what is the quantum and its types?
Answer:
It is the physics that explains how everything works, the nature of the particles that make up matter and the forces with which they interact.
Its types: Electromagnetism, the strong nuclear force, and the weak nuclear force.
Hope this help :)
Which two options are forms of kinetic energy?
A. Thermal energy
OO
O B. Nuclear energy
c. Sound energy
O D. Gravitational energy
O E. Chemical energy
Answer:
Themal energy is the best answer from you
Your friend said that the star in this picture with the highest apparent magnitude must definitely have the highest absolute brightness as well.
Answer:
A white dwarf, also called a degenerate dwarf
Explanation:
sorry if im wrong im kind of du-m
12. An organ pipe that is 1.75 m long and open at both ends produces sound of
frequency 303 Hz when resonating in its second overtone. What is the speed of
sound in the room?
295 m/s
328 m/s
354 m/s
389 m/s
401 m/s
Answer:
354 m/s
Explanation:
For the second overtune (Third harmonic) of an open pipe,
λ = 2L/3................................ Equation 1
Where L = Length of the open pipe, λ = Wave length.
Given: L = 1.75 m.
Substitute into equation 1
λ = 2(1.75)/3
λ = 1.17 m.
From the question,
V = λf.......................... Equation 2
V = speed of sound in the room, f = frequency
Given: f = 303 Hz.
Substitute into equation 2
V = 1.17(303)
V = 353.5
V ≈ 354 m/s
Hence the right answer is 354 m/s
A plane takes off at St. Louis, flies straight to Denver, and then returns the same way. The plane flies at the same speed with respect to the ground during the entire flight, and there are no head winds or tail winds. Since the earth revolves around its axis once a day, you might expect that the times for the outbound trip and the return trip differ, depending on whether the plane flies against the earth's rotation or with it. Is this expectation true or false
Answer:
In the Both time
Explanation:
A plane takes off at St.Louis, flies straight to Denver, and then returns the same way. The plane flies at the same speed with respect to the ground during ...
Depending on whether the plane flies against the earth's rotation or with it. Is this expectation is true statement.
What is Plane?Physical quantities such as work, temperature, and distance can all be completely represented in daily life by their magnitude. The laws of arithmetic can, however, be used to explain how these physical values relate to one another.
Motion in two dimensions is another name for motion in a plane. For instance, a projectile moving in a circle. The origin, along with the two coordinate axes X and Y, will serve as the reference point for the investigation of this kind of motion.
Therefore, Depending on whether the plane flies against the earth's rotation or with it. Is this expectation is true statement.
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A solenoid of length 0.35 m and diameter 0.040 m carries a current of 5.0 A through its windings. If the magnetic field in the center of the solenoid is 2.8 ×102 T, what is the number of turns per meter for this solenoid?
Correct question:
A solenoid of length 0.35 m and diameter 0.040 m carries a current of 5.0 A through its windings. If the magnetic field in the center of the solenoid is 2.8 x 10⁻² T, what is the number of turns per meter for this solenoid?
Answer:
the number of turns per meter for the solenoid is 4.5 x 10³ turns/m.
Explanation:
Given;
length of solenoid, L= 0.35 m
diameter of the solenoid, d = 0.04 m
current through the solenoid, I = 5.0 A
magnetic field in the center of the solenoid, 2.8 x 10⁻² T
The number of turns per meter for the solenoid is calculated as follows;
[tex]B =\mu_o I(\frac{N}{L} )\\\\B = \mu_o I(n)\\\\n = \frac{B}{\mu_o I} \\\\n = \frac{2.8 \times 10^{-2}}{4 \pi \times 10^{-7} \times 5.0} \\\\n = 4.5 \times 10^3 \ turns/m[/tex]
Therefore, the number of turns per meter for the solenoid is 4.5 x 10³ turns/m.
Which type of biological molecule would contain fats?
A) Amino Acids.
C) Nucleic Acids.
B) Lipids.
D) Carbohydrates.
