τ = 0.3606 s is the time constant for this damped oscillation.
What is Damped oscillation?
True SHM is rarely followed by oscillations in the real world. Any kind of friction usually dampens the motion, causing it to stop or require more energy to proceed. In this section, we look at some instances of damped harmonic motion and see how the equations of motion might be changed to account for this increasingly prevalent situation.
A few seconds after being plucked, a guitar string stops vibrating. On a playground swing, you have to keep pushing in order to keep swinging. Completely undamped motion is uncommon, despite the fact that we may frequently reduce friction and other nonconservative forces to a low level. In fact, dampening oscillations might even be desirable, like with automotive shock absorbers.
Estimation of frequency:
Time taken to complete one oscillation, T = 0.5 s
Hence,
Frequency, f = 1/T = 1/0.5 s = 2 Hz
Estimation of time constant:
1 = 4e(-0.5/τ)
Amplitude = 1 cm
e(-0.5/τ) = 0.25
Do reciprocal,
e(0.5/τ) = 4
On taking natural log both side and solving for τ,
τ = 0.3606 s
This is time constant.
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How many joules of work are done on a box when a force of 25 N pushes it 3 m?
Answer:
75joules
Explanation:
Workdone = force x distance
workdone = 25 x 3
workdone = 75joules
Two blocks with masses 2.00 kg and 4.00 kg are placed side-by-side on a frictionless horizontal surface. A horizontal force with a magnitude of 5.20 N is applied to the 2.00-kg block perpendicular to its surface. What is the magnitude of the force on the 4.00-kg block?
a. 2.60
b. 3.11
c. 3.47
d. 3.89
The magnitude of the force on the 4.00-kg block is 3.47 N.
The given parameters:
Mass of first block, m1 = 2 kgMass of second block, m2 = 4 kgHorizontal force, F = 5.2 NAcceleration of both blocks;The acceleration of both blocks is calculated by applying Newton's second law of motion.\(a = \frac{F}{m_1 + m_2} \\\\a = \frac{5.2}{2 + 4} \\\\a = \frac{5.2}{6} \\\\a = 0.867 \ m/s^2\)
The magnitude of the force on the 4.00-kg block is calculated as follows;
F = ma
F = 4 x 0.867
F = 3.47 N
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Solar System - Scaling. When you look at Neptune in a telescope, you are actually looking into the past as the light has to travel from Neptune to your eyes. If the speed of light is ~300,000 km/s, how far back into the past are you looking (or put another way, how long does it take light to travel from Neptune to your eyes on Earth)
Answer:
Distance from sun to Neptune = 4.495E9 km
Time for light to travel = 4.495E9 / 3E5 sec = 14,980 sec
That is from sun to Neptune time fof light = 250 min
Time for light to travel from sun to earth is about 8 min
So the time from Neptune would be 242 to 258 min depending on position of Neptune - Note that Neptune is about 30X as far from the sun as earth and
250 min / 8 min is roughly 30
The uniform motion of kinematics allows us to find the time it takes for light to arrive from Neptune to Earth, which varies between:
t₁ = 1.45 10⁴ s and t₂₂= 1.55 10⁴ s
depending on the relative distance of the two planets
given parameters
The speed of light c = 300,000 km / s = 3 10⁸ m / s The distance from Neptune to Sum
to find
The time when light arrives from Neptune to Earth
They velocit of an electromagnetic wave is constant, so we can use the uniform motion relationships
v = d / t
t = d / v
where v is the speed of light, d the distance and y time, in this case the speed of the wave is the speed of light (v = c)
We look in the tables for the distances and the rotation periods around the sun
distance ( m) period (s)
Sun Neptunium 4.50 10¹² 5.2 10⁹
Sun - Earth 1.5 10¹¹ 3.2 10⁷
With the data of the period it is observed that the rotation of Neptune is much greater than that of Eart rotation around the sun, for which we will assume that Neptunium is fixed in space and the Earth may be in its aphelion or perihelion, maximum approach o away distance from the sun, consequently we calculate the time for the two cases:
