The circuit having a parallel combination of resistance will have more current and series combination have less current.
In today's high-tech world, electrical engineers are faced with the design and analysis of an ever-increasing variety of circuits and systems. However, all of these systems are based on his DC circuit behavior at a basic level. In fact, the ability to analyze and simplify such circuits is critical to understanding and designing more complex circuits.
In addition, measurement of DC circuit quantity i. H. Voltage, current, resistance, the most basic and basic measurements an electrician can perform. In this experiment, students learn the use and limitations of modern digital multimeters and experimentally verify the validity of Thevenin's Theorem, his of key concepts in circuit theory.
It is often desirable to measure electricity without using an electricity meter. This situation is most common when examining current waveforms on the oscilloscope. The problem that arises here is that the oscilloscope cannot measure the current directly (this will become apparent in further experiments), so we have to devise a way to measure the current indirectly.
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Two ropes have equal length and are stretched the same way. The speed of a pulse on rope 1 is 1.4 times the speed on rope 2.
Determine the ratio of the masses of the two ropes (m1/m2).
The ratio of the masses of the two ropes (m1/m2) is 0.51.
The square root of the tension divided by the mass per length, or the linear density determines the speed of a wave on a string.
The formula used to relate the speed of a pulse on the rope (the speed of a wave on a string under tension) is given by |v| = √(F/μ) where v is the speed of a wave, F is tension in the string/rope, and μ is linear density (μ=m(mass)/L(length)).
Then, the speed of pulse on the first rope is, v₁ = √(F/μ₁). Given v₁ is 1.4 times v₂. Therefore, 1.4 v₂ = √(F/μ₁).
The speed of pulse on the second rope is, v₂ = √(F/μ₂). Substitute this value in the previous equation, and we get,
1.4√(F/μ₂) = √(F/μ₁)
Squaring on both sides of this equation,
[tex]\begin{aligned}1.96\left(\frac{F}{\mu_2}\right)&=\frac{F}{\mu_1}\\\frac{\mu_1}{\mu_2}&=\frac{1}{1.96}\\\frac{\frac{m_1}{L}}{\frac{m_2}{L}}&=\frac{1}{1.96}\\\frac{m_1}{m_2}&=0.510\end{aligned}[/tex]
The answer is 0.51.
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How do you dilate a figure with a center of dilation?
The point where the image gradually dilated is known as the center of dilatation (or point of dilation).
The dilation's centre is at the centre point. To calculate the lengths to the preimage as well as the dilation image, utilise the centre point. The scale factor was determined by these distances. Dilation A dilation would be a transformation that increases or decreases a figure's size.
The center of dilation as well as figure could be translated so that the origin will become the center, and then translated back to create a dilation with just about any point other than origin as the center.
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which of these stars will take the shortest time to go from the earliest protostar stage to the main sequence?
The stars which will take the shortest time to go from the earliest protostar stage to the main sequence is: the star ten times the mass of our sun.
The correct answer choice is option b.
What is meant by the stars?A star can simply be defined as a luminous body which is usually, frequently and most of the time found in the night sky. However, a star is a also an incandescent body just the like the sun and it is in a fixed point. Below are some few characteristics of the star:
A star is massiveA star is a celestial bodyA star shines by radiation A star is held together by gravityA star also produce Its own lightIt also produce its own heat energy.In conclusion, it can be deduced from the explanation given above that stars are luminous and highly massive bodies.
Complete question:
Which of these stars will take the shortest time to go from the earliest protostar stage to the main sequence?
a. the star which has the same mass as our sun
b. the star ten times the mass of our sun
c. the star six times the mass of our sun
d. none of the above
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finally, remind yourself the properties of light from the first lecture and qs1. if instead of pulses, the astronaut had sent a continuous beam of green light. how would that light appear to those on the spaceship?
Although the green light flashes first, you are moving away from it, so it illuminates you at the same moment as the red light.
A green laser is frequently used by amateur astronomers because its visible beam may be used to point out stars and galaxies because it penetrates the atmosphere so far. Red lasers are less adaptable than green lasers. Each wave or component of light in a laser has been precisely aligned to form a single, strong beam.
Green DPSS lasers are the most popular and first emerged on the market about 2000. (also called diode-pumped solid-state frequency-doubled, DPSSFD). Due to the rarity of laser diodes in this wavelength region, they are more complicated than regular red laser pointers.