B
Explanation:
lipids contains fat
hope it helps
The parallel plates in a capacitor, with a plate area of 9.30 cm2 and an air-filled separation of 4.50 mm, are charged by a 7.80 V battery. They are then disconnected from the battery and pulled apart (without discharge) to a separation of 9.60 mm. Neglecting fringing, find (a) the potential difference between the plates, (b) the initial stored energy, (c) the final stored energy, and (d) the work required to separate the plates.
Answer:
a) ΔV ’= 1.66 10¹ V= 16.6 V, b) U = 55.64 10⁻¹² J, c) U_f = 1.18 10⁻¹⁰ J
d) W = 6.236 10⁻¹¹ J
Explanation:
Capacitance can be found for a parallel plate capacitor
C = ε₀ [tex]\frac{A}{d}[/tex]
Let's reduce the magnitudes to the SI system
A = 9.30 cm² (1 m / 10² cm) 2 = 9.30 10⁻⁴ m²
c = 4.50 mm (1 m / 1000 mm) = 4.50 10⁻³ m
Co = 8.85 10⁻¹² 9.30 10⁻⁴ /4.50 10⁻³
Co = 1.829 10⁻¹² F
when the plates separate at d = 9.60 10⁻³ m, the capcitance changes to
C = ε₀ \frac{A}{d_1}
C = 8.85 10⁻¹² 9.30 10⁻⁴/9.60 10⁻³
C = 8.57 10⁻¹³ F
a) the potential difference
C =
since the capacitor is not discharged, let's look for the initial charge
Co = \frac{Q}{ \Delta V}
Q = C₀ ΔV
Q = 1.829 10⁻¹² 7.80
Q = 14.2662 10⁻¹² C
when the condensate plates are separated
C = \frac{Q}{ \Delta V' }
ΔV ’= Q / C
ΔV ’= 14.266 10⁻¹² / 8.57 10⁻¹³
ΔV ’= 1.66 10¹ V= 16.6 V
b) the stored energy is
U = ½ C ΔV²
for initial separation
U = ½ C₀ ΔV²
U = ½ 1.829 10⁻¹² 7.80²
U = 55.64 10⁻¹² J
c) The energy for end separation;
U_f = ½ C DV’2
U_f = ½ 8.57 10⁻¹³ 16,6²2
U_f = 1.18 10⁻¹⁰ J
d) The work
as there are no losses, the work is equal to the variation of the energy
W = ΔU = U_f -U₀
W = 1.18 10⁻¹⁰ - 55.64 10-12
W = 6.236 10⁻¹¹ J
what happens to the loudness of the sound as the amplitude increases
it increases and gets louder
and it amplifies the noise
Suppose that you'd like to find out if a distant star is moving relative to the earth. The star is much too far away to detect any change in its brightness as it moves toward or away from the earth. Instead we can use the Doppler effect to determine its relative speed. For this problem we are going to look at the spectral lines from hydrogen, specifically the one with a wavelength of 656.46 nm.
The hydrogen atoms in a star are also moving at high velocity because of the random motions caused by their high temperature. As a result, each atom is Doppler shifted a little bit differently, leading to a finite width of each spectral line, such as the 656.46-nm line we were just discussing. For a star like our sun, this leads to a finite width of the spectral lines of roughly Δλ=0.04nm.
If our instruments can only resolve to this accuracy, what is the lowest speed V, greater than 0, that we can measure a star to be moving?
Answer:
The answer is "[tex]\bold{18 \ \frac{km}{s}}[/tex]"
Explanation:
Its concern is not whether star speed is significantly lower than the light speed. Taking into consideration the relativistic tempo (small speed star)
[tex]\to \frac{\Delta \lambda}{\lambda} = \frac{v}{c}\\\\\to v = \frac{\Delta \lambda}{\lambda} \left (c \right ) \\\\[/tex]
[tex]= \left ( \frac{0.04}{656.46} \right ) (3 \times 10^8)\\\\ = 18280 \ \frac{m}{s} \approx 18 \ \frac{km}{s}[/tex]
Which molecules are in put in photosynthesis
Answer:
During the process of photosynthesis, cells use carbon dioxide and energy from the Sun to make sugar molecules and oxygen. These sugar molecules are the basis for more complex molecules made by the photosynthetic cell, such as glucose.