Maximum approach
positions relative distance from the dos Plantetas is
Δd = \(x_{Neptuno - Sum} - x_{Earth - Sum}\)d
Δd = 4.50 10¹² - 1.5 10¹¹
Δd = 43.5 10¹¹ m
the time it takes for Neptune's light to reach Earth is
Δt = \(\frac{ 43.5 \ 10^{11} }{3 \ 10^8}\)
Δt = 14.5 10³ s
Δt = 1.45 10⁴ s
We reduce to hours
Δt = 1.45 10⁴ s (1 h / 3600 s) = 4.03 h
Maximum away
Δd = \(x_{Neptune - Sum} + x_{Neptune-Sum}\)
Δd = 4.50 10¹² + 1.5 10¹¹
Δd = 46.5 10¹¹
The time is
Δt = \(\frac{46.5 \ 10^{11}}{ 3 \ 10^8}\)
Δt = 15.5 10³
Δt = 1.55 10⁴ s
We reduce to hours
Δt = 1.55 10⁴ s (1 h / 3600 s) = 4.31 h
In conclusion, the time it takes for light to arrive from Neptune to Earth varies between:
t₁ = 1.45 10⁴ s and t₂ = 1.55 10⁴ s
depending on the relative distance of the two plants
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Brainlist!! Help!! The nucleus contains protons and neutrons. Being positively charged, the protons repel each other. The nucleus should fly apart due to the repulsive force. Yet, the nuclei of most atoms are stable - explain.
Answer:
The reason that the nucleus of most atoms does not fall apart despite the oppositely charged protons exerting a repulsive force on each other is the strong nuclear force.
What is the strong nuclear force?The strong nuclear force is one of the fundamental forces in nature that acts between protons and neutrons in the atomic nucleus. It is a short-range force that is much stronger than the electromagnetic force (which produces the repulsion between protons).
The strong nuclear force is responsible for holding the nucleus together.
Additionally, the ratio of protons to neutrons in a nucleus also affects its stability. Therefore, if there is an imbalance in this ratio, the repulsive force between the protons can become too strong, causing the nucleus to become unstable and undergo radioactive decay.
Overall, the nucleus remains stable due to the balance between the strong nuclear force and the repulsive force between the protons.
calculate the weight of 100kg object onthe surface of planet with mass and diametre of 4.8×10^24 k and 12000km respectively
The weight of the 100 kg object on the surface of the planet is approximately 8.9 × 10¹² Newtons.
To calculate the weight of a 100 kg object on the surface of a planet, we need to use the formula for gravitational force:
F = (G * M * m) / r²
Where:
F is the gravitational force (weight)
G is the gravitational constant (approximately 6.67430 × 10⁻¹¹ N m²/kg²)
M is the mass of the planet (4.8 × 10²⁴ kg)
m is the mass of the object (100 kg)
r is the radius of the planet (diameter / 2)
Given that the diameter of the planet is 12,000 km, we can find the radius by dividing it by 2:
r = 12,000 km / 2 = 6,000 km = 6,000,000 m
Now, we can substitute the values into the formula:
F = (6.67430 × 10⁻¹¹ N m²/kg² * 4.8 × 10²⁴ kg * 100 kg) / (6,000,000 m)²
Simplifying the expression:
F = (6.67430 × 10⁻¹¹ N m²/kg² * 4.8 × 10²⁶ kg) / 36,000,000,000 m²
F ≈ 8.9 × 10¹² N
Therefore, the weight of the 100 kg object on the surface of the planet is approximately 8.9 × 10¹² Newtons.
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Which statement best describes how resources are distributed?
according to it's nature
The half-life of a radioactive isotope is 210 d. How many days would it take for the decay rate of a sample of this isotope to fall to 0.58 of its initial rate?
It would take approximately 546 days for the decay rate of the sample of this radioactive isotope to fall to 0.58 of its initial rate.
1. The decay rate of a radioactive isotope is proportional to the number of radioactive atoms present in the sample at any given time.
2. The decay rate can be expressed as a function of time using the formula: R(t) = R₀ * \(e^{(-\lambda t\)), where R(t) is the decay rate at time t, R₀ is the initial decay rate, λ is the decay constant, and e is the base of the natural logarithm.
3. The half-life of a radioactive isotope is the time it takes for half of the radioactive atoms in a sample to decay. In this case, the half-life is given as 210 days.