Green lasers have been used in head-mounted displays and laser guides for defense and security purposes. Amateur astronomers can also direct and track stars using green laser guides.
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when the displacement of a mass on a spring is 1212 aa , what percentage of the energy is kinetic energy?
The percentage of kinetic energy is not determined by the displacement of the mass; it is determined by the velocity of the mass. To calculate the percentage of kinetic energy, you would need to know the velocity of the mass.
The percentage of energy that is kinetic energy depends on the velocity of the mass. If the mass is at rest, then there is no kinetic energy and all of the energy is potential energy. If the mass is moving, then the percentage of kinetic energy depends on the speed of the mass.
At any given displacement, the potential energy of the mass on a spring is equal to 1/2 kx^2, where k is the spring constant and x is the displacement. The kinetic energy of the mass is equal to 1/2 mv^2, where m is the mass and v is the velocity.
In the given example, if the displacement is 1212 aa, then the potential energy of the mass is (1/2)k(1212)^2. The kinetic energy of the mass is (1/2)mv^2. Since the velocity is not given, we cannot calculate the percentage of kinetic energy.
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Do you think the pitch of the waves sounds significantly different when both speakers are on?
Whether this difference is significant would depend on the specific frequencies of the sound waves and on the listener's perception of the pitch.
The pitch of a sound is determined by the frequency of the sound waves, which is the number of cycles of the wave per second. If the frequency of the sound waves produced by the two speakers is the same, then the pitch of the combined sound will be the same as the pitch of each speaker individually.
However, if the two speakers produce sound waves with different frequencies, then the pitch of the combined sound will be different from the pitch of each speaker individually.
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An unused flashbulb contains magnesium and oxygen. After use, the contents are changed to magnesium oxide but the total mass does not change. This observation can best be explained by the.
After the use the content of the unused flash bulb does not change although the contents are changed to magnesium oxide. This observation can be explained by the law of conservation of mass.
The chemical equation of the reaction of magnesium and oxygen to give magnesium oxide as product is,
2Mg + O₂ → 2MgO
As we can see from the above reaction, the total number of atoms on the product side are equal to the total number of atom on the react inside.
This observation can be explained by the law of conservation of mass because the mass on both side of the reaction are equal to each other.
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what event made it possible for photons to begin to travel freely through the universe when the universe was about 380,000 years old?
When the universe was about 380,000 years old universe had expanded and cooled enough for electrons to combine with nuclei to form neutral atoms.
As the universe cooled, conditions became simply proper to provide rise to the building blocks of matter – the quarks and electrons of which we're all made. some millionths of a second later, quarks aggregated to supply protons and neutrons. within minutes, these protons and neutrons blended into nuclei. because the universe continued to amplify and funky, matters started to occur extra slowly.
It took 380,000 years for electrons to be trapped in orbits around nuclei, forming the primary atoms. these were mainly helium and hydrogen, that is still by means of far the maximum abundant factors in the universe. gift observations propose that the primary stars fashioned from clouds of gasoline around hundred and fifty–200 million years after the Big Bang. Heavier atoms such as carbon, oxygen, and iron, have in view that been constantly produced in the hearts of stars and catapulted at some stage in the universe in marvelous stellar explosions known as supernovae.
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When a spring is stretched by 2cm It stores 100 J of energy?
The spring constant is 50000 N/m.
The spring constant is a measure of the stiffness of a spring.
Springs with higher stiffness are more difficult to stretch. Under the action of external forces, the spring is deformed and returns to its original position when the force is removed. Spring constant is denoted by 'k' and has a SI unit as N/m.It has different values for different springs and different material.Electric potential energy of a spring can be calculated as P.E. = (1/2)*k*x²
where, k is the spring constant,
x is the displacement caused by the spring
The spring stores energy = 100 J
Displacement suffered by the spring = 2 cm = 0.02 m
P.E. = (1/2)*k*x²
k = (2*P.E.) / x²
k = (2 * 100) / (0.02)²
k = 200 / 0.0004
k = 50000 N/m
The spring constant for a spring which stores 100 J energy is 50000 N/m.
The question is incomplete the complete question is "When a spring is stretched by 2cm It stores 100 J of energy? What is the spring constant".
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A 2. 0 kg body is initially traveling at a velocity of 40 m/s east. If a constant force of 10 n due east is applied to the body for 5. 0 s, the final speed of the body is.