Explanation:
yes.
Answer:
Sunlight, Carbon Dioxide, and Water
Explanation:
Technically minerals are in there too but when I learned this it was just Sunlight, Carbon Dioxide, and Water
a ford explorer traveled 100 miles the next day for 5 hours. What was the average speed of this vehicle?
Answer:
25 miles per hour
Explanation:
It was 20 miles per hour the next day. We don't have enough information to calculate the average speed for the whole trip.
Equipotential lines are usually shown in a manner similar to topographical contour lines, in which the difference in the value of consecutive lines is constant. Clear the equipotential lines using the Clear button on the voltage tool. Place the first equipotential line 1 m\rm m away from the charge. It should have a value of roughly 9 V\rm V. Now, produce several additional equipotential lines, increasing and decreasing by an interval of 3 V\rm V (e.g., one with 12 V\rm V, one with 15 V\rm V, and one with 6 V\rm V). Don�t worry about getting these exact values. You can be off by a few tenths of a volt.Which statement best describes the distribution of the equipotential lines?1-The equipotential lines are closer together in regions where the electric field is weaker.2-The equipotential lines are closer together in regions where the electric field is stronger.3-The equipotential lines are equally spaced. The distance between each line is the same for all adjacent lines.
Answer:
the correct one is 2. the equipotential lines must be closer together where the field has more intensity
Explanation:
The equipotential line concept is a line or surface where a test charge can move without doing work, therefore the potential in this line is constant and they are perpendicular to the electric field lines.
In this exercise we have a charge and a series of equipotential lines, if this is a point charge the lines are circles around the charge, where the potential is given by
V = k q / r
also the electric field and the electuary potential are related
E = [tex]- \frac{dV}{dr}[/tex]
therefore the equipotential lines must be closer together where the field has more intensity
When checking the answers, the correct one is 2
A wet towel spread out and hung outside on a day without wind dries faster than an identical wet towel left rolled up in a plastic bag.
Explain why.
Even without wind the sun help evaporate the water so it would dry faster. The other towel is in a bag so the moist and water has nowhere to go, therefor staying in the towel.
Consider identical spherical conducting space ships in deep space where gravitational fields from other bodies are negligible compared to the gravitational attraction between the ships. Construct a problem in which you place identical excess charges on the space ships to exactly counter their gravitational attraction. Calculate the amount of excess charge needed. Examine whether that charge depends on the distance between the centers of the ships, the masses of the ships, or any other factors. Discuss whether this would be an easy, difficult, or even impossible thing to do in practice.
Answer:
q = 8.61 10⁻¹¹ m
charge does not depend on the distance between the two ships.
it is a very small charge value so it should be easy to create in each one
Explanation:
In this exercise we have two forces in balance: the electric force and the gravitational force
F_e -F_g = 0
F_e = F_g
Since the gravitational force is always attractive, the electric force must be repulsive, which implies that the electric charge in the two ships must be of the same sign.
Let's write Coulomb's law and gravitational attraction
[tex]k \frac{q_1q_2}{r^2} = G \frac{m_1m_2}{r^2}[/tex]
In the exercise, indicate that the two ships are identical, therefore the masses of the ships are the same and we will place the same charge on each one.
k q² = G m²
q = [tex]\sqrt{ \frac{G}{k} }[/tex] m
we substitute
q = [tex]\sqrt{ \frac{ 6.67 \ 10^{-11}}{8.99 \ 10^{9}} }[/tex] m
q = [tex]\sqrt{0.7419 \ 10^{-20}}[/tex] m
q = 0.861 10⁻¹⁰ m
q = 8.61 10⁻¹¹ m
This amount of charge does not depend on the distance between the two ships.