4. Using the half-life, we can find the decay constant (λ) using the formula: λ = ln(2) / T₁/₂, where ln(2) is the natural logarithm of 2 and T₁/₂ is the half-life.
5. Substituting the given half-life into the formula, we have: λ = ln(2) / 210.
6. Now, we need to find the time it takes for the decay rate to fall to 0.58 of its initial rate. Let's call this time "t".
7. Using the formula for the decay rate, we can write: 0.58 * R₀ = R₀ * e^(-λt).
8. Simplifying the equation, we get: 0.58 = \(e^{(-\lambda t\)).
9. Taking the natural logarithm of both sides, we have: ln(0.58) = -λt.
10. Substituting the value of λ from step 5, we get: ln(0.58) = -(ln(2) / 210) * t.
11. Solving for t, we have: t = (ln(0.58) * 210) / ln(2).
12. Evaluating the expression, we find: t ≈ 546.
13. Therefore, it would take approximately 546 days for the decay rate of the sample of this radioactive isotope to fall to 0.58 of its initial rate.
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1. una ambulancia que se mueve con una velocidad de 120 km/h, necesita recorre un tramo recto de 60 km. Calcula el tiempo necesario para que la ambulancia llegue a su destino.
2. Una abeja vuela en línea recta hacia el oeste durante 30 s. Si posee una velocidad de 15 m/s, calcular la distancia total recorrida por la abeja.
3. Una pelota se desplaza en línea recta y recorre una distancia de 10 m en 5 s ¿cuál es su rapidez?
4. Un objeto vuela con una velocidad de 150 m/s durante 60s, calcular la distancia que se desplaza durante ese tiempo.
5. Calcular la distancia que recorre un tren durante 5 horas si la magnitud de su velocidad es de 120 km/h
6. ¿Cuál es el tiempo que tarda en automóvil en recorrer 120 km con una rapidez de 60 km/h?
7. Un auto viaja a 27 m/s y cambia su velocidad cambia a 35 m/s, en un tiempo de 6 segundos. Calcula la aceleración del automóvil.
8. Una persona camina 2 m/s y cambia su velocidad cambia a 7 m/s, en un tiempo de 4 segundos. Calcula la aceleración del automóvil.
9. Un avión viaja se mueve con una velocidad a 185 m/s y reduce su velocidad a 80 m/s, en un tiempo de 5 segundos. Calcula la aceleración del avión.
10. Un auto parte del reposo y alcanza una velocidad de 27.7 m/s en un tiempo de 3.2 segundos. Calcula su aceleración.
The kinematic relation of the uniform motion we can find for these short exercises the answers are:
1. Δt = 1800 s, 2. Δx = 450 m, 3. v = 2 m / s, 4. Δx = 9000 m
5. Δx = 600 km , 6. Δt = 2 h, 7. a = 1.33 m / s², 8. a = 1.25 m / s²
9. a = - 21 m / s², 10. a = 8.66 m / s²
Kinematics analyzes the movement of bodies, looking for relationships between position, velocity and acceleration
This exercise is a series of short problems
1. The ambulance travel time is requested
The ambulance goes at a constant speed, which is why it is a problem of uniform movement
v = \(\frac{\Delta x}{\Delta t}\)
Δt = \(\frac{\Delta x}{v}\)
Where v is the average velocity, Δx and Δt are the displacement and time of the body
Δt = 60/120
Δt = 0.5 h
The international system of units (SI) was created to exchange magnitudes safely and if there are errors, in this system the unit of time is the second. Let's reduce to seconds
Δt = 0.5 h (3600 s / 1h)
Δt = 1800 s
2. They ask for the distance traveled by the bee going at constant speed
Δx = v Δt
Δx = 15 30
Δx = 450 m
3. The average speed of the ball is requested
v = \(\frac{\Delta x}{\Delta t}\)
v = 10/5
v = 2 m / s
4. The distance traveled by an airplane
Δx = v Δt
Δx = 150 60
Δx = 9000 m
5. The distance traveled by the train
Δx = v ΔDt
Δx = 120 5
Δx = 600 km
6. The time it takes to travel a distance
Δt = \(\frac{\Delta x}{v}\)
Δt = 120/60
Δt = 2 h
7. The average acceleration is requested
Average acceleration is defined as the change in velocity per time interval
a = \(\frac{\Delta v}{ \Delta t}\)
a = 35-27 / 6
a = 1.33 m / s²
8. The average acceleration of the car
a = 7-2 / 4
a = 1.25 m / s²
9. The average acceleration of the plane
a = 80-185 / 5
a = - 21 m / s²
The negative sign indicates that the acceleration is stopping the plane
10. The average acceleration
a = 27.7 -0 / 3.2
a = 8.66 m / s²
In conclusion with the kinematic relation of the uniform motion we can find for these short exercises the answers are:
1. Δt = 1800 s, 2. Δx = 450 m, 3. v = 2 m / s, 4. Δx = 9000 m
5. Δx = 600 km , 6. Δt = 2 h, 7. a = 1.33 m / s², 8. a = 1.25 m / s²
9. a = - 21 m / s², 10. a = 8.66 m / s²
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A falling ball has potential energy of 5 J and a kinetic energy of 10 J. What is the ball's mechanical energy?