The Final speed of the body is = 65 m/s
The values that are given in the question are,
Mass of body = 2.0 kg
Velocity of the body = 40 m/s
Force = 10 N
Time = 5.0 s
Firstly we need to calculate the momentum change of the body,
The formula for change in momentum,
⇒ Change in momentum = Force x Time
⇒ Change in momentum = 10 x 5.0
⇒ Change in momentum = 50 kg-m/s
Now we have to calculate the change in velocity,
⇒ Change in velocity = Change in momentum / Mass
⇒ Change in velocity = 50 / 2.0
⇒ Change in velocity = 25 m/s towards east
Now to calculate the final velocity
⇒ Final velocity = initial velocity + change in velocity
⇒ Final Velocity = 40 + 25 = 65 m/s
Therefore, the final speed of the body is = 65 m/s
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i need to know 2 types of matter and descriptions of them
When light shines on a sample, each element emits specific wavelengths producing a unique fingerprint called its ______ spectra.
When light shines on a sample, each element emit specific wavelength producing a fingerprint called its line spectra.
When light falls on a shining surface it makes a specific pattern.
These patterns are either in the form of bright lights are in the form of dark lights which are uniform in nature providing a particular pattern for each and every light.
This line spectra is the result of the light of different frequencies emitted by the excitation and releasing of electrons from there respective orbitals in an atom.
Basically these line spectra provides a fingerprint of the specific wavelength emitted by each element.
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What is the wavelength of a Colour of frequency 4 x 10 14 Hz in a medium of refractive index 1.5 *?
Answer:
Approximately [tex]5.00 \times 10^{-7}\; {\rm m}[/tex] ([tex]500\; {\rm nm}[/tex].)
Explanation:
Look up the speed of light in vacuum: [tex]c \approx 3.00 \times 10^{8}\; {\rm m\cdot s^{-1}}[/tex].
If the speed of light in a medium is [tex]v[/tex], the refractive index of that medium will be [tex]n = (c / v)[/tex]. For the medium in this question, it is given that [tex]n = 1.5[/tex]. Rearrange the equation [tex]n = (c / v)\![/tex] to find [tex]v[/tex]:
[tex]\begin{aligned}v &= \frac{c}{n} \\ &\approx \frac{3.00 \times 10^{8}\; {\rm m\cdot s^{-1}}}{1.5} \\ &\approx 2.00 \times 10^{8}\; {\rm m\cdot s^{-1}}\end{aligned}[/tex].
Let [tex]f[/tex] denote the frequency of this wave. When the speed of this wave is [tex]v[/tex], the wavelength [tex]\lambda[/tex] of this wave will be [tex]\lambda = (v / f)[/tex]. Therefore:
[tex]\begin{aligned}\lambda &= \frac{v}{f} \\ &\approx \frac{2.00\times 10^{8}\; {\rm m\cdot s^{-1}}}{4 \times 10^{14}\; {\rm s^{-1}}} \\ &\approx 5.00 \times 10^{-7}\; {\rm m} \\ &= 500\; {\rm nm}\end{aligned}[/tex].
(Note that [tex]1\; {\rm Hz} = 1\; {\rm s^{-1}}[/tex].)
The energy from the core creates convection currents which push magma through the lithosphere at the
crust is moved towards the
Word Bank:
mid-oceanic ridge old subduction zones
Blank 1:
Blank 2:
Blank 3:
Question 3 (1 point)
and is recycled as it is pushed down by gravity into the mantle.
to form new crust.
The energy from the core creates convection currents which push magma through the lithosphere at the mid-oceanic ridge to form new crust. Old crust is moved towards the subduction zones and is recycled as it is pushed down by gravity into the mantle.
What is lithosphere?Lithosphere is defined as the layer of the earth that is located at the outermost part which is made up of earth crust and upper mantle.
The core of the earth is centrally located and is filled with hot magma. The energy from the magma is being pushed through the lithosphere at the mid-oceanic ridge to form new crust.
The mid-oceanic ridge is defined as the ridges that form along divergent plate boundaries.
Old crust is moved towards the subduction zones and is recycled as it is pushed down by gravity into the mantle.
The subduction zones are the zones where old crusts are destroyed through a process called subduction.
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a plank having mass 6.6 kg rides on top of two identical solid cylindrical rollers each having radius 4.6 cm and mass 1.9 kg. the plank is pulled by a constant horizontal force 6 n applied to its end and perpendicular to the axes of the cylinders (which are parallel). the cylinders roll without slipping on a flat surface. there is also no slipping between cylinders and plank. m r m r f m find the acceleration of the plank. answer in units of m/s 2 .