It is also proportional to the mass of the ships with the proportionality factor found.
Suppose the ships have a mass of m = 1000 kg, let's find the cargo
q = 8.61 10⁻¹¹ 10³
q = 8.61 10⁻⁸ C
this is a very small charge value so it should be easy to create in each one
a toy of mass 600 is whirled by a child in a horizontal circle using a string of length 2m with a linear speed of 5 m/s determine the centripetal force experience by the toy?
there you go, 15N. I hope this helps
An old mining tunnel disappears into a hillside. You would like to know how long the tunnel is, but it's too dangerous to go inside. Recalling your recent physics class, you decide to try setting up standing-wave resonances inside the tunnel. Using your subsonic amplifier and loudspeaker, you find resonances at 5.0 Hz and 6.4 Hz , and at no frequencies between these. It's rather chilly inside the tunnel, so you estimate the sound speed to be 333 m/s .
Answer:
L = 116.6 m
Explanation:
For this exercise we approximate the tunnel as a tube with one end open and the other closed, at the open end there is a belly and at the closed end a node, therefore the resonances occur at
λ = 4L 1st harmonic
λ = 4L / 3 third harmonic
λ = 4L / 5 fifth harmonic
General term
λ = 4L / n n = 1, 3, 5,... odd
n = (2n + 1) n are all integers
They indicate that two consecutive resonant frequencies were found, the speed of the wave is related to the wavelength and its frequency
v = λ f
λ = v / f
we substitute
[tex]\frac{v}{f} = \frac{4L}{n}[/tex]
L = [tex]n \frac{ v}{4f}[/tex]
for the first resonance n = n
L = (2n + 1) [tex]\frac{v}{4f_1}[/tex]
for the second resonance n = n + 1
L = (2n + 3) [tex]\frac{v}{4f_2}[/tex]
we have two equations with two unknowns, let's solve by equating
(2n + 1) \frac{v}{4f_1}= (2n + 3) \frac{v}{4f_2}
(2n + 1) f₂ = (2n +3) f₁
2n + 1 = (2n + 3) [tex]\frac{f_1}{f_2}[/tex]
2n (1 - \frac{f_1}{f_2}) = 3 \frac{f_1}{f_2} -1
we substitute the values
2n (1- [tex]\frac{5}{6.4}[/tex]) = 3 [tex]\frac{5}{6.4}[/tex] -1
2n 0.21875 = 1.34375
n = 1.34375 / 2 0.21875
n = 3
remember that n must be an integer.
We use one of the equations to find the length of the Tunal
L = (2n + 1) \frac{v}{4f_1}
L = (2 3 + 1) [tex]\frac{333}{4 \ 5.0}[/tex]
L = 116.55 m
A 5kg object accelerates from 3m/s to 7m/s in 5 seconds. Calculate the force required
to provide this acceleration.
Answer:
4N
Explanation:
a = (7-3)/5 = 0.8m/s^2
F = ma = (5)(0.8) = 4 Newtons
A runner taking part in the 200-m dash must run around the end of a track that has a circular arc with a radius of curvature of 29.5 m. The runner starts the race at a constant speed. If she completes the 200-m dash in 24.4 s and runs at constant speed throughout the race, what is her centripetal acceleration as she runs the curved portion of the track
Answer: [tex]2.27\ m/s^2[/tex]
Explanation:
Given
Length of the race track [tex]L=200\ m[/tex]
the radius of curvature of the track [tex]r=29.5\ m[/tex]
time taken to run on track is [tex]t=24.4\ s[/tex]
Speed of runner is
[tex]\Rightarrow v=\dfrac{L}{t}=\dfrac{200}{24.4}\\\\\Rightarrow v=8.196\ m/s[/tex]
Centripetal acceleration is
[tex]\Rightarrow a_c=\dfrac{v^2}{r}=\dfrac{8.196^2}{29.5}\\\\\Rightarrow a_c=2.27\ m/s^2[/tex]
. An object 8.5 cm high is placed 28 cm from a converging lens. The focal length of the lens is 12 cm. Calculate the image distance, di. Calculate the image height, hi.