does wi-fi, cellphone or microwave radiation cause serious damage to our bodies? why or why not?
Answer: No. Wifi, Cellphone or Microwave radiations does not affect human health.
Explanation:
There is no scientific studies that prove that non iodizing radiation have an adverse effect on human health. Any forms of cellular or satellite communications use non iodizing radiations. So even we use them for extended period of time they seems to be quite safe. As for microwave as well as radio waves and visible light , they all are non iodizing radiation. The one that may cause any ill effect on human body will be the UV radiations which is not in any of these. So all these are quite safe to use and amazing technology to make human lives better.
1) Should I take vitamin D supplements in the winter?
23) What is the magnitude of the electric field intensity at a point where a proton experiences an
electrostatic force of magnitude 2.30 x 10-25 newton?
Answer: I am so sorry i hadn't learned this Yet~!
Explanation:
Explain how the energy from the sun is transferred to fossil fuel?
Answer:
Fossil fuels are energy sources that come directly from nature. First, the sun's radiant energy is stored as chemical energy in plants by photosynthesis. When the plants die, they start decaying. After many years, the plants are turned into fossil fuels.
Explanation:
Fossil fuels rely on the Sun's energy because the energy in fossil fuels came from plants and algae as they performed photosynthesis, which requires sunlight.
When sunlight strikes a plant, some of the energy is trapped through photosynthesis and is stored in chemical bonds as the plant grows.
(u should paraphrase it)
An average person is 175 cm tall. How
many people could you stack one on top of
another to reach the top of the CN Tower
(553 m)?
The number of people you will stack to reach the top of the CN Tower (553 m) is 316 people
Hor to convert 175 centimeters to metersWe'll begin by converting 175 cm to m. This can be obtained as illustrated below:
100 cm = 1 m
Therefore,
175 cm = (175 cm × 1 m) / 100 cm
175 cm = 1.75 m
Thus, 175 cm is equivalent to 1.75 m
How to determine the number of people neededThe number of people needed to be stacked to get to the top of the CN tower can be o btained asfollow:
Height of tower = 553 mHeight of a person = 1.75 mNumber of people needed =?Number of people needed = Height of tower / height of a person
Number of people needed = 553 / 1.75
Number of people needed = 316 people
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A basketball player jumps straight up for a ball. To do this, he lowers his body 0.310 m and then accelerates through this distance by forcefully straightening his legs. This player leaves the floor with a vertical velocity sufficient to carry him 0.920 m above the floor.(a) Calculate his velocity when he leaves the floor. (b) Calculate his acceleration while he is straightening his legs. He goes from zero to the velocity found in (a) in a distance of 0.300 m. (c) Calculate the force he exerts on the floor to do this, given that his mass is 110.0 kg.