The acceleration of plank which is weighing 6.6 Kg will be 0.90 m/s² to ride it with a force of 6 N.
What is acceleration?Acceleration is the rate of change in velocity. It is a vector quantity and thus having both magnitude and direction. According to Newton's second law of motion, the force acting on a moving body is the product of its mass and acceleration.
Thus, a force is needed to accelerate the body having a mass m and the relation is written as:
f = m a
Given that, mass of the plank is 6.6 kg
horizontal force acting on the plank = 6N
hence, acceleration = force/mass = 6 N/6.6 Kg = 0.9 m/s².
Therefore, the acceleration of the plank is 0.9 m/s².
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an object of mass m is rotating about a fixed axis with angular momentum l. its moment of inertia about this axis is i. what is its kinetic energy?
The Kinetic energy would be 1/2IL².
What is Rotational Kinetic energy ?Rotational energy also known as angular kinetic energy is defined as: The kinetic energy due to the rotation of an object and is part of its total kinetic energy. Rotational kinetic energy is directly proportional to the rotational inertia and the square of the magnitude of the angular velocity.As we know linear Kinetic energy = 1/2mv²
where m= mass and v= velocity.
Similarly rotational kinetic energy is given by = 1/2IL²
where I- moment of inertia and L=angular momentum.
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using a cable with a tension of 1470 n, a tow truck pulls a car 5.00 km along a horizontal roadway. part a how much work does the cable do on the car if it pulls horizontally? express your answer with the appropriate units.
The work done by the string on the auto is 7350J.
In each case the forces are constant and the relegation(displacement) is along a straight line, so
Work = F.s cos θ
Where,
F = force
S = relegation (displacement)
Since, it's vertical to the road
so,
θ = 0 °
All mass-bearing objects are attracted by gravitational force. Because it consistently attempts to bring masses together rather than push them apart, the gravitational force is referred to as attractive.
Given,
force = 1470 N
s = 5 km
Now,
Work = 1470.5. cos 0
Work = 7350 J
Thus, the work done by the string on the auto is 7350J.
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in a binary-star system that produces a nova, the white dwarf pulls matter from the companion star. the matter forms an accretion disk that orbits the white dwarf. then a specific sequence of events must take place for a nova event to occur.a. Material accumulates onto white dwarf's surface increasing in temperature and density b. Nuclear fusion reactions cause the an enormous but temporary increase in luminosity c. At a temperature of 10 million K, the accumulated surface hydrogen begins nuclear fusion d. As nuclear fuel is burned up or blown into space, fusion ceases and the star dims 1. first step2. last step
The correct order is:
Material accumulates onto the white dwarf's surface, increasing in temperature and density.
At a temperature of 10 million K, the accumulated surface hydrogen begins nuclear fusion.
Nuclear fusion reactions cause an enormous but temporary increase in luminosity
As nuclear fuel is burned up or blown into space, fusion ceases and the star dims.
What occurs when a white dwarf is in a binary star system?
A tight binary star system allows the companion star of a white dwarf or neutron star to give mass to the other star. A white dwarf's surface material has the potential to suddenly burst into fusion, creating a nova.
What type of binary system is needed for a nova to occur?
An eruption from the surface of a white dwarf star in a binary star system is known as a nova. When a white dwarf, which is the dense core of a former normal star, "steals" gas from a neighboring companion star, a nova is a result. When the white dwarf's surface accumulates enough gas, an explosion results.
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A wheel 31 cm in diameter accelerates uniformly from 240 rpm to 360 rpm in 6. 8 s. How far will a point on the edge of the wheel have traveled in this time?.
Within 6.8 s, a point on the edge of the wheel have traveled a distance of 6,619.12 cm.
The problem can be solved using the equation of rotational motion.
α = (ω_t - ω_o) / t
Where:
α = angular acceleration
ω_t = final rotational speed
ω_o = initial rotational speed
t = time period
The angle travelled is given by:
∅ = ω_o x t + 1/2 x αt²
∅ = ω_o x t + 1/2 x [ (ω_t - ω_o) / t] t²
∅ = 1/2 (ω_o + ω_t ) x t
Parameters given:
ω_o = 240 rpm = 25.13 rad/s
ω_t = 360 rpm = 37.7 rad/s
t = 6.8 s
Hence,
∅ = 1/2 (25.13 + 37.7 ) x 6.8 = 213.2 rad
The wheel radius:
r = 31 cm
The wheel perimeter:
C = 2 x π x 31 = 194.68 cm
Since the whole angle of a circle is 2π rad, then the travelled angle is equal to:
213.2 / 2π = 34 wheel perimeter.