The converging lens is also called a concave lens. The height of the image formed by the lens is 2.55 cm.
Using the lens formula;
1/f = 1/u + 1/v
f = focal length of the lens
u = object distance
v = image distance
Note that the focal length of a converging lens is positive
Substituting values;
1/12 = 1/28 + 1/v
1/v = 1/12 - 1/28
v = 8.4 cm
Magnification= image height/object height = image distance/object distance
image height = ?
object height = 8.5 cm
image distance = 8.4 cm
object distance = 28 cm
So
image height/8.5 = 8.4/28
image height = 8.5 × 8.4/28
image height = 2.55 cm
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This figure shows a sinusoidal wave that is traveling from left to right, in the +x-direction. Assume that it is described by a frequency of 57.1 cycles per second, or hertz (Hz).
7.60 cm4.80 cm
A sinusoidal wave lies on an unlabeled coordinate system. One of the wave's maxima lies on the vertical axis. The horizontal distance from the first maximum to the first minimum is labeled 4.80 cm and the vertical distance between a maximum and a minimum is labeled 7.60 cm.
(a)
What is the wave's amplitude (in cm)?
cm
(b)
What is the wavelength (in cm)?
cm
(c)
Calculate the wave's period (in s).
s
(d)
Compute the speed of this wave (in m/s).
m/s
Answer:
a) A = 3.80 cm, b) λ = 9.60 cm, c) T = 1.75 10⁻² s, d) v = 5.48 m / s
Explanation:
The wave is a way of transporting energy and moment without the need to transport the material. They are described by expressions of the type
x = A sin (kx - wt)
where the amplitude A is the distance from the point of zero intensity to the maximum.
Frequency is the number of times the wave oscillates per unit of time
the wavelength is the distance necessary for the wave to start repeating.
a) In the exercise it tells us that the vertical distance from a machismo to a minimum that is worth 7.60 cm
when checking the definition of amplitude is from zero to a maximum, therefore the value given is twice the amplitude
2A = 7.60
A = 3.80 cm
b) the distance between a minimum and the next maximum is 4.80 cm
Using the definition of wavelength the given value corresponds to half wavelength
λ/ 2 = 4.80
λ = 9.60 cm
c) frequency and period are related
f = 1 / T
T = 1 / f
we calculate
T = 1 / 57.1
T = 0.0175 s
T = 1.75 10⁻² s
d) the speed of the wave is related to the frequency and the wavelength
v = λ f
v = 0.0960 57.1
v = 5.48 m / s
Where do animals such as snakes and frogs, which do not make their own body heat, usually get their heat?
fuel
the sun
hot lava
friction
plssssssss answer correctly
The answer to this question is the Sun.
As im writing this, 12,826 people vote the Sun.
animals such as snakes and frogs, which do not make their own body heat. They usually get their heat from the sun. Hence option B is correct.
What is Cold-blooded animals ?A body temperature that is only slightly higher than the ambient temperature. This condition differentiates cold-blooded, or homoiothermic, animals from fishes, amphibians, reptiles, and invertebrates (birds and mammals). Due of their reliance on external warmth for metabolic activity, terrestrial cold-blooded species are restricted to locations with temperatures ranging from 5-10° to 35-40° C (41-50° to 95-104° F).
Cold-blooded creatures cannot create their own body heat, but they may control it by modifying their surroundings. Alligators and other reptiles frequently lay in the sun to warm up. They, on the other hand, cool off by swimming, going into a burrow in the earth, or moving inside the sade of a rock.
Hence option B is correct.
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What is the kinetic energy of a 10kg object that is moving with a speed of 60m/s.
The answer is 18000 J
I hope this helps!^^ , if you need the work to be shown please tell me, I hope you have a great day!^^
Please help! Due in 5 min! I will pick brainiest! Thanks! YOU ROCK!