Answer:
a) v₀ = 4.25 m / s , b) a = 30.1 m / s², c) F = 3311 N
Explanation:
a) to calculate the speed with which it leaves the ground we use the kinematic relations
v² = v₀² - 2 g y
where the speed at the highest point is zero (v = 0) and the height is y = 0.920m, this implies that our reference system is on the ground
v₀ = √ 2gy
let's calculate
v₀ = √(2 9.8 0.920)
v₀ = 4.25 m / s
b) to find the acceleration to reach the speed of v = 4.25 m over a distance of y = 0.300 m
v² = v₀² + 2 a y
in this case it starts from an initial velocity of zero
v² = 2 a y
a = v² / 2y
let's calculate
a = 4.25² / (2 0.300)
a = 30.1 m / s²
c) to calculate the force we use Newton's second law
F = m a
let's calculate
F = 110.0 30.1
F = 3311 N
While flying due east at 33 m/s, an airplane is also being carried due north at 12 m/s by the wind. What is the plane’s resultant velocity?
Answer:
35.11 m/s
Explanation:
While flying due east at 33 meters/second, an airplane is also being carried due north at 12 meters/second by the wind, then the plane’s resultant velocity would be 35.11 meters/second.
What is Velocity?The total displacement covered by any object per unit of time is known as velocity.
the mathematical expression for velocity is given by
velocity = total displacement /total time
As given in the problem , while flying due east at 33 m/s, an airplane is also being carried due north at 12 m/s by the wind and we have to find the resultant velocity of the plane,
Resultant velocity = √( 33² + 12²)
= 35.11 meters/second
Thus, the plane’s resultant velocity would be 35.11 meters/second
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State Newton’s Universal Law of Gravitation.
Answer:
This law states that every particle attract every other particle in the universe with a force that is directly proportional to the product of their mass and inversely proportional to the square of the distance between the center
A compound microscope has a barrel length of =159.0 mm and an objective with a 4.070 mm focal length. The total angular magnification of the microscope is −411.0. Using the approximation that the barrel length is large relative to the focal lengths, determine the angular magnification of the eyepiece.
A microscope's objective lens has a focal length of 4 cm, whereas the eye lens's focal length is 8 cm. if 24 cm is the minimum distance at which a person can see clearly,
How does a microscope use magnification?When an object is magnified, it appears in a microscope image at a scale bigger than its true size. Only when it is able to see more specifics of an item in the photograph than when studying the thing without the use of a magnifying glass does magnification serve a beneficial purpose.
What magnification do 10x and 4x represent?Scanning objective lenses typically have a magnification of 4x, and when paired with a 10x eyepiece lens, they have a total magnification of 40x.
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the very act of observing a particle has a dramatic effect on its behaviour why do you think this is the case
Answer:
Explanation:
In the microscopic world of quantum mechanics, particles don't behave like familiar everyday objects. They can exist in multiple states simultaneously and behave as both particles and waves. When we try to measure or observe a particle, we typically use light or other particles to interact with it. However, this interaction can disturb the particle's state. Imagine trying to measure the position of an electron using light. Light consists of photons, and when photons interact with the electron, they transfer energy to it. This energy exchange causes the electron's position and momentum to become uncertain. The more precisely we try to measure its position, the more uncertain its momentum becomes, and vice versa. This is known as the Heisenberg uncertainty principle.
So, the act of observing a particle disturbs its state because the interaction between the observer and the particle affects its properties. The very act of measurement or observation introduces a level of uncertainty and alters the particle's behavior. It's important to note that this behavior is specific to the quantum world and doesn't directly translate to the macroscopic world we experience in our daily lives. Quantum mechanics operates at extremely small scales and involves probabilities and uncertainties that are not typically noticeable in our macroscopic observations.
19. Assume this process is 10% efficient, how much work is done by the gas expanding into the atmosphere?
The amount of work done by the gas is proportional to the pressure and the change in volume, as well as the efficiency of the process. If the pressure and volume are known, the work done by the gas can be calculated by multiplying these values by the efficiency of the process.
The amount of work done by a gas when it expands is proportional to the change in volume, pressure, and temperature. According to the first law of thermodynamics, the energy of a closed system is conserved, so the work done by the expanding gas is equal to the energy transferred from the gas to the environment in the form of work. Therefore, the work done by the gas is equal to the change in energy of the system. Assume that the process is 10% efficient. Then, only 10% of the energy available to the system is converted into work. This means that the remaining 90% of the energy is lost to the environment in the form of heat. As a result, the amount of work done by the gas expanding into the atmosphere is given by the formula
W = E x η, where W is the work done by the gas, E is the energy available to the system, and η is the efficiency of the process. The energy available to the system is determined by the difference between the internal energy of the gas before and after the expansion. The internal energy of a gas is determined by its temperature, pressure, and volume.