Hence,
the distance travelled = 34 x 194.68 cm = 6,619.12 cm
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an airplane traveling at half the speed of sound emits a sound of frequency 4.90 khz. (a) at what frequency does a stationary listener hear the sound as the plane approaches? khz
The frequency does a desk bound listener pay attention the sound because the aircraft approaches is 10.0kHz
Calculation :
velocity of the of sound (air plane) v= 172m/s
frequency emmited n=four.9 kHz
= four.9*[tex]10^{three}[/tex]
n' =[tex](frac{V-Vo}{V-V,} )*n\[/tex]
[tex]V,[/tex]= [tex]frac{V}{2}[/tex]
Then the obvious frequency heard wiil be
n' =[tex](four.nine*10^{three} Hz)(frac{V}{V-frac{V}{2} } )[/tex]
= (4.nine[tex]10^{3}Hz[/tex]*)(2)
=10.0kHz
Frequency is the quantity of occurrences of a repeating event according to unit time. [1] For readability, additionally called temporal frequency, which is different from angular frequency. Frequency is measured in hertz (Hz) and corresponds to 1 event in keeping with 2d. length is the reciprocal of frequency due to the fact period is the time c language between events. [2]
For example, if the coronary heart beats 120 instances according to minute (2 hertz), the length T (the c language between repeated beats) is zero.5 seconds (60 seconds divided by way of one hundred twenty beats). cost). Frequency is an crucial parameter used in technological know-how and engineering to explain the speed of vibrations and vibrational phenomena inclusive of mechanical vibrations, audio indicators (sound), radio waves, and light.
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anderson video - ampere's law 2 of 5 review | constants first, launch the video ampere's law. after watching the video, answer the follow-up question below. select to launch video part a a solenoid has 15 turns per centimeter of its length. the solenoid is twisted into a circle so that it becomes shaped like a toroid. what is the magnetic field at the center of each turn of the toroid? the current is 57 ma.
A magnetic field's influence is perceived in the area of space known as the magnetic field. The following formula is used to determine the magnetic field at the center of each rotation of the toroid:
B = μnI
where as;
μ is permeability of empty space = 12.56x 10⁻⁷ Tm/A
There are 18 rotations each centimeter, which equates to 1,800 turns per meter.
I = 57 m is the solenoid's current.
A = 0.057A
B = (12.56 x 10⁻⁷) x (1,500) x (0.057)
B = 1.08 x 10⁻⁴ T
Is the magnetic field at the center zero?At the core of a bar magnet, the resultant magnetic field is zero.
What does a toroid's magnetic field at its core look like?Since there is no net current flowing through this region, the magnetic field is also zero. The toroid therefore has no magnetic moment.
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what speed must a neutron have if its de broglie wavelength is to be equal to the interionic spacing of table salt (0.282 nm )?
The speed of the neutron have if its de broglie wavelength is to be equal to the interionic spacing of table salt is 1.40 km/s
Given that the mass of the neutron is m 1.67 × 10-27 kg and has a
the interionic distance of table salt, which is equivalent to the de Broglie wavelength, is
λ = 0.282 nm.
The de Broglie wavelength is expressed as
h p h mv -
where, h is the Planck's Constant h above equation in terms of v, we get = 6.63 × 10-34 Js. Rearranging the (1)
h mλ v =
By changing every value in the equation (1),
we get 6.63 × 10-34 J. s (1.67 × 10-27 kg) (0.282 × 10-9 m) km S
= 1.40 km/s
Therefore, the speed is v = 1.40 km/s
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What is the wave speed if the wavelength is 5 meters and the frequency is 4 Hz?
The speed of wave is 20m/sec if the wavelength of wave is 5 meters and the frequency of the wave is 4Hz.
At the point when a wave goes through reflection, it simply stays in the medium and just switches its heading of movement. The smooth wave has voyaged twofold its distance. This reflection peculiarity of waves is regularly seen in sound waves.