The resistance of an electric stove burner element is 11 ohms. What current flows through this
element when it runs off a 220 volt line?
Answer:
Current flow I = 20 ampere
Explanation:
Given:
Resistance R = 11 ohms
Voltage V = 220 volts
Find:
Current flow I
Computation:
Current flow I = V / R
Current flow I = 220 / 11
Current flow I = 20 ampere
The amount of current flow through the element is of 20 A.
Given data:
The magnitude of resistance of Electric stove is, R = 11 ohms.
The magnitude of potential difference in a line is, V' = 220 V.
Here we can simple go for Ohm's law. As per the Ohm's law, the potential difference across the element is proportional to the current flow and the resistance of the element.
The expression is,
V' = I × R
here, I is the amount of current flowing through the element.
Solving as,
220 = I × 11
I = 20 A
Thus, we can conclude that the amount of current flow through the element is of 20 A.
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3. The car's mass is 400 kg. It moves at a velocity of 20 m/s. Calculate the car's momentum. *
(10 Points)
0.05 kg.m/s
8000 kg.m/s
80,000 kg.m/s
20 kg.m/s
Answer:
momentum=mass×velocity
momentum =400kg×20m/s=8000kg.m/s
PLEASE HELP ME I WILL GIVE BRAINLY
Select five short rope exercises and describe how they are done.
Answer:
Jumping battle slams - just move the rope up and down
Alternating jump wave - jump and move the rope side to side
Alternating wide circles - move the rope in a circle position
Jumping jacks
Squat to sholder
Explanation:
The guy above me is correct give him Brainliest
You use 350 W of power to move a 7.0 N object 5 m.
How long did it take?
Answer:
0.1 second
Explanation:
We are given;
Power; P = 350 W
Force; F = 7 N
Distance; d = 5 m
Formula for power is;
P = workdone/time taken
Workdone = F × d
Thus;
350 = (7 × 5)/t
t = 35/350
t = 0.1 second
Score
A 4400 Kg Track Travelling at Intial speed 52m\s can be stopeed in 42 sec. By Gently Break, also the track can be stoped
in 7.6 m\s if the driver hit the wall.
a. What is the Impulse Excerted on the Vehicle?
b. Whta is the Averge force is exeted on the Track in each stops ?
Answer:
(a) J = 10560 kg-m/s (b) 251.42 N
Explanation:
Given that,
Mass, m = 4400 kg
Initial speed, u = 5.2 m/s
Final speed, v = 7.6 m/s
Time, t = 42 s
(a) Let J be the impulse exerted on the vehicle. Impuse is equal to the change in momentum such as :
J = m(v-u)
J = 4400 (7.6-5.2)
J = 10560 kg-m/s
(b) Impulse = Force × t
[tex]F=\dfrac{J}{t}\\\\F=\dfrac{10560}{42}\\F=251.42\ N[/tex]
Hence, this is the required solution.
What is the wavelength of microwaves with a frequency of 3x10^10 Hz?
Answer:
0.01 m
Explanation:
Since the speed of light is 3.0×10^8 m/s
Use the equation,
Wavelength = speed ÷ frequency
Wavelength = 3.0×10^8 ÷ 3×10^10
Wavelength = 0.01m
The following statements are related to the force of a magnetic field on a current-carrying wire. Indicate whether each statement is true or false.
1) The magnetic force on the wire is independent of the direction of the current.
A) True
B) False
2) The force on the wire is directed perpendicular to both the wire and the magnetic field.
A) True
B) False
3) The force takes its largest value when the magnetic field is parallel to the wire.
A) True
B) False
Answer:
1) B: False
2) A: True
3) B: False
Explanation:
1) Statement is false because the force is not independent of the current but rather depends on the direction of the field and current.
2) Statement is true as per right hand thumb rule.
3) The statement is false because force takes its largest value when the magnetic field direction and electric current direction are perpendicular to each other.