Assuming that the temperature and pressure are constant, the change in internal energy is proportional to the change in volume. Therefore, the energy available to the system is equal to the product of the pressure and the change in volume: E = P x ΔV, where P is the pressure of the gas and ΔV is the change in volume during the expansion. Substituting this equation into the formula for work, we get W = P x ΔV x η.
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13. A solid has a mass of 200 g in air and when partly immersed in a certain liquid it has a mass of 160 g. Given that the relative density of the liquid is 0.80, calculate the volume of the solid immersed in the liquid.
The volume of the solid immersed in the liquid is 5 x 10⁻⁵ m³.
What is the volume of the solid?The volume of the solid is calculated as follows;
V = (Ws - Wa) / (ρg)
where;
Ws is the weight of the solid in airWa is the weight of the solid in liquidρ is the density of the solidg is gravityWs = 0.2 kg x 9.8 m/s²
Ws = 1.96 N
Wa = 0.16 kg x 9.8 m/s²
Wa = 1.568 N
ρ = 0.8 x 1000 g/km³ = 800 kg/m³
The volume is calculated as;
V = (1.96 - 1.568 )/(800 x 9.8)
V = 5 x 10⁻⁵ m³
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If a green ball is rolled from an initial position of 2m and travels to a final
position of 10m - what distance does it travel? (just type the number, no
need for units).
Answer:
10m-2m = 8m
8
Hope this helps!
difference between speed and velocity
Answer:
Speed is the time rate at which an object is moving along a path, while velocity is the rate and direction of an object's movement.
Explanation:
What is the illuminance on an object that is 1.50 m away from a light source with a luminous flux of 665 lumens?1) 1.87 lx2) 3.53 lx3) 23.5 lx4) 425 lx
Given data:
* The illuminance of an object is 1.5 m.
* The luminous flux is 665 lumens.
Solution:
The illuminance in terms of the flux and distance is,
\(E=\frac{P}{4\pi\times d^2}\)where P is the flux, and d is the distance of the object,
Substituting the known values,
\(E=\frac{665}{4\times\pi\times(1.5)^2}\)By solving the above values,
\(E=23.5\text{ Ix}\)Thus, the illuminance is 23.5 Ix.
Hence, 3rd option is the correct answer.
9 The friction between the car and the mud is equivalent to the__________________.
(a) Resultant force
(b) Drag force
(c) Equivalent force
(d) Stretch force
(e) None
A flat sheet of paper of area 0.450 m2 is oriented so that the normal to the sheet is at an angle of 600 to a uniform electric field of magnitude 18 N C-1. What is the magnitude of the electric flux through the sheet? A. 3.22 N m2 C-1 B. 21.42 N m2 C-1 C. 5.04 N m2 C-1 D. 11.72 N m2 C-1 E. 4.05 N m2 C
The magnitude of the electric flux through the sheet is 4.05 N m² C⁻¹ (Option E).
The electric flux through a surface is given by the product of the electric field strength and the area of the surface projected perpendicular to the electric field.
In this case, the electric field strength is 18 N C⁻¹, and the area of the sheet projected perpendicular to the electric field is 0.450 m²
(since the normal to the sheet makes an angle of 60° with the electric field). Multiplying these values gives the electric flux:
Electric flux = Electric field strength × Area
Electric flux = 18 N C⁻¹ × 0.450 m²
Electric flux = 8.1 N m² C⁻¹
In summary, the magnitude of the electric flux through the sheet is 4.05 N m² C⁻¹. This value is obtained by multiplying the given electric field strength by the projected area of the sheet perpendicular to the electric field.
The angle of 60° is taken into account to determine the effective area for calculating the flux.(Option E).
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Which of the following statements is NOT true about sunspots?
А
Sunspots appear and disappear in a cycle that lasts approxi-
mately 15 years.
B
Sunspots occur in pairs.