At the point when you let out a noisy cry inside a ravine, you frequently hear the reverberation of the holler. The sound waves travel through the medium; for this situation, the air and bounce off the gulch wall and return to the beginning of the sound (you). The outcome is that you hear the reverberation of your holler.
We know that when a electromagnetic wave is travelling through speed v passing in a medium, it contains certain speed and wavelength which is given by the formula
c=ν × λ
where c is the speed of light in vacuum,
ν is the frequency of the wave and
λ is the wavelength of the wave.
Now, we have frequency(ν) = 4Hz, and wavelength(λ)=5meters.
So, using the above formula, we get
c=4 × 5
=>c=20m-Hz
or c=20m/sec as 1Hz=1 /sec
Hence, wave speed is 20m/sec.
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Deep surveys of the universe indicate that the largest structures in space are no larger than about 50 mpc in size.t/f
Deep surveys of the universe indicate that the largest structures in space are about 50 Mpc in size is false.
Deep surveys of the universe indicate that the largest structures in space are no larger than about 50 Mpc in size. If the universe had an edge, that fact would violate the assumption of isotropy in the cosmological principle. Hubble's law implies that the universe will expand forever.
The cosmological principle is based on two observations and an assumption, that the universe is the same everywhere, it is the same in every direction, and that the laws of physics are the same everywhere. The statement that the universe is homogeneous regards the universe on very large scales.
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a tire has a tread pattern with a crevice every 4.00 cm. each crevice makes a single vibration as the tire moves. what is the frequency (in hz) of these vibrations if the car moves at 24.2 m/s?
the frequency (in hz) of these vibrations if the car moves at 24.2 m/s is 605 HZ .
Calculation :
frequency = [tex]\frac{number of contacts}{second}[/tex]
frequency = [tex]\frac{24.2 m/s}{1}*100cm*\frac{1}{4cm}[/tex]
= 605 HZ
Frequency describes the number of waves passing through a particular location in a particular time. So if the wave takes 1/2 second to travel, the frequency is 2 per second. If it takes 1/100th of an hour, the frequency is 100 per hour.
Frequency is the number of occurrences of a repeating event per unit time. ... sometimes called time-frequency for clarity,
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a photographer uses his camera, whose lens has a 50 mm focal length, to focus on an object 1.5 m away. he then wants to take a picture of an object that is 30 cm away. how far
The distance the lens needed to be moved so that it can focus on the second object is 0.008 m.
From the question, it is said that focal length f = 50 mm = 50* 10⁻³ m
The lens equation is mathematically represented as
1/u + 1/v = 1/f
Given that, u = 1.5 m
1/1.5 + 1/v = 1/(50* 10⁻³)
1/v = 1/(50* 10⁻³) - 1/1.5
v = 0.052 m
Object distance u = 30 cm = 0.3 m
1/0.3 + 1/ v₁ = 1/(50* 10⁻³)
v₁ = 0.06 m
The distance the lens needed to be moved is evaluated as
z = | v - v₁ | = | 0.052 - 0.06 | = 0.008 m
The question is incomplete. The continuation of the question is 'how far must the lens move to focus on this second object?'
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Astrology, that unlikely and vague pseudoscience, makes much of the position of the planets at the moment of one’s birth. the only known force that a planet could exerts on us is gravitational, so if there is anything to astrology we should expect this force to be significant.
Calculate the magnitude of the gravitational force (in N) exerted on a 3.00 kg baby by a 110 kg father 0.150 m away at birth (he is assisting, so he is close to the child.)
The magnitude of the gravitational force exerted on baby by the father is 9724 N * 10⁻⁷ N.
Gravitational force formula is given by:
F = GMm/r²
where G = gravitational constant
M and m = masses given
r = distance given in radius
Concerning this question, we can make a substitute of the formula to be:
F(father) = G*M(father)*m(baby)/r²
F = G * (110 kg) * (3kg) / (0.150)²
Where; G = gravitational constant ⇒ 6.63 * 10⁻¹¹ Nm²/kg²
M = mass of the father ⇒ 110 kg
m = mass of the child ⇒ 3 kg
r = distance between father and child ⇒ 0.2m
Force = 6.63 * 10⁻¹¹ * 110 * 3 / (0.150)²
Force = 9724 N * 10⁻⁷ N
We can now conclude that the gravitational force exerted on the baby by the father at birth is 9724 N * 10⁻⁷ N.
Therefore, the magnitude of the gravitational force exerted on baby by the father is 9724 N * 10⁻⁷ N.
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a loop antenna with loop radius of 0.4 m is made of copper wire of radius 4 mm. if the loop radiates at 6 mhz and carries a current of 50 a, find (a) the radiation resistance of the loop (b) the power radiated
The radiation resistance of the loop is 99.9 % and the power radiated 4.97 KW.
The real resistance = 8 π² (L/λ)²
L = 2πr
λ = C/f
power radiated = [tex]I^{2}[/tex]/r = 80 × 3.14 × 3.14 × 4× 3.14 × 3.14 × 0.4 × 0.4/2500
Power radiated = 4.97 KW
Radiation efficiency = Radiation resistance / radiation resistance + ohm resistance
ρL/A = 8 × [tex]10^{-8}[/tex] = 1.99 / 8 × [tex]10^{-8}[/tex] + 1.99
99.9 %
The definition of radiation is energy that emanates from a source and moves at the speed of light through space. This energy has wave-like characteristics and is surrounded by an electric and magnetic field. Electromagnetic waves are another name for radiation. In nature, there are many different types of electromagnetic radiation. One example is visible light. The highest energy radiation comprises x-rays, gamma rays, and ultraviolet light. Both X-rays and gamma rays are extremely energetic. They can take electrons out of atoms when they engage with them, ionizing the atom in the process. Our cells' DNA can be harmed by radiation. High radiation exposures can result in cutaneous radiation injuries (CRI) or acute radiation syndrome (ARS) (CRI).
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refer to question 12 on page 6/7: a type ii cepheid has an apparent magnitude of 12 and a pulsation period of 3 days. what is the approximate distance to this cepheid variable star?
Is the approximate distance to this cepheid variable star 4,000 parsecs
The distance modulus equation will be used for distance to the sapid variable. The formula will be m. M is equivalent to 5 log d - 5 minus m. Now that we need to determine d, we can rearrange the equation. by writing, then We will discover that d is equal to 10 when we multiply m by m minus m plus 5 and divide by 5. Substitute now.
The values are: d equals 10 equals 12 less 1 less plus 5 divided by 5. When you solve for d, you will find that it is equivalent to 3090.1 per since d is equal to 10 to 18, divided by five, which is equal to 10, is to 3.6. As a result, the cepheid variable star is 3090.1 per second.
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Since the beginning of time some people have believe that the earth is flat beginning around 300 BC however some began to wonder if the earth is actually spherical in shape
Since the beginning of time, some people have believed that the earth is flat, then the example demonstrates that argumentation is a necessary part of scientific inquiry. Hence, option B is correct.
What is Earth?In terms of size and mass, Earth is the third planet from the Sun and the fifth-largest planetary in the solar system. The only site in the cosmos known to support life is in its close-surface conditions, which is its most notable characteristic. The emblem serves as a designation. The word "Earth" is derived from the Old English and Germanic meanings for "ground" and "earth," and it is the only title for a planet in our solar system that does not originate from Greco-Roman mythology. English is the world's most widely used astronomical language.
This reasoning is a crucial part of every scientific research since it offers a variety of hypotheses about the scientific concerns that must be supported. Among these hypotheses, only one provides a thorough explanation of the phenomenon.
As a result, it can be seen in the given question that ideas about the form of the earth over time have demonstrated that argumentation is a component of the scientific process.
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Your question is not complete, the complete might be like this:
Since the beginning of time, some people have believed that the Earth is flat. Beginning around 300 B.C., however, some began to wonder if the Earth is actually spherical in shape. Those that favored the flat Earth theory made the following observations: The Earth is stable because it is a flat landmass sandwiched between a flat layer of air and a flat layer of water. If the Earth is spherical, then oceans should be curved on top, but they aren't. Some rivers flow for hundreds of miles but only fall a few feet. The rivers should fall farther if the Earth is spherical. If the Earth is spherical, people and animals that inhabit the Earth would notice as they walked along its surface. Since mariners are able to use flat compasses to navigate their ships without crashing, the Earth must not be a sphere. Those that disputed the flat Earth theory countered the claim by observing that Christopher Columbus was able to sail around the Earth without falling off the edge and later, by presenting satellite imagery taken from space of a spherical Earth. This example demonstrates that
A. scientific knowledge is unattainable.
B. argumentation is a necessary part of scientific inquiry.
C. it is easy to resolve a scientific argument.
D. it is not possible to settle a dispute.