                v ’= 21.44 m / s

What are the approximate dimensions of the smallest object on Earth that astronauts can resolve by eye when they are orbiting 275 km above the Earth

Answers

Answer:

   s_400 = 16.5 m , s_700 = 29.4 m

Explanation:

The limit of the human eye's solution is determined by the diffraction limit that is given by the expression

                   θ = 1.22 λ / D

where you lick the wavelength and D the mediator of the circular aperture.

In our case, the dilated pupil has a diameter of approximately 8 mm = 8 10-3 m and the eye responds to a wavelength between 400 nm and 700 nm.

by introducing these values ​​into the formula

                 

λ = 400 nm      θ = 1.22 400 10⁻⁹ / 8 10⁻³ = 6 10⁻⁵ rad

λ = 700 nm     θ = 1.22 700 10⁻⁹ / 8 10⁻³-3 = 1.07 10⁻⁴ rad

Now we can use the definition radians

          θ= s / R

where s is the supported arc and R is the radius. Let's find the sarcos for each case

λ = 400 nm       s_400 = θ R

                         S_400 = 6 10⁻⁵ 275 10³

                         s_400 = 16.5 m

λ = 700 nm s_ 700 = 1.07 10⁻⁴ 275 10³

                          s_700 = 29.4 m

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