с
Sunspots are darker because they are cooler than the surround
area.
D
none of the above
Given an average cell is ten times the mass of a bacterium. The mass of a bacterium is 10-15 kg. And a person's mass is 65 kg.
Given the mass of the cell is 1 * 10⁻¹⁴ kg, the number of cells in the 65 kg person is 6.5 * 10¹⁵ cells.
What is the number of cells in a person weighing 65 kg?The number of cells in a human is calculated as follows:
The mass of an average cell is ten times the mass of a bacterium.
The mass of a bacterium = 1 * 10⁻¹⁵ kg
mass of an average cell = 10 * 1 * 10⁻¹⁵ kg = 1 * 10⁻¹⁴ kg
Number of cells = mass of person/mass of cell
Number of cells = 65 kg/1 * 10⁻¹⁴ kg
Number of cells = 6.5 * 10¹⁵ cells.
In conclusion, the number of cells is obtained by dividing the mass of the person by the mass of a average cell.
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Note that the complete question is given below:
Assuming the mass of an average cell is ten times the mass of a bacterium (which is 10⁻¹⁵ kg): Calculate the number of cells in a human assuming the mass of the person is 10² kg.
2O POINTS!!! Easy Question. Will Mark Brainiest
1. What evidence do you see that Newton's third law applies to electrostatic forces?
Answer: The Electrostatic force of attraction or repulsion between two charges shows that the Newton's third law applies to electrostatic forces.
Explanation: Consider two Oppositely charged charges separated by distance d.
The electrostatic force exerted by charge 1 on charge 2 is.
By Coulomb's Law :
F1 = k \(\frac{Q1 Q2}{d2}\).....................................(1)
The electrostatic force exerted by charge 2 on charge 1 is.
F2 = - k \(\frac{Q1 Q2}{d2}\) ................................. (2)
negative sign shows that force are in opposite direction.
From Equation 1 and 2
F1 = - F2
Which implies Newton Third law.
The evidence observed from Newton's third law is that the electrostatic force of attraction exerted by two different charges will be equal in magnitude by acting opposite direction.
Newton's third law states that action and reaction are equal and opposite.
For different charges, say a positive charge (+ Q) and negative charge (-Q), the force of attraction exerted by both charges is equal and opposite.
let the positive charge, +Q = Q₁let the negative charge, -Q = Q₂From Coulomb's law, the force of attraction the two charges will exert on each other is equal and can be calculated as follows;
\(F_{12} = \frac{kQ_1Q_2}{r^2}\)
Thus, the evidence observed from Newton's third law is that the electrostatic force of attraction exerted by two different charges will be equal in magnitude by acting opposite direction.
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Clark Grizwald put a neon green blinking tree that weights 2.3 kg on top of his house 50 meters off the ground but he didn't tie it down right. How much velocity does the blinking tree have when it is 2 meters from the ground? Solve
The magnitude of the velocity the blinking tree has when it is 2 meters from the ground is 21.8 m/s.
Conservation of energyTo calculate the velocity of the blinking tree when it is 2 meters from the ground, we can use the principles of conservation of energy.
First, let's determine the potential energy of the tree when it is 50 meters off the ground. The potential energy (PE) can be calculated using the formula:
PE = mgh
Where:
m = mass of the tree (2.3 kg)
g = acceleration due to gravity (9.8 m/s^2)
h = height (50 m)
PE = 2.3 x 9.8 x 50
PE = 1131 J
Next, we can calculate the potential energy of the tree when it is 2 meters from the ground:
PE = mgh
Thus:
PE = 2.3 x 9.8 x 2
PE = 45.04 J
According to the conservation of energy principle:
PE(initial) = PE(final) + KE(final)
1131 J = 45.04 J + KE(final)
Now, let's solve for the kinetic energy (KE) at the final position:
KE(final) = PE(initial) - PE(final)
KE(final) = 1131 J - 45.04 J
KE(final) = 1085.96 J
KE = (1/2)mv^2
1085.96 J = (1/2) x 2.3 kg x v^2
Thus
v^2 = (2 x 1085.96 J) / 2.3 kg
v^2 = 474.76 m^2/s^2
v ≈ 21.8 m/s